Chapter 4: Motion in a Plane

Introduction

Overview

• Motion in 1D has two directions: positive (+) and negative (-).
• For motion in 2D or more, vectors are essential to describe displacement, velocity, and acceleration.
• Study includes vectors, motion in a plane with constant acceleration, projectile motion, and uniform circular motion.

Solved Examples

• Ex 1: Describe motion in 2D using vectors.
  • Use vector components for displacement, velocity, acceleration.
  • Apply vector addition for resultant quantities.

Practice Set

• Q1: How many directions are possible in one-dimensional motion?
  • A) One
  • B) Two
  • C) Three
  • D) Infinite
• Q2: Which quantity requires vectors for complete description in 2D motion?
  • A) Speed
  • B) Displacement
  • C) Temperature
  • D) Distance
• Q3: What type of motion requires vector analysis?
  • A) Motion along a straight line
  • B) Motion in a plane
  • C) Motion at rest
  • D) None of these

Answer Key

• Q1: B - One-dimensional motion has two possible directions: + and -.
• Q2: B - Displacement in 2D requires both magnitude and direction, hence vectors.
• Q3: B - Motion in a plane involves directions in two dimensions, requiring vectors.

Memory Aids

• 1D motion: Two directions (+, -)
• 2D motion: Use vectors for direction and magnitude
• Vectors needed for velocity and acceleration in plane

Scalars and Vectors

Scalars

• Have magnitude only, no direction.
• Examples: distance, speed, temperature.
• Follow ordinary algebra rules for addition, subtraction (same units), multiplication, division.
• Examples:
  • Perimeter = 2 × (length + breadth)
  • Density = mass / volume

Vectors

• Have both magnitude and direction.
• Obey triangle law of addition.
• Examples: velocity, acceleration.
• Represented by arrowed line; length = magnitude, arrowhead = direction.
• Notation: bold letters or letters with arrow, e.g., vector F.
• Magnitude (modulus) denoted as |vector A| or A.

Solved Examples

• Ex 1: Identify scalar or vector: Speed = 30 m/s
  • Speed is scalar (magnitude only).
• Ex 2: Represent vector vector A with magnitude 5 units
  • Draw arrow of length proportional to 5 units with direction.

Practice Set

• Q1: Which is a scalar quantity?
  • A) Velocity
  • B) Acceleration
  • C) Speed
  • D) Force
• Q2: Vector magnitude is represented by?
  • A) vector A
  • B) |vector A|
  • C) vector A²
  • D) None
• Q3: Which law do vectors obey?
  • A) Distributive law
  • B) Triangle law of addition
  • C) Commutative law only
  • D) None

Answer Key

• Q1: C - Speed has magnitude only, no direction.
• Q2: B - Magnitude is denoted by |vector A|.
• Q3: B - Vectors obey triangle law of addition.

Memory Aids

• Scalars = magnitude only
• Vectors = magnitude + direction
• Vector notation: arrow or bold letter

Scalars

• Have magnitude only, no direction.
• Examples: distance, speed, temperature.
• Follow ordinary algebra rules for addition, subtraction (same units), multiplication, division.
• Examples:
  • Perimeter = 2 × (length + breadth)
  • Density = mass / volume

Vectors

• Have both magnitude and direction.
• Obey triangle law of addition.
• Examples: velocity, acceleration.
• Represented by arrowed line; length = magnitude, arrowhead = direction.
• Notation: bold letters or letters with arrow, e.g., vector F.
• Magnitude (modulus) denoted as |vector A| or A.

Solved Examples

• Ex 1: Identify scalar or vector: Speed = 30 m/s
  • Speed is scalar (magnitude only).
• Ex 2: Represent vector vector A with magnitude 5 units
  • Draw arrow of length proportional to 5 units with direction.

Practice Set

• Q1: Which is a scalar quantity?
  • A) Velocity
  • B) Acceleration
  • C) Speed
  • D) Force
• Q2: Vector magnitude is represented by?
  • A) vector A
  • B) |vector A|
  • C) vector A²
  • D) None
• Q3: Which law do vectors obey?
  • A) Distributive law
  • B) Triangle law of addition
  • C) Commutative law only
  • D) None

Answer Key

• Q1: C - Speed has magnitude only, no direction.
• Q2: B - Magnitude is denoted by |vector A|.
• Q3: B - Vectors obey triangle law of addition.

Memory Aids

• Scalars = magnitude only
• Vectors = magnitude + direction
• Vector notation: arrow or bold letter

Scalars

• Have magnitude only, no direction.
• Examples: distance, speed, temperature.
• Follow ordinary algebra rules for addition, subtraction (same units), multiplication, division.
• Examples:
  • Perimeter = 2 × (length + breadth)
  • Density = mass / volume

Vectors

• Have both magnitude and direction.
• Obey triangle law of addition.
• Examples: velocity, acceleration.
• Represented by arrowed line; length = magnitude, arrowhead = direction.
• Notation: bold letters or letters with arrow, e.g., vector F.
• Magnitude (modulus) denoted as |vector A| or A.

Solved Examples

• Ex 1: Identify scalar or vector: Speed = 30 m/s
  • Speed is scalar (magnitude only).
• Ex 2: Represent vector vector A with magnitude 5 units
  • Draw arrow of length proportional to 5 units with direction.

Practice Set

• Q1: Which is a scalar quantity?
  • A) Velocity
  • B) Acceleration
  • C) Speed
  • D) Force
• Q2: Vector magnitude is represented by?
  • A) vector A
  • B) |vector A|
  • C) vector A²
  • D) None
• Q3: Which law do vectors obey?
  • A) Distributive law
  • B) Triangle law of addition
  • C) Commutative law only
  • D) None

Answer Key

• Q1: C - Speed has magnitude only, no direction.
• Q2: B - Magnitude is denoted by |vector A|.
• Q3: B - Vectors obey triangle law of addition.

Memory Aids

• Scalars = magnitude only
• Vectors = magnitude + direction
• Vector notation: arrow or bold letter

Scalars

• Have magnitude only, no direction.
• Examples: distance, speed, temperature.
• Follow ordinary algebra rules for addition, subtraction (same units), multiplication, division.
• Examples:
  • Perimeter = 2 × (length + breadth)
  • Density = mass / volume

Vectors

• Have both magnitude and direction.
• Obey triangle law of addition.
• Examples: velocity, acceleration.
• Represented by arrowed line; length = magnitude, arrowhead = direction.
• Notation: bold letters or letters with arrow, e.g., vector F.
• Magnitude (modulus) denoted as |vector A| or A.

Solved Examples

• Ex 1: Identify scalar or vector: Speed = 30 m/s
  • Speed is scalar (magnitude only).
• Ex 2: Represent vector vector A with magnitude 5 units
  • Draw arrow of length proportional to 5 units with direction.

Practice Set

• Q1: Which is a scalar quantity?
  • A) Velocity
  • B) Acceleration
  • C) Speed
  • D) Force
• Q2: Vector magnitude is represented by?
  • A) vector A
  • B) |vector A|
  • C) vector A²
  • D) None
• Q3: Which law do vectors obey?
  • A) Distributive law
  • B) Triangle law of addition
  • C) Commutative law only
  • D) None

Answer Key

• Q1: C - Speed has magnitude only, no direction.
• Q2: B - Magnitude is denoted by |vector A|.
• Q3: B - Vectors obey triangle law of addition.

Memory Aids

• Scalars = magnitude only
• Vectors = magnitude + direction
• Vector notation: arrow or bold letter

Scalars

• Have magnitude only, no direction.
• Examples: distance, speed, temperature.
• Follow ordinary algebra rules for addition, subtraction (same units), multiplication, division.
• Examples:
  • Perimeter = 2 × (length + breadth)
  • Density = mass / volume

Vectors

• Have both magnitude and direction.
• Obey triangle law of addition.
• Examples: velocity, acceleration.
• Represented by arrowed line; length = magnitude, arrowhead = direction.
• Notation: bold letters or letters with arrow, e.g., vector F.
• Magnitude (modulus) denoted as |vector A| or A.

Solved Examples

• Ex 1: Identify scalar or vector: Speed = 30 m/s
  • Speed is scalar (magnitude only).
• Ex 2: Represent vector vector A with magnitude 5 units
  • Draw arrow of length proportional to 5 units with direction.

Practice Set

• Q1: Which is a scalar quantity?
  • A) Velocity
  • B) Acceleration
  • C) Speed
  • D) Force
• Q2: Vector magnitude is represented by?
  • A) vector A
  • B) |vector A|
  • C) vector A²
  • D) None
• Q3: Which law do vectors obey?
  • A) Distributive law
  • B) Triangle law of addition
  • C) Commutative law only
  • D) None

Answer Key

• Q1: C - Speed has magnitude only, no direction.
• Q2: B - Magnitude is denoted by |vector A|.
• Q3: B - Vectors obey triangle law of addition.

Memory Aids

• Scalars = magnitude only
• Vectors = magnitude + direction
• Vector notation: arrow or bold letter

Scalars

• Have magnitude only, no direction.
• Examples: distance, speed, temperature.
• Follow ordinary algebra rules for addition, subtraction (same units), multiplication, division.
• Examples:
  • Perimeter = 2 × (length + breadth)
  • Density = mass / volume

Vectors

• Have both magnitude and direction.
• Obey triangle law of addition.
• Examples: velocity, acceleration.
• Represented by arrowed line; length = magnitude, arrowhead = direction.
• Notation: bold letters or letters with arrow, e.g., vector F.
• Magnitude (modulus) denoted as |vector A| or A.

Solved Examples

• Ex 1: Identify scalar or vector: Speed = 30 m/s
  • Speed is scalar (magnitude only).
• Ex 2: Represent vector vector A with magnitude 5 units
  • Draw arrow of length proportional to 5 units with direction.

Practice Set

• Q1: Which is a scalar quantity?
  • A) Velocity
  • B) Acceleration
  • C) Speed
  • D) Force
• Q2: Vector magnitude is represented by?
  • A) vector A
  • B) |vector A|
  • C) vector A²
  • D) None
• Q3: Which law do vectors obey?
  • A) Distributive law
  • B) Triangle law of addition
  • C) Commutative law only
  • D) None

Answer Key

• Q1: C - Speed has magnitude only, no direction.
• Q2: B - Magnitude is denoted by |vector A|.
• Q3: B - Vectors obey triangle law of addition.

Memory Aids

• Scalars = magnitude only
• Vectors = magnitude + direction
• Vector notation: arrow or bold letter

Scalars

• Have magnitude only, no direction.
• Examples: distance, speed, temperature.
• Follow ordinary algebra rules for addition, subtraction (same units), multiplication, division.
• Examples:
  • Perimeter = 2 × (length + breadth)
  • Density = mass / volume

Vectors

• Have both magnitude and direction.
• Obey triangle law of addition.
• Examples: velocity, acceleration.
• Represented by arrowed line; length = magnitude, arrowhead = direction.
• Notation: bold letters or letters with arrow, e.g., vector F.
• Magnitude (modulus) denoted as |vector A| or A.

Solved Examples

• Ex 1: Identify scalar or vector: Speed = 30 m/s
  • Speed is scalar (magnitude only).
• Ex 2: Represent vector vector A with magnitude 5 units
  • Draw arrow of length proportional to 5 units with direction.

Practice Set

• Q1: Which is a scalar quantity?
  • A) Velocity
  • B) Acceleration
  • C) Speed
  • D) Force
• Q2: Vector magnitude is represented by?
  • A) vector A
  • B) |vector A|
  • C) vector A²
  • D) None
• Q3: Which law do vectors obey?
  • A) Distributive law
  • B) Triangle law of addition
  • C) Commutative law only
  • D) None

Answer Key

• Q1: C - Speed has magnitude only, no direction.
• Q2: B - Magnitude is denoted by |vector A|.
• Q3: B - Vectors obey triangle law of addition.

Memory Aids

• Scalars = magnitude only
• Vectors = magnitude + direction
• Vector notation: arrow or bold letter

Scalars

• Have magnitude only, no direction.
• Examples: distance, speed, temperature.
• Follow ordinary algebra rules for addition, subtraction (same units), multiplication, division.
• Examples:
  • Perimeter = 2 × (length + breadth)
  • Density = mass / volume

Vectors

• Have both magnitude and direction.
• Obey triangle law of addition.
• Examples: velocity, acceleration.
• Represented by arrowed line; length = magnitude, arrowhead = direction.
• Notation: bold letters or letters with arrow, e.g., vector F.
• Magnitude (modulus) denoted as |vector A| or A.

Solved Examples

• Ex 1: Identify scalar or vector: Speed = 30 m/s
  • Speed is scalar (magnitude only).
• Ex 2: Represent vector vector A with magnitude 5 units
  • Draw arrow of length proportional to 5 units with direction.

Practice Set

• Q1: Which is a scalar quantity?
  • A) Velocity
  • B) Acceleration
  • C) Speed
  • D) Force
• Q2: Vector magnitude is represented by?
  • A) vector A
  • B) |vector A|
  • C) vector A²
  • D) None
• Q3: Which law do vectors obey?
  • A) Distributive law
  • B) Triangle law of addition
  • C) Commutative law only
  • D) None

Answer Key

• Q1: C - Speed has magnitude only, no direction.
• Q2: B - Magnitude is denoted by |vector A|.
• Q3: B - Vectors obey triangle law of addition.

Memory Aids

• Scalars = magnitude only
• Vectors = magnitude + direction
• Vector notation: arrow or bold letter

Scalars

• Have magnitude only, no direction.
• Examples: distance, speed, temperature.
• Follow ordinary algebra rules for addition, subtraction (same units), multiplication, division.
• Examples:
  • Perimeter = 2 × (length + breadth)
  • Density = mass / volume

Vectors

• Have both magnitude and direction.
• Obey triangle law of addition.
• Examples: velocity, acceleration.
• Represented by arrowed line; length = magnitude, arrowhead = direction.
• Notation: bold letters or letters with arrow, e.g., ector F.
• Magnitude (modulus) denoted as |vector A| or A.

Solved Examples

• Ex 1: Identify scalar or vector: Speed = 30 m/s
  • Speed is scalar (magnitude only).
• Ex 2: Represent vector vector A with magnitude 5 units
  • Draw arrow of length proportional to 5 units with direction.

Practice Set

• Q1: Which is a scalar quantity?
  • A) Velocity
  • B) Acceleration
  • C) Speed
  • D) Force
• Q2: Vector magnitude is represented by?
  • A) vector A
  • B) |vector A|
  • C) vector A²
  • D) None
• Q3: Which law do vectors obey?
  • A) Distributive law
  • B) Triangle law of addition
  • C) Commutative law only
  • D) None

Answer Key

• Q1: C - Speed has magnitude only, no direction.
• Q2: B - Magnitude is denoted by |vector A|.
• Q3: B - Vectors obey triangle law of addition.

Memory Aids

• Scalars = magnitude only
• Vectors = magnitude + direction
• Vector notation: arrow or bold letter

Scalars

• Have magnitude only, no direction.
• Examples: distance, speed, temperature.
• Follow ordinary algebra rules for addition, subtraction (same units), multiplication, division.
• Examples:
  • Perimeter = 2 × (length + breadth)
  • Density = mass / volume

Vectors

• Have both magnitude and direction.
• Obey triangle law of addition.
• Examples: velocity, acceleration.
• Represented by arrowed line; length = magnitude, arrowhead = direction.
• Notation: bold letters or letters with arrow, e.g., vector F.
• Magnitude (modulus) denoted as |vector A| or A.

Solved Examples

• Ex 1: Identify scalar or vector: Speed = 30 m/s
  • Speed is scalar (magnitude only).
• Ex 2: Represent vector vector A with magnitude 5 units
  • Draw arrow of length proportional to 5 units with direction.

Practice Set

• Q1: Which is a scalar quantity?
  • A) Velocity
  • B) Acceleration
  • C) Speed
  • D) Force
• Q2: Vector magnitude is represented by?
  • A) vector A
  • B) |vector A|
  • C) vector A²
  • D) None
• Q3: Which law do vectors obey?
  • A) Distributive law
  • B) Triangle law of addition
  • C) Commutative law only
  • D) None

Answer Key

• Q1: C - Speed has magnitude only, no direction.
• Q2: B - Magnitude is denoted by |vector A|.
• Q3: B - Vectors obey triangle law of addition.

Memory Aids

• Scalars = magnitude only
• Vectors = magnitude + direction
• Vector notation: arrow or bold letter

Scalars

• Have magnitude only, no direction.
• Examples: distance, speed, temperature.
• Follow ordinary algebra rules for addition, subtraction (same units), multiplication, division.
• Examples:
  • Perimeter = 2 × (length + breadth)
  • Density = mass / volume

Vectors

• Have both magnitude and direction.
• Obey triangle law of addition.
• Examples: velocity, acceleration.
• Represented by arrowed line; length = magnitude, arrowhead = direction.
• Notation: bold letters or letters with arrow, e.g., ector F.
• Magnitude (modulus) denoted as |vector A| or A.

Solved Examples

• Ex 1: Identify scalar or vector: Speed = 30 m/s
  • Speed is scalar (magnitude only).
• Ex 2: Represent vector vector A with magnitude 5 units
  • Draw arrow of length proportional to 5 units with direction.

Practice Set

• Q1: Which is a scalar quantity?
  • A) Velocity
  • B) Acceleration
  • C) Speed
  • D) Force
• Q2: Vector magnitude is represented by?
  • A) vector A
  • B) |vector A|
  • C) vector A²
  • D) None
• Q3: Which law do vectors obey?
  • A) Distributive law
  • B) Triangle law of addition
  • C) Commutative law only
  • D) None

Answer Key

• Q1: C - Speed has magnitude only, no direction.
• Q2: B - Magnitude is denoted by |vector A|.
• Q3: B - Vectors obey triangle law of addition.

Memory Aids

• Scalars = magnitude only
• Vectors = magnitude + direction
• Vector notation: arrow or bold letter

Scalars

• Have magnitude only, no direction.
• Examples: distance, speed, temperature.
• Follow ordinary algebra rules for addition, subtraction (same units), multiplication, division.
• Examples:
  • Perimeter = 2 × (length + breadth)
  • Density = mass / volume

Vectors

• Have both magnitude and direction.
• Obey triangle law of addition.
• Examples: velocity, acceleration.
• Represented by arrowed line; length = magnitude, arrowhead = direction.
• Notation: bold letters or letters with arrow, e.g., \(\vec{F}\).
• Magnitude (modulus) denoted as \(|\vec{A}|\) or \(A\).

Solved Examples

• Ex 1: Identify scalar or vector: Speed = 30 m/s
  • Speed is scalar (magnitude only).
• Ex 2: Represent vector \(\vec{A}\) with magnitude 5 units
  • Draw arrow of length proportional to 5 units with direction.

Practice Set

• Q1: Which is a scalar quantity?
  • A) Velocity
  • B) Acceleration
  • C) Speed
  • D) Force
• Q2: Vector magnitude is represented by?
  • A) \(\vec{A}\)
  • B) \(|\vec{A}|\)
  • C) \(\vec{A}^2\)
  • D) None
• Q3: Which law do vectors obey?
  • A) Distributive law
  • B) Triangle law of addition
  • C) Commutative law only
  • D) None

Answer Key

• Q1: C - Speed has magnitude only, no direction.
• Q2: B - Magnitude is denoted by \(|\vec{A}|\).
• Q3: B - Vectors obey triangle law of addition.

Memory Aids

• Scalars = magnitude only
• Vectors = magnitude + direction
• Vector notation: arrow or bold letter

Scalars

• Have magnitude only, no direction.
• Examples: distance, speed, temperature.
• Follow ordinary algebra rules for addition, subtraction (same units), multiplication, division.
• Examples:
  • Perimeter = 2 × (length + breadth)
  • Density = mass / volume

Vectors

• Have both magnitude and direction.
• Obey triangle law of addition.
• Examples: velocity, acceleration.
• Represented by arrowed line; length = magnitude, arrowhead = direction.
• Notation: bold letters or letters with arrow, e.g., vector F.
• Magnitude (modulus) denoted as |vector A| or A.

Solved Examples

• Ex 1: Identify scalar or vector: Speed = 30 m/s
  • Speed is scalar (magnitude only).
• Ex 2: Represent vector vector A with magnitude 5 units
  • Draw arrow of length proportional to 5 units with direction.

Practice Set

• Q1: Which is a scalar quantity?
  • A) Velocity
  • B) Acceleration
  • C) Speed
  • D) Force
• Q2: Vector magnitude is represented by?
  • A) vector A
  • B) |vector A|
  • C) vector A²
  • D) None
• Q3: Which law do vectors obey?
  • A) Distributive law
  • B) Triangle law of addition
  • C) Commutative law only
  • D) None

Answer Key

• Q1: C - Speed has magnitude only, no direction.
• Q2: B - Magnitude is denoted by |vector A|.
• Q3: B - Vectors obey triangle law of addition.

Memory Aids

• Scalars = magnitude only
• Vectors = magnitude + direction
• Vector notation: arrow or bold letter

Scalars

• Have magnitude only, no direction.
• Examples: distance, speed, temperature.
• Follow ordinary algebra rules for addition, subtraction (same units), multiplication, division.
• Examples:
  • Perimeter = 2 × (length + breadth)
  • Density = mass / volume

Vectors

• Have both magnitude and direction.
• Obey triangle law of addition.
• Examples: velocity, acceleration.
• Represented by arrowed line; length = magnitude, arrowhead = direction.
• Notation: bold letters or letters with arrow, e.g., vector F.
• Magnitude (modulus) denoted as |vector A| or A.

Solved Examples

• Ex 1: Identify scalar or vector: Speed = 30 m/s
  • Speed is scalar (magnitude only).
• Ex 2: Represent vector vector A with magnitude 5 units
  • Draw arrow of length proportional to 5 units with direction.

Practice Set

• Q1: Which is a scalar quantity?
  • A) Velocity
  • B) Acceleration
  • C) Speed
  • D) Force
• Q2: Vector magnitude is represented by?
  • A) vector A
  • B) |vector A|
  • C) vector A²
  • D) None
• Q3: Which law do vectors obey?
  • A) Distributive law
  • B) Triangle law of addition
  • C) Commutative law only
  • D) None

Answer Key

• Q1: C - Speed has magnitude only, no direction.
• Q2: B - Magnitude is denoted by |vector A|.
• Q3: B - Vectors obey triangle law of addition.

Memory Aids

• Scalars = magnitude only
• Vectors = magnitude + direction
• Vector notation: arrow or bold letter

Scalars

• Have magnitude only, no direction.
• Examples: distance, speed, temperature.
• Follow ordinary algebra rules for addition, subtraction (same units), multiplication, division.
• Examples:
  • Perimeter = 2 × (length + breadth)
  • Density = mass / volume

Vectors

• Have both magnitude and direction.
• Obey triangle law of addition.
• Examples: velocity, acceleration.
• Represented by arrowed line; length = magnitude, arrowhead = direction.
• Notation: bold letters or letters with arrow, e.g., vector F.
• Magnitude (modulus) denoted as |vector A| or A.

Solved Examples

• Ex 1: Identify scalar or vector: Speed = 30 m/s
  • Speed is scalar (magnitude only).
• Ex 2: Represent vector vector A with magnitude 5 units
  • Draw arrow of length proportional to 5 units with direction.

Practice Set

• Q1: Which is a scalar quantity?
  • A) Velocity
  • B) Acceleration
  • C) Speed
  • D) Force
• Q2: Vector magnitude is represented by?
  • A) vector A
  • B) |vector A|
  • C) vector A²
  • D) None
• Q3: Which law do vectors obey?
  • A) Distributive law
  • B) Triangle law of addition
  • C) Commutative law only
  • D) None

Answer Key

• Q1: C - Speed has magnitude only, no direction.
• Q2: B - Magnitude is denoted by |vector A|.
• Q3: B - Vectors obey triangle law of addition.

Memory Aids

• Scalars = magnitude only
• Vectors = magnitude + direction
• Vector notation: arrow or bold letter

Scalars

• Have magnitude only, no direction.
• Examples: distance, speed, temperature.
• Follow ordinary algebra rules for addition, subtraction (same units), multiplication, division.
• Examples:
  • Perimeter = 2 × (length + breadth)
  • Density = mass / volume

Vectors

• Have both magnitude and direction.
• Obey triangle law of addition.
• Examples: velocity, acceleration.
• Represented by arrowed line; length = magnitude, arrowhead = direction.
• Notation: bold letters or letters with arrow, e.g., \(\vec{F}\).
• Magnitude (modulus) denoted as \(|\vec{A}|\) or \(A\).

Solved Examples

• Ex 1: Identify scalar or vector: Speed = 30 m/s
  • Speed is scalar (magnitude only).
• Ex 2: Represent vector \(\vec{A}\) with magnitude 5 units
  • Draw arrow of length proportional to 5 units with direction.

Practice Set

• Q1: Which is a scalar quantity?
  • A) Velocity
  • B) Acceleration
  • C) Speed
  • D) Force
• Q2: Vector magnitude is represented by?
  • A) \(\vec{A}\)
  • B) \(|\vec{A}|\)
  • C) \(\vec{A}^2\)
  • D) None
• Q3: Which law do vectors obey?
  • A) Distributive law
  • B) Triangle law of addition
  • C) Commutative law only
  • D) None

Answer Key

• Q1: C - Speed has magnitude only, no direction.
• Q2: B - Magnitude is denoted by \(|\vec{A}|\).
• Q3: B - Vectors obey triangle law of addition.

Memory Aids

• Scalars = magnitude only
• Vectors = magnitude + direction
• Vector notation: arrow or bold letter

Position and Displacement Vectors

Position Vector

• Gives position of object relative to origin in coordinate system.
• Denoted as vector r or vector OP.
• Provides:
  • Distance from origin (magnitude)
  • Direction from origin

Displacement Vector

• Vector showing change in position from initial to final point.
• Denoted as vector PP' for motion from P to P'.
• Is the straight line joining initial and final positions.
• Independent of actual path taken.
• Magnitude of displacement ≤ path length.

Equality of Vectors

• Two vectors are equal if they have same magnitude and direction.
• Position of vector does not affect equality.

Solved Examples

• Ex 1: Find displacement vector if object moves from P(2,3) to P'(5,7).
  • Displacement = vector PP' = (5-2, 7-3) = (3,4).
  • Magnitude = √(3² + 4²) = 5 units.
• Ex 2: Check equality of vectors vector A and vector B with same length and direction but different positions.
  • Vectors are equal if magnitude and direction match.

Practice Set

• Q1: Displacement vector is
  • A) Path length
  • B) Straight line from initial to final position
  • C) Distance traveled
  • D) None
• Q2: Two vectors are equal if
  • A) Same magnitude only
  • B) Same direction only
  • C) Same magnitude and direction
  • D) Same initial point
• Q3: Magnitude of displacement compared to path length is
  • A) Always greater
  • B) Always equal
  • C) Less than or equal
  • D) None

Answer Key

• Q1: B - Displacement is the straight line joining initial and final points.
• Q2: C - Equality requires same magnitude and direction.
• Q3: C - Displacement magnitude is less than or equal to path length.

Memory Aids

• Displacement = shortest path vector
• Equal vectors = same length + direction
• Position vector from origin to point

Position Vector

• Gives position of object relative to origin in coordinate system.
• Denoted as vector r or vector OP.
• Provides:
  • Distance from origin (magnitude)
  • Direction from origin

Displacement Vector

• Vector showing change in position from initial to final point.
• Denoted as vector PP' for motion from P to P'.
• Is the straight line joining initial and final positions.
• Independent of actual path taken.
• Magnitude of displacement ≤ path length.

Equality of Vectors

• Two vectors are equal if they have same magnitude and direction.
• Position of vector does not affect equality.

Solved Examples

• Ex 1: Find displacement vector if object moves from P(2,3) to P'(5,7).
  • Displacement = vector PP' = (5-2, 7-3) = (3,4).
  • Magnitude = √(3² + 4²) = 5 units.
• Ex 2: Check equality of vectors vector A and vector B with same length and direction but different positions.
  • Vectors are equal if magnitude and direction match.

Practice Set

• Q1: Displacement vector is
  • A) Path length
  • B) Straight line from initial to final position
  • C) Distance traveled
  • D) None
• Q2: Two vectors are equal if
  • A) Same magnitude only
  • B) Same direction only
  • C) Same magnitude and direction
  • D) Same initial point
• Q3: Magnitude of displacement compared to path length is
  • A) Always greater
  • B) Always equal
  • C) Less than or equal
  • D) None

Answer Key

• Q1: B - Displacement is the straight line joining initial and final points.
• Q2: C - Equality requires same magnitude and direction.
• Q3: C - Displacement magnitude is less than or equal to path length.

Memory Aids

• Displacement = shortest path vector
• Equal vectors = same length + direction
• Position vector from origin to point

Position Vector

• Gives position of object relative to origin in coordinate system.
• Denoted as vector r or vector OP.
• Provides:
  • Distance from origin (magnitude)
  • Direction from origin

Displacement Vector

• Vector showing change in position from initial to final point.
• Denoted as vector PP' for motion from P to P'.
• Is the straight line joining initial and final positions.
• Independent of actual path taken.
• Magnitude of displacement ≤ path length.

Equality of Vectors

• Two vectors are equal if they have same magnitude and direction.
• Position of vector does not affect equality.

Solved Examples

• Ex 1: Find displacement vector if object moves from P(2,3) to P'(5,7).
  • Displacement = vector PP' = (5-2, 7-3) = (3,4).
  • Magnitude = √(3² + 4²) = 5 units.
• Ex 2: Check equality of vectors vector A and vector B with same length and direction but different positions.
  • Vectors are equal if magnitude and direction match.

Practice Set

• Q1: Displacement vector is
  • A) Path length
  • B) Straight line from initial to final position
  • C) Distance traveled
  • D) None
• Q2: Two vectors are equal if
  • A) Same magnitude only
  • B) Same direction only
  • C) Same magnitude and direction
  • D) Same initial point
• Q3: Magnitude of displacement compared to path length is
  • A) Always greater
  • B) Always equal
  • C) Less than or equal
  • D) None

Answer Key

• Q1: B - Displacement is the straight line joining initial and final points.
• Q2: C - Equality requires same magnitude and direction.
• Q3: C - Displacement magnitude is less than or equal to path length.

Memory Aids

• Displacement = shortest path vector
• Equal vectors = same length + direction
• Position vector from origin to point

Position Vector

• Gives position of object relative to origin in coordinate system.
• Denoted as ector r or ector OP.
• Provides:
  • Distance from origin (magnitude)
  • Direction from origin

Displacement Vector

• Vector showing change in position from initial to final point.
• Denoted as oector PP' for motion from P to P'.
• Is the straight line joining initial and final positions.
• Independent of actual path taken.
• Magnitude of displacement ≤ path length.

Equality of Vectors

• Two vectors are equal if they have same magnitude and direction.
• Position of vector does not affect equality.

Solved Examples

• Ex 1: Find displacement vector if object moves from P(2,3) to P'(5,7).
  • Displacement = oector PP' = (5-2, 7-3) = (3,4).
  • Magnitude = √(3² + 4²) = 5 units.
• Ex 2: Check equality of vectors vector A and vector B with same length and direction but different positions.
  • Vectors are equal if magnitude and direction match.

Practice Set

• Q1: Displacement vector is
  • A) Path length
  • B) Straight line from initial to final position
  • C) Distance traveled
  • D) None
• Q2: Two vectors are equal if
  • A) Same magnitude only
  • B) Same direction only
  • C) Same magnitude and direction
  • D) Same initial point
• Q3: Magnitude of displacement compared to path length is
  • A) Always greater
  • B) Always equal
  • C) Less than or equal
  • D) None

Answer Key

• Q1: B - Displacement is the straight line joining initial and final points.
• Q2: C - Equality requires same magnitude and direction.
• Q3: C - Displacement magnitude is less than or equal to path length.

Memory Aids

• Displacement = shortest path vector
• Equal vectors = same length + direction
• Position vector from origin to point

Position Vector

• Gives position of object relative to origin in coordinate system.
• Denoted as ector r or ector OP.
• Provides:
  • Distance from origin (magnitude)
  • Direction from origin

Displacement Vector

• Vector showing change in position from initial to final point.
• Denoted as oector PP' for motion from P to P'.
• Is the straight line joining initial and final positions.
• Independent of actual path taken.
• Magnitude of displacement ≤ path length.

Equality of Vectors

• Two vectors are equal if they have same magnitude and direction.
• Position of vector does not affect equality.

Solved Examples

• Ex 1: Find displacement vector if object moves from P(2,3) to P'(5,7).
  • Displacement = oector PP' = (5-2, 7-3) = (3,4).
  • Magnitude = √(3² + 4²) = 5 units.
• Ex 2: Check equality of vectors vector A and vector B with same length and direction but different positions.
  • Vectors are equal if magnitude and direction match.

Practice Set

• Q1: Displacement vector is
  • A) Path length
  • B) Straight line from initial to final position
  • C) Distance traveled
  • D) None
• Q2: Two vectors are equal if
  • A) Same magnitude only
  • B) Same direction only
  • C) Same magnitude and direction
  • D) Same initial point
• Q3: Magnitude of displacement compared to path length is
  • A) Always greater
  • B) Always equal
  • C) Less than or equal
  • D) None

Answer Key

• Q1: B - Displacement is the straight line joining initial and final points.
• Q2: C - Equality requires same magnitude and direction.
• Q3: C - Displacement magnitude is less than or equal to path length.

Memory Aids

• Displacement = shortest path vector
• Equal vectors = same length + direction
• Position vector from origin to point

Position Vector

• Gives position of object relative to origin in coordinate system.
• Denoted as ector r or ector OP.
• Provides:
  • Distance from origin (magnitude)
  • Direction from origin

Displacement Vector

• Vector showing change in position from initial to final point.
• Denoted as oector PP' for motion from P to P'.
• Is the straight line joining initial and final positions.
• Independent of actual path taken.
• Magnitude of displacement ≤ path length.

Equality of Vectors

• Two vectors are equal if they have same magnitude and direction.
• Position of vector does not affect equality.

Solved Examples

• Ex 1: Find displacement vector if object moves from P(2,3) to P'(5,7).
  • Displacement = oector PP' = (5-2, 7-3) = (3,4).
  • Magnitude = √(3² + 4²) = 5 units.
• Ex 2: Check equality of vectors vector A and vector B with same length and direction but different positions.
  • Vectors are equal if magnitude and direction match.

Practice Set

• Q1: Displacement vector is
  • A) Path length
  • B) Straight line from initial to final position
  • C) Distance traveled
  • D) None
• Q2: Two vectors are equal if
  • A) Same magnitude only
  • B) Same direction only
  • C) Same magnitude and direction
  • D) Same initial point
• Q3: Magnitude of displacement compared to path length is
  • A) Always greater
  • B) Always equal
  • C) Less than or equal
  • D) None

Answer Key

• Q1: B - Displacement is the straight line joining initial and final points.
• Q2: C - Equality requires same magnitude and direction.
• Q3: C - Displacement magnitude is less than or equal to path length.

Memory Aids

• Displacement = shortest path vector
• Equal vectors = same length + direction
• Position vector from origin to point

Position Vector

• Gives position of object relative to origin in coordinate system.
• Denoted as ector r or ector OP.
• Provides:
  • Distance from origin (magnitude)
  • Direction from origin

Displacement Vector

• Vector showing change in position from initial to final point.
• Denoted as vector PP' for motion from P to P'.
• Is the straight line joining initial and final positions.
• Independent of actual path taken.
• Magnitude of displacement ≤ path length.

Equality of Vectors

• Two vectors are equal if they have same magnitude and direction.
• Position of vector does not affect equality.

Solved Examples

• Ex 1: Find displacement vector if object moves from P(2,3) to P'(5,7).
  • Displacement = vector PP' = (5-2, 7-3) = (3,4).
  • Magnitude = √(3² + 4²) = 5 units.
• Ex 2: Check equality of vectors vector A and vector B with same length and direction but different positions.
  • Vectors are equal if magnitude and direction match.

Practice Set

• Q1: Displacement vector is
  • A) Path length
  • B) Straight line from initial to final position
  • C) Distance traveled
  • D) None
• Q2: Two vectors are equal if
  • A) Same magnitude only
  • B) Same direction only
  • C) Same magnitude and direction
  • D) Same initial point
• Q3: Magnitude of displacement compared to path length is
  • A) Always greater
  • B) Always equal
  • C) Less than or equal
  • D) None

Answer Key

• Q1: B - Displacement is the straight line joining initial and final points.
• Q2: C - Equality requires same magnitude and direction.
• Q3: C - Displacement magnitude is less than or equal to path length.

Memory Aids

• Displacement = shortest path vector
• Equal vectors = same length + direction
• Position vector from origin to point

Position Vector

• Gives position of object relative to origin in coordinate system.
• Denoted as ector r or ector OP.
• Provides:
  • Distance from origin (magnitude)
  • Direction from origin

Displacement Vector

• Vector showing change in position from initial to final point.
• Denoted as ector PP' for motion from P to P'.
• Is the straight line joining initial and final positions.
• Independent of actual path taken.
• Magnitude of displacement ≤ path length.

Equality of Vectors

• Two vectors are equal if they have same magnitude and direction.
• Position of vector does not affect equality.

Solved Examples

• Ex 1: Find displacement vector if object moves from P(2,3) to P'(5,7).
  • Displacement = ector PP' = (5-2, 7-3) = (3,4).
  • Magnitude = √(3² + 4²) = 5 units.
• Ex 2: Check equality of vectors vector A and vector B with same length and direction but different positions.
  • Vectors are equal if magnitude and direction match.

Practice Set

• Q1: Displacement vector is
  • A) Path length
  • B) Straight line from initial to final position
  • C) Distance traveled
  • D) None
• Q2: Two vectors are equal if
  • A) Same magnitude only
  • B) Same direction only
  • C) Same magnitude and direction
  • D) Same initial point
• Q3: Magnitude of displacement compared to path length is
  • A) Always greater
  • B) Always equal
  • C) Less than or equal
  • D) None

Answer Key

• Q1: B - Displacement is the straight line joining initial and final points.
• Q2: C - Equality requires same magnitude and direction.
• Q3: C - Displacement magnitude is less than or equal to path length.

Memory Aids

• Displacement = shortest path vector
• Equal vectors = same length + direction
• Position vector from origin to point

Position Vector

• Gives position of object relative to origin in coordinate system.
• Denoted as vector r or vector OP.
• Provides:
  • Distance from origin (magnitude)
  • Direction from origin

Displacement Vector

• Vector showing change in position from initial to final point.
• Denoted as vector PP' for motion from P to P'.
• Is the straight line joining initial and final positions.
• Independent of actual path taken.
• Magnitude of displacement ≤ path length.

Equality of Vectors

• Two vectors are equal if they have same magnitude and direction.
• Position of vector does not affect equality.

Solved Examples

• Ex 1: Find displacement vector if object moves from P(2,3) to P'(5,7).
  • Displacement = vector PP' = (5-2, 7-3) = (3,4).
  • Magnitude = √(3² + 4²) = 5 units.
• Ex 2: Check equality of vectors vector A and vector B with same length and direction but different positions.
  • Vectors are equal if magnitude and direction match.

Practice Set

• Q1: Displacement vector is
  • A) Path length
  • B) Straight line from initial to final position
  • C) Distance traveled
  • D) None
• Q2: Two vectors are equal if
  • A) Same magnitude only
  • B) Same direction only
  • C) Same magnitude and direction
  • D) Same initial point
• Q3: Magnitude of displacement compared to path length is
  • A) Always greater
  • B) Always equal
  • C) Less than or equal
  • D) None

Answer Key

• Q1: B - Displacement is the straight line joining initial and final points.
• Q2: C - Equality requires same magnitude and direction.
• Q3: C - Displacement magnitude is less than or equal to path length.

Memory Aids

• Displacement = shortest path vector
• Equal vectors = same length + direction
• Position vector from origin to point

Position Vector

• Gives position of object relative to origin in coordinate system.
• Denoted as vector r or vector OP.
• Provides:
  • Distance from origin (magnitude)
  • Direction from origin

Displacement Vector

• Vector showing change in position from initial to final point.
• Denoted as vector PP' for motion from P to P'.
• Is the straight line joining initial and final positions.
• Independent of actual path taken.
• Magnitude of displacement ≤ path length.

Equality of Vectors

• Two vectors are equal if they have same magnitude and direction.
• Position of vector does not affect equality.

Solved Examples

• Ex 1: Find displacement vector if object moves from P(2,3) to P'(5,7).
  • Displacement = vector PP' = (5-2, 7-3) = (3,4).
  • Magnitude = √(3² + 4²) = 5 units.
• Ex 2: Check equality of vectors vector A and vector B with same length and direction but different positions.
  • Vectors are equal if magnitude and direction match.

Practice Set

• Q1: Displacement vector is
  • A) Path length
  • B) Straight line from initial to final position
  • C) Distance traveled
  • D) None
• Q2: Two vectors are equal if
  • A) Same magnitude only
  • B) Same direction only
  • C) Same magnitude and direction
  • D) Same initial point
• Q3: Magnitude of displacement compared to path length is
  • A) Always greater
  • B) Always equal
  • C) Less than or equal
  • D) None

Answer Key

• Q1: B - Displacement is the straight line joining initial and final points.
• Q2: C - Equality requires same magnitude and direction.
• Q3: C - Displacement magnitude is less than or equal to path length.

Memory Aids

• Displacement = shortest path vector
• Equal vectors = same length + direction
• Position vector from origin to point

Position Vector

• Gives position of object relative to origin in coordinate system.
• Denoted as vector r or vector OP.
• Provides:
  • Distance from origin (magnitude)
  • Direction from origin

Displacement Vector

• Vector showing change in position from initial to final point.
• Denoted as vector PP' for motion from P to P'.
• Is the straight line joining initial and final positions.
• Independent of actual path taken.
• Magnitude of displacement ≤ path length.

Equality of Vectors

• Two vectors are equal if they have same magnitude and direction.
• Position of vector does not affect equality.

Solved Examples

• Ex 1: Find displacement vector if object moves from P(2,3) to P'(5,7).
  • Displacement = vector PP' = (5-2, 7-3) = (3,4).
  • Magnitude = √(3² + 4²) = 5 units.
• Ex 2: Check equality of vectors vector A and vector B with same length and direction but different positions.
  • Vectors are equal if magnitude and direction match.

Practice Set

• Q1: Displacement vector is
  • A) Path length
  • B) Straight line from initial to final position
  • C) Distance traveled
  • D) None
• Q2: Two vectors are equal if
  • A) Same magnitude only
  • B) Same direction only
  • C) Same magnitude and direction
  • D) Same initial point
• Q3: Magnitude of displacement compared to path length is
  • A) Always greater
  • B) Always equal
  • C) Less than or equal
  • D) None

Answer Key

• Q1: B - Displacement is the straight line joining initial and final points.
• Q2: C - Equality requires same magnitude and direction.
• Q3: C - Displacement magnitude is less than or equal to path length.

Memory Aids

• Displacement = shortest path vector
• Equal vectors = same length + direction
• Position vector from origin to point

Position Vector

• Gives position of object relative to origin in coordinate system.
• Denoted as \(\vec{r}\) or \(\overrightarrow{OP}\).
• Provides:
  • Distance from origin (magnitude)
  • Direction from origin

Displacement Vector

• Vector showing change in position from initial to final point.
• Denoted as \(\overrightarrow{PP'}\) for motion from \(P\) to \(P'\).
• Is the straight line joining initial and final positions.
• Independent of actual path taken.
• Magnitude of displacement \(\leq\) path length.

Equality of Vectors

• Two vectors are equal if they have same magnitude and direction.
• Position of vector does not affect equality.

Solved Examples

• Ex 1: Find displacement vector if object moves from \(P(2,3)\) to \(P'(5,7)\).
  • Displacement \(= \overrightarrow{PP'} = (5-2, 7-3) = (3,4)\).
  • Magnitude \(= \sqrt{3^2 + 4^2} = 5\) units.
• Ex 2: Check equality of vectors \(\vec{A}\) and \(\vec{B}\) with same length and direction but different positions.
  • Vectors are equal if magnitude and direction match.

Practice Set

• Q1: Displacement vector is
  • A) Path length
  • B) Straight line from initial to final position
  • C) Distance traveled
  • D) None
• Q2: Two vectors are equal if
  • A) Same magnitude only
  • B) Same direction only
  • C) Same magnitude and direction
  • D) Same initial point
• Q3: Magnitude of displacement compared to path length is
  • A) Always greater
  • B) Always equal
  • C) Less than or equal
  • D) None

Answer Key

• Q1: B - Displacement is the straight line joining initial and final points.
• Q2: C - Equality requires same magnitude and direction.
• Q3: C - Displacement magnitude is less than or equal to path length.

Memory Aids

• Displacement = shortest path vector
• Equal vectors = same length + direction
• Position vector from origin to point

Position Vector

• Gives position of object relative to origin in coordinate system.
• Denoted as vector r or vector OP.
• Provides:
  • Distance from origin (magnitude)
  • Direction from origin

Displacement Vector

• Vector showing change in position from initial to final point.
• Denoted as vector PP' for motion from P to P'.
• Is the straight line joining initial and final positions.
• Independent of actual path taken.
• Magnitude of displacement ≤ path length.

Equality of Vectors

• Two vectors are equal if they have same magnitude and direction.
• Position of vector does not affect equality.

Solved Examples

• Ex 1: Find displacement vector if object moves from P(2,3) to P'(5,7).
  • Displacement = vector PP' = (5-2, 7-3) = (3,4).
  • Magnitude = √(3² + 4²) = 5 units.
• Ex 2: Check equality of vectors vector A and vector B with same length and direction but different positions.
  • Vectors are equal if magnitude and direction match.

Practice Set

• Q1: Displacement vector is
  • A) Path length
  • B) Straight line from initial to final position
  • C) Distance traveled
  • D) None
• Q2: Two vectors are equal if
  • A) Same magnitude only
  • B) Same direction only
  • C) Same magnitude and direction
  • D) Same initial point
• Q3: Magnitude of displacement compared to path length is
  • A) Always greater
  • B) Always equal
  • C) Less than or equal
  • D) None

Answer Key

• Q1: B - Displacement is the straight line joining initial and final points.
• Q2: C - Equality requires same magnitude and direction.
• Q3: C - Displacement magnitude is less than or equal to path length.

Memory Aids

• Displacement = shortest path vector
• Equal vectors = same length + direction
• Position vector from origin to point

Position Vector

• Gives position of object relative to origin in coordinate system.
• Denoted as vector r or vector OP.
• Provides:
  • Distance from origin (magnitude)
  • Direction from origin

Displacement Vector

• Vector showing change in position from initial to final point.
• Denoted as vector PP' for motion from P to P'.
• Is the straight line joining initial and final positions.
• Independent of actual path taken.
• Magnitude of displacement ≤ path length.

Equality of Vectors

• Two vectors are equal if they have same magnitude and direction.
• Position of vector does not affect equality.

Solved Examples

• Ex 1: Find displacement vector if object moves from P(2,3) to P'(5,7).
  • Displacement = vector PP' = (5-2, 7-3) = (3,4).
  • Magnitude = √(3² + 4²) = 5 units.
• Ex 2: Check equality of vectors vector A and vector B with same length and direction but different positions.
  • Vectors are equal if magnitude and direction match.

Practice Set

• Q1: Displacement vector is
  • A) Path length
  • B) Straight line from initial to final position
  • C) Distance traveled
  • D) None
• Q2: Two vectors are equal if
  • A) Same magnitude only
  • B) Same direction only
  • C) Same magnitude and direction
  • D) Same initial point
• Q3: Magnitude of displacement compared to path length is
  • A) Always greater
  • B) Always equal
  • C) Less than or equal
  • D) None

Answer Key

• Q1: B - Displacement is the straight line joining initial and final points.
• Q2: C - Equality requires same magnitude and direction.
• Q3: C - Displacement magnitude is less than or equal to path length.

Memory Aids

• Displacement = shortest path vector
• Equal vectors = same length + direction
• Position vector from origin to point

Position Vector

• Gives position of object relative to origin in coordinate system.
• Denoted as vector r or vector OP.
• Provides:
  • Distance from origin (magnitude)
  • Direction from origin

Displacement Vector

• Vector showing change in position from initial to final point.
• Denoted as vector PP' for motion from P to P'.
• Is the straight line joining initial and final positions.
• Independent of actual path taken.
• Magnitude of displacement ≤ path length.

Equality of Vectors

• Two vectors are equal if they have same magnitude and direction.
• Position of vector does not affect equality.

Solved Examples

• Ex 1: Find displacement vector if object moves from P(2,3) to P'(5,7).
  • Displacement = vector PP' = (5-2, 7-3) = (3,4).
  • Magnitude = √(3² + 4²) = 5 units.
• Ex 2: Check equality of vectors vector A and vector B with same length and direction but different positions.
  • Vectors are equal if magnitude and direction match.

Practice Set

• Q1: Displacement vector is
  • A) Path length
  • B) Straight line from initial to final position
  • C) Distance traveled
  • D) None
• Q2: Two vectors are equal if
  • A) Same magnitude only
  • B) Same direction only
  • C) Same magnitude and direction
  • D) Same initial point
• Q3: Magnitude of displacement compared to path length is
  • A) Always greater
  • B) Always equal
  • C) Less than or equal
  • D) None

Answer Key

• Q1: B - Displacement is the straight line joining initial and final points.
• Q2: C - Equality requires same magnitude and direction.
• Q3: C - Displacement magnitude is less than or equal to path length.

Memory Aids

• Displacement = shortest path vector
• Equal vectors = same length + direction
• Position vector from origin to point

Position Vector

• Gives position of object relative to origin in coordinate system.
• Denoted as \(\vec{r}\) or \(\overrightarrow{OP}\).
• Provides:
  • Distance from origin (magnitude)
  • Direction from origin

Displacement Vector

• Vector showing change in position from initial to final point.
• Denoted as \(\overrightarrow{PP'}\) for motion from \(P\) to \(P'\).
• Is the straight line joining initial and final positions.
• Independent of actual path taken.
• Magnitude of displacement \(\leq\) path length.

Equality of Vectors

• Two vectors are equal if they have same magnitude and direction.
• Position of vector does not affect equality.

Solved Examples

• Ex 1: Find displacement vector if object moves from \(P(2,3)\) to \(P'(5,7)\).
  • Displacement \(= \overrightarrow{PP'} = (5-2, 7-3) = (3,4)\).
  • Magnitude \(= \sqrt{3^2 + 4^2} = 5\) units.
• Ex 2: Check equality of vectors \(\vec{A}\) and \(\vec{B}\) with same length and direction but different positions.
  • Vectors are equal if magnitude and direction match.

Practice Set

• Q1: Displacement vector is
  • A) Path length
  • B) Straight line from initial to final position
  • C) Distance traveled
  • D) None
• Q2: Two vectors are equal if
  • A) Same magnitude only
  • B) Same direction only
  • C) Same magnitude and direction
  • D) Same initial point
• Q3: Magnitude of displacement compared to path length is
  • A) Always greater
  • B) Always equal
  • C) Less than or equal
  • D) None

Answer Key

• Q1: B - Displacement is the straight line joining initial and final points.
• Q2: C - Equality requires same magnitude and direction.
• Q3: C - Displacement magnitude is less than or equal to path length.

Memory Aids

• Displacement = shortest path vector
• Equal vectors = same length + direction
• Position vector from origin to point

Multiplication of Vectors by Real Numbers

Concept Explanation

• Multiplying vector vector A by a real number n gives vector nvector A.
• Magnitude: |nvector A| = |n| × |vector A|.
• Direction: Same as vector A if n > 0, opposite if n < 0.
• Example: 2vector A doubles length, same direction; -1vector A reverses direction, same length.
• Negative scalar changes direction and scales magnitude.

Solved Examples

Ex 1: Find magnitude and direction of -3vector A if |vector A| = 5 units.

• Magnitude: |-3| × 5 = 15 units.
• Direction: Opposite to vector A (since n = -3 < 0).

Practice Set

Q1: If |vector A| = 4 units, what is the magnitude of 2vector A?

• A) 2 units
• B) 4 units
• C) 6 units
• D) 8 units

Q2: Multiplying vector A by -1 results in a vector that is:

• A) Same magnitude, same direction
• B) Same magnitude, opposite direction
• C) Double magnitude, same direction
• D) Half magnitude, opposite direction

Q3: If vector B = -1.5vector A and |vector A| = 10 units, find |vector B| and direction.

• A) 15 units, same direction
• B) 15 units, opposite direction
• C) 6.67 units, same direction
• D) 6.67 units, opposite direction

Answer Key

Q1: D) 8 units — Magnitude doubles: 2 × 4 = 8.

Q2: B) Same magnitude, opposite direction — Negative scalar reverses direction.

Q3: B) 15 units, opposite direction — Magnitude 1.5 × 10 = 15, direction reversed due to negative sign.

Memory Aids

• Positive scalar: length changes, direction same.
• Negative scalar: length changes, direction reverses.
• Magnitude formula: |nvector A| = |n| × |vector A|.

Concept Explanation

• Multiplying vector ܀A by a real number n gives vector n܀A.
• Magnitude: |n܀A| = |n| × |܀A|.
• Direction: Same as ܀A if n > 0, opposite if n < 0.
• Example: 2܀A doubles length, same direction; -1܀A reverses direction, same length.
• Negative scalar changes direction and scales magnitude.

Solved Examples

Ex 1: Find magnitude and direction of -3܀A if |܀A| = 5 units.

• Magnitude: |-3| × 5 = 15 units.
• Direction: Opposite to ܀A (since n = -3 < 0).

Practice Set

Q1: If |܀A| = 4 units, what is the magnitude of 2܀A?

• A) 2 units
• B) 4 units
• C) 6 units
• D) 8 units

Q2: Multiplying ܀A by -1 results in a vector that is:

• A) Same magnitude, same direction
• B) Same magnitude, opposite direction
• C) Double magnitude, same direction
• D) Half magnitude, opposite direction

Q3: If ܀B = -1.5܀A and |܀A| = 10 units, find |܀B| and direction.

• A) 15 units, same direction
• B) 15 units, opposite direction
• C) 6.67 units, same direction
• D) 6.67 units, opposite direction

Answer Key

Q1: D) 8 units — Magnitude doubles: 2 × 4 = 8.

Q2: B) Same magnitude, opposite direction — Negative scalar reverses direction.

Q3: B) 15 units, opposite direction — Magnitude 1.5 × 10 = 15, direction reversed due to negative sign.

Memory Aids

• Positive scalar: length changes, direction same.
• Negative scalar: length changes, direction reverses.
• Magnitude formula: |n܀A| = |n| × |܀A|.

Concept Explanation

• Multiplying vector vector A by a real number n gives vector nvector A.
• Magnitude: |nvector A| = |n| × |vector A|.
• Direction: Same as vector A if n > 0, opposite if n < 0.
• Example: 2vector A doubles length, same direction; -1vector A reverses direction, same length.
• Negative scalar changes direction and scales magnitude.

Solved Examples

Ex 1: Find magnitude and direction of -3vector A if |vector A| = 5 units.

• Magnitude: |-3| × 5 = 15 units.
• Direction: Opposite to vector A (since n = -3 < 0).

Practice Set

Q1: If |vector A| = 4 units, what is the magnitude of 2vector A?

• A) 2 units
• B) 4 units
• C) 6 units
• D) 8 units

Q2: Multiplying vector A by -1 results in a vector that is:

• A) Same magnitude, same direction
• B) Same magnitude, opposite direction
• C) Double magnitude, same direction
• D) Half magnitude, opposite direction

Q3: If vector B = -1.5vector A and |vector A| = 10 units, find |vector B| and direction.

• A) 15 units, same direction
• B) 15 units, opposite direction
• C) 6.67 units, same direction
• D) 6.67 units, opposite direction

Answer Key

Q1: D) 8 units — Magnitude doubles: 2 × 4 = 8.

Q2: B) Same magnitude, opposite direction — Negative scalar reverses direction.

Q3: B) 15 units, opposite direction — Magnitude 1.5 × 10 = 15, direction reversed due to negative sign.

Memory Aids

• Positive scalar: length changes, direction same.
• Negative scalar: length changes, direction reverses.
• Magnitude formula: |nvector A| = |n| × |vector A|.

Concept Explanation

• Multiplying vector vector A by a real number n gives vector nvector A.
• Magnitude: |nvector A| = |n| × |vector A|.
• Direction: Same as vector A if n > 0, opposite if n < 0.
• Example: 2vector A doubles length, same direction; -1vector A reverses direction, same length.
• Negative scalar changes direction and scales magnitude.

Solved Examples

Ex 1: Find magnitude and direction of -3vector A if |vector A| = 5 units.

• Magnitude: |-3| × 5 = 15 units.
• Direction: Opposite to vector A (since n = -3 < 0).

Practice Set

Q1: If |vector A| = 4 units, what is the magnitude of 2vector A?

• A) 2 units
• B) 4 units
• C) 6 units
• D) 8 units

Q2: Multiplying vector A by -1 results in a vector that is:

• A) Same magnitude, same direction
• B) Same magnitude, opposite direction
• C) Double magnitude, same direction
• D) Half magnitude, opposite direction

Q3: If vector B = -1.5vector A and |vector A| = 10 units, find |vector B| and direction.

• A) 15 units, same direction
• B) 15 units, opposite direction
• C) 6.67 units, same direction
• D) 6.67 units, opposite direction

Answer Key

Q1: D) 8 units — Magnitude doubles: 2 × 4 = 8.

Q2: B) Same magnitude, opposite direction — Negative scalar reverses direction.

Q3: B) 15 units, opposite direction — Magnitude 1.5 × 10 = 15, direction reversed due to negative sign.

Memory Aids

• Positive scalar: length changes, direction same.
• Negative scalar: length changes, direction reverses.
• Magnitude formula: |nvector A| = |n| × |vector A|.

Concept Explanation

• Multiplying vector A by a real number n gives vector nA.
• Magnitude: |nA| = |n| × |A|.
• Direction: Same as A if n > 0, opposite if n < 0.
• Example: 2A doubles length, same direction; -1A reverses direction, same length.
• Negative scalar changes direction and scales magnitude.

Solved Examples

Ex 1: Find magnitude and direction of -3A if |A| = 5 units.

• Magnitude: |-3| × 5 = 15 units.
• Direction: Opposite to A (since n = -3 < 0).

Practice Set

Q1: If |A| = 4 units, what is the magnitude of 2A?

• A) 2 units
• B) 4 units
• C) 6 units
• D) 8 units

Q2: Multiplying A by -1 results in a vector that is:

• A) Same magnitude, same direction
• B) Same magnitude, opposite direction
• C) Double magnitude, same direction
• D) Half magnitude, opposite direction

Q3: If B = -1.5A and |A| = 10 units, find |B| and direction.

• A) 15 units, same direction
• B) 15 units, opposite direction
• C) 6.67 units, same direction
• D) 6.67 units, opposite direction

Answer Key

Q1: D) 8 units — Magnitude doubles: 2 × 4 = 8.

Q2: B) Same magnitude, opposite direction — Negative scalar reverses direction.

Q3: B) 15 units, opposite direction — Magnitude 1.5 × 10 = 15, direction reversed due to negative sign.

Memory Aids

• Positive scalar: length changes, direction same.
• Negative scalar: length changes, direction reverses.
• Magnitude formula: |nA| = |n| × |A|.

Concept Explanation

• Multiplying vector vector A by a real number n gives vector nvector A.
• Magnitude: |nvector A| = |n| × |vector A|.
• Direction: Same as vector A if n > 0, opposite if n < 0.
• Example: 2vector A doubles length, same direction; -1vector A reverses direction, same length.
• Negative scalar changes direction and scales magnitude.

Solved Examples

Ex 1: Find magnitude and direction of -3vector A if |vector A| = 5 units.

• Magnitude: |-3| × 5 = 15 units.
• Direction: Opposite to vector A (since n = -3 < 0).

Practice Set

Q1: If |vector A| = 4 units, what is the magnitude of 2vector A?

• A) 2 units
• B) 4 units
• C) 6 units
• D) 8 units

Q2: Multiplying vector A by -1 results in a vector that is:

• A) Same magnitude, same direction
• B) Same magnitude, opposite direction
• C) Double magnitude, same direction
• D) Half magnitude, opposite direction

Q3: If vector B = -1.5vector A and |vector A| = 10 units, find |vector B| and direction.

• A) 15 units, same direction
• B) 15 units, opposite direction
• C) 6.67 units, same direction
• D) 6.67 units, opposite direction

Answer Key

Q1: D) 8 units — Magnitude doubles: 2 × 4 = 8.

Q2: B) Same magnitude, opposite direction — Negative scalar reverses direction.

Q3: B) 15 units, opposite direction — Magnitude 1.5 × 10 = 15, direction reversed due to negative sign.

Memory Aids

• Positive scalar: length changes, direction same.
• Negative scalar: length changes, direction reverses.
• Magnitude formula: |nvector A| = |n| × |vector A|.

Concept Explanation

• Multiplying vector A by a real number n gives vector nA.
• Magnitude: |nA| = |n| × |A|.
• Direction: Same as A if n > 0, opposite if n < 0.
• Example: 2A doubles length, same direction; -1A reverses direction, same length.
• Negative scalar changes direction and scales magnitude.

Solved Examples

Ex 1: Find magnitude and direction of -3A if |A| = 5 units.

• Magnitude: |-3| × 5 = 15 units.
• Direction: Opposite to A (since n = -3 < 0).

Practice Set

Q1: If |A| = 4 units, what is the magnitude of 2A?

• A) 2 units
• B) 4 units
• C) 6 units
• D) 8 units

Q2: Multiplying A by -1 results in a vector that is:

• A) Same magnitude, same direction
• B) Same magnitude, opposite direction
• C) Double magnitude, same direction
• D) Half magnitude, opposite direction

Q3: If B = -1.5A and |A| = 10 units, find |B| and direction.

• A) 15 units, same direction
• B) 15 units, opposite direction
• C) 6.67 units, same direction
• D) 6.67 units, opposite direction

Answer Key

Q1: D) 8 units — Magnitude doubles: 2 × 4 = 8.

Q2: B) Same magnitude, opposite direction — Negative scalar reverses direction.

Q3: B) 15 units, opposite direction — Magnitude 1.5 × 10 = 15, direction reversed due to negative sign.

Memory Aids

• Positive scalar: length changes, direction same.
• Negative scalar: length changes, direction reverses.
• Magnitude formula: |nA| = |n| × |A|.

Concept Explanation

• Multiplying vector vector A by a real number n gives vector nvector A.
• Magnitude: |nvector A| = |n| × |vector A|.
• Direction: Same as vector A if n > 0, opposite if n < 0.
• Example: 2vector A doubles length, same direction; -1vector A reverses direction, same length.
• Negative scalar changes direction and scales magnitude.

Solved Examples

Ex 1: Find magnitude and direction of -3vector A if |vector A| = 5 units.

• Magnitude: |-3| × 5 = 15 units.
• Direction: Opposite to vector A (since n = -3 < 0).

Practice Set

Q1: If |vector A| = 4 units, what is the magnitude of 2vector A?

• A) 2 units
• B) 4 units
• C) 6 units
• D) 8 units

Q2: Multiplying vector A by -1 results in a vector that is:

• A) Same magnitude, same direction
• B) Same magnitude, opposite direction
• C) Double magnitude, same direction
• D) Half magnitude, opposite direction

Q3: If vector B = -1.5vector A and |vector A| = 10 units, find |vector B| and direction.

• A) 15 units, same direction
• B) 15 units, opposite direction
• C) 6.67 units, same direction
• D) 6.67 units, opposite direction

Answer Key

Q1: D) 8 units — Magnitude doubles: 2 × 4 = 8.

Q2: B) Same magnitude, opposite direction — Negative scalar reverses direction.

Q3: B) 15 units, opposite direction — Magnitude 1.5 × 10 = 15, direction reversed due to negative sign.

Memory Aids

• Positive scalar: length changes, direction same.
• Negative scalar: length changes, direction reverses.
• Magnitude formula: |nvector A| = |n| × |vector A|.

Concept Explanation

• Multiplying vector vector A by a real number n gives vector nvector A.
• Magnitude: |nvector A| = |n| × |vector A|.
• Direction: Same as vector A if n > 0, opposite if n < 0.
• Example: 2vector A doubles length, same direction; -1vector A reverses direction, same length.
• Negative scalar changes direction and scales magnitude.

Solved Examples

Ex 1: Find magnitude and direction of -3vector A if |vector A| = 5 units.

• Magnitude: |-3| × 5 = 15 units.
• Direction: Opposite to vector A (since n = -3 < 0).

Practice Set

Q1: If |vector A| = 4 units, what is the magnitude of 2vector A?

• A) 2 units
• B) 4 units
• C) 6 units
• D) 8 units

Q2: Multiplying vector A by -1 results in a vector that is:

• A) Same magnitude, same direction
• B) Same magnitude, opposite direction
• C) Double magnitude, same direction
• D) Half magnitude, opposite direction

Q3: If vector B = -1.5vector A and |vector A| = 10 units, find |vector B| and direction.

• A) 15 units, same direction
• B) 15 units, opposite direction
• C) 6.67 units, same direction
• D) 6.67 units, opposite direction

Answer Key

Q1: D) 8 units — Magnitude doubles: 2 × 4 = 8.

Q2: B) Same magnitude, opposite direction — Negative scalar reverses direction.

Q3: B) 15 units, opposite direction — Magnitude 1.5 × 10 = 15, direction reversed due to negative sign.

Memory Aids

• Positive scalar: length changes, direction same.
• Negative scalar: length changes, direction reverses.
• Magnitude formula: |nvector A| = |n| × |vector A|.

Concept Explanation

• Multiplying vector ܀A by a real number n gives vector n܀A.
• Magnitude: |n܀A| = |n| × |܀A|.
• Direction: Same as ܀A if n > 0, opposite if n < 0.
• Example: 2܀A doubles length, same direction; -1܀A reverses direction, same length.
• Negative scalar changes direction and scales magnitude.

Solved Examples

Ex 1: Find magnitude and direction of -3܀A if |܀A| = 5 units.

• Magnitude: |-3| × 5 = 15 units.
• Direction: Opposite to ܀A (since n = -3 < 0).

Practice Set

Q1: If |܀A| = 4 units, what is the magnitude of 2܀A?

• A) 2 units
• B) 4 units
• C) 6 units
• D) 8 units

Q2: Multiplying ܀A by -1 results in a vector that is:

• A) Same magnitude, same direction
• B) Same magnitude, opposite direction
• C) Double magnitude, same direction
• D) Half magnitude, opposite direction

Q3: If ܀B = -1.5܀A and |܀A| = 10 units, find |܀B| and direction.

• A) 15 units, same direction
• B) 15 units, opposite direction
• C) 6.67 units, same direction
• D) 6.67 units, opposite direction

Answer Key

Q1: D) 8 units — Magnitude doubles: 2 × 4 = 8.

Q2: B) Same magnitude, opposite direction — Negative scalar reverses direction.

Q3: B) 15 units, opposite direction — Magnitude 1.5 × 10 = 15, direction reversed due to negative sign.

Memory Aids

• Positive scalar: length changes, direction same.
• Negative scalar: length changes, direction reverses.
• Magnitude formula: |n܀A| = |n| × |܀A|.

Concept Explanation

• Multiplying vector vector A by a real number n gives vector nvector A.
• Magnitude: |nvector A| = |n| × |vector A|.
• Direction: Same as vector A if n > 0, opposite if n < 0.
• Example: 2vector A doubles length, same direction; -1vector A reverses direction, same length.
• Negative scalar changes direction and scales magnitude.

Solved Examples

Ex 1: Find magnitude and direction of -3vector A if |vector A| = 5 units.

• Magnitude: |-3| × 5 = 15 units.
• Direction: Opposite to vector A (since n = -3 < 0).

Practice Set

Q1: If |vector A| = 4 units, what is the magnitude of 2vector A?

• A) 2 units
• B) 4 units
• C) 6 units
• D) 8 units

Q2: Multiplying vector A by -1 results in a vector that is:

• A) Same magnitude, same direction
• B) Same magnitude, opposite direction
• C) Double magnitude, same direction
• D) Half magnitude, opposite direction

Q3: If vector B = -1.5vector A and |vector A| = 10 units, find |vector B| and direction.

• A) 15 units, same direction
• B) 15 units, opposite direction
• C) 6.67 units, same direction
• D) 6.67 units, opposite direction

Answer Key

Q1: D) 8 units — Magnitude doubles: 2 × 4 = 8.

Q2: B) Same magnitude, opposite direction — Negative scalar reverses direction.

Q3: B) 15 units, opposite direction — Magnitude 1.5 × 10 = 15, direction reversed due to negative sign.

Memory Aids

• Positive scalar: length changes, direction same.
• Negative scalar: length changes, direction reverses.
• Magnitude formula: |nvector A| = |n| × |vector A|.

Concept Explanation

• Multiplying vector \( \vec{A} \) by a real number \( n \) gives vector \( n\vec{A} \).
• Magnitude: \( |n\vec{A}| = |n| \times |\vec{A}| \).
• Direction: Same as \( \vec{A} \) if \( n > 0 \), opposite if \( n < 0 \).
• Example: \( 2\vec{A} \) doubles length, same direction; \( -1\vec{A} \) reverses direction, same length.
• Negative scalar changes direction and scales magnitude.

Solved Examples

Ex 1: Find magnitude and direction of \( -3\vec{A} \) if \( |\vec{A}| = 5 \) units.

• Magnitude: \( |-3| \times 5 = 15 \) units.
• Direction: Opposite to \( \vec{A} \) (since \( n = -3 < 0 \)).

Practice Set

Q1: If \( |\vec{A}| = 4 \) units, what is the magnitude of \( 2\vec{A} \)?

• A) 2 units
• B) 4 units
• C) 6 units
• D) 8 units

Q2: Multiplying \( \vec{A} \) by \( -1 \) results in a vector that is:

• A) Same magnitude, same direction
• B) Same magnitude, opposite direction
• C) Double magnitude, same direction
• D) Half magnitude, opposite direction

Q3: If \( \vec{B} = -1.5\vec{A} \) and \( |\vec{A}| = 10 \) units, find \( |\vec{B}| \) and direction.

• A) 15 units, same direction
• B) 15 units, opposite direction
• C) 6.67 units, same direction
• D) 6.67 units, opposite direction

Answer Key

Q1: D) 8 units — Magnitude doubles: \( 2 \times 4 = 8 \).

Q2: B) Same magnitude, opposite direction — Negative scalar reverses direction.

Q3: B) 15 units, opposite direction — Magnitude \( 1.5 \times 10 = 15 \), direction reversed due to negative sign.

Memory Aids

• Positive scalar: length changes, direction same.
• Negative scalar: length changes, direction reverses.
• Magnitude formula: \( |n\vec{A}| = |n| \times |\vec{A}| \).

Concept Explanation

• Multiplying vector vector A by a real number n gives vector nvector A.
• Magnitude: |nvector A| = |n| × |vector A|.
• Direction: Same as vector A if n > 0, opposite if n < 0.
• Example: 2vector A doubles length, same direction; -1vector A reverses direction, same length.
• Negative scalar changes direction and scales magnitude.

Solved Examples

Ex 1: Find magnitude and direction of -3vector A if |vector A| = 5 units.

• Magnitude: |-3| × 5 = 15 units.
• Direction: Opposite to vector A (since n = -3 < 0).

Practice Set

Q1: If |vector A| = 4 units, what is the magnitude of 2vector A?

• A) 2 units
• B) 4 units
• C) 6 units
• D) 8 units

Q2: Multiplying vector A by -1 results in a vector that is:

• A) Same magnitude, same direction
• B) Same magnitude, opposite direction
• C) Double magnitude, same direction
• D) Half magnitude, opposite direction

Q3: If vector B = -1.5vector A and |vector A| = 10 units, find |vector B| and direction.

• A) 15 units, same direction
• B) 15 units, opposite direction
• C) 6.67 units, same direction
• D) 6.67 units, opposite direction

Answer Key

Q1: D) 8 units — Magnitude doubles: 2 × 4 = 8.

Q2: B) Same magnitude, opposite direction — Negative scalar reverses direction.

Q3: B) 15 units, opposite direction — Magnitude 1.5 × 10 = 15, direction reversed due to negative sign.

Memory Aids

• Positive scalar: length changes, direction same.
• Negative scalar: length changes, direction reverses.
• Magnitude formula: |nvector A| = |n| × |vector A|.

Concept Explanation

• Multiplying vector A by a real number n gives vector nA.
• Magnitude: |nA| = |n| × |A|.
• Direction: Same as A if n > 0, opposite if n < 0.
• Example: 2A doubles length, same direction; -1A reverses direction, same length.
• Negative scalar changes direction and scales magnitude.

Solved Examples

Ex 1: Find magnitude and direction of -3A if |A| = 5 units.

• Magnitude: |-3| × 5 = 15 units.
• Direction: Opposite to A (since n = -3 < 0).

Practice Set

Q1: If |A| = 4 units, what is the magnitude of 2A?

• A) 2 units
• B) 4 units
• C) 6 units
• D) 8 units

Q2: Multiplying A by -1 results in a vector that is:

• A) Same magnitude, same direction
• B) Same magnitude, opposite direction
• C) Double magnitude, same direction
• D) Half magnitude, opposite direction

Q3: If B = -1.5A and |A| = 10 units, find |B| and direction.

• A) 15 units, same direction
• B) 15 units, opposite direction
• C) 6.67 units, same direction
• D) 6.67 units, opposite direction

Answer Key

Q1: D) 8 units — Magnitude doubles: 2 × 4 = 8.

Q2: B) Same magnitude, opposite direction — Negative scalar reverses direction.

Q3: B) 15 units, opposite direction — Magnitude 1.5 × 10 = 15, direction reversed due to negative sign.

Memory Aids

• Positive scalar: length changes, direction same.
• Negative scalar: length changes, direction reverses.
• Magnitude formula: |nA| = |n| × |A|.

Concept Explanation

• Multiplying vector vector A by a real number n gives vector nvector A.
• Magnitude: |nvector A| = |n| × |vector A|.
• Direction: Same as vector A if n > 0, opposite if n < 0.
• Example: 2vector A doubles length, same direction; -1vector A reverses direction, same length.
• Negative scalar changes direction and scales magnitude.

Solved Examples

Ex 1: Find magnitude and direction of -3vector A if |vector A| = 5 units.

• Magnitude: |-3| × 5 = 15 units.
• Direction: Opposite to vector A (since n = -3 < 0).

Practice Set

Q1: If |vector A| = 4 units, what is the magnitude of 2vector A?

• A) 2 units
• B) 4 units
• C) 6 units
• D) 8 units

Q2: Multiplying vector A by -1 results in a vector that is:

• A) Same magnitude, same direction
• B) Same magnitude, opposite direction
• C) Double magnitude, same direction
• D) Half magnitude, opposite direction

Q3: If vector B = -1.5vector A and |vector A| = 10 units, find |vector B| and direction.

• A) 15 units, same direction
• B) 15 units, opposite direction
• C) 6.67 units, same direction
• D) 6.67 units, opposite direction

Answer Key

Q1: D) 8 units — Magnitude doubles: 2 × 4 = 8.

Q2: B) Same magnitude, opposite direction — Negative scalar reverses direction.

Q3: B) 15 units, opposite direction — Magnitude 1.5 × 10 = 15, direction reversed due to negative sign.

Memory Aids

• Positive scalar: length changes, direction same.
• Negative scalar: length changes, direction reverses.
• Magnitude formula: |nvector A| = |n| × |vector A|.

Concept Explanation

• Multiplying vector \( \vec{A} \) by a real number \( n \) gives vector \( n\vec{A} \).
• Magnitude: \( |n\vec{A}| = |n| \times |\vec{A}| \).
• Direction: Same as \( \vec{A} \) if \( n > 0 \), opposite if \( n < 0 \).
• Example: \( 2\vec{A} \) doubles length, same direction; \( -1\vec{A} \) reverses direction, same length.
• Negative scalar changes direction and scales magnitude.

Solved Examples

Ex 1: Find magnitude and direction of \( -3\vec{A} \) if \( |\vec{A}| = 5 \) units.

• Magnitude: \( |-3| \times 5 = 15 \) units.
• Direction: Opposite to \( \vec{A} \) (since \( n = -3 < 0 \)).

Practice Set

Q1: If \( |\vec{A}| = 4 \) units, what is the magnitude of \( 2\vec{A} \)?

• A) 2 units
• B) 4 units
• C) 6 units
• D) 8 units

Q2: Multiplying \( \vec{A} \) by \( -1 \) results in a vector that is:

• A) Same magnitude, same direction
• B) Same magnitude, opposite direction
• C) Double magnitude, same direction
• D) Half magnitude, opposite direction

Q3: If \( \vec{B} = -1.5\vec{A} \) and \( |\vec{A}| = 10 \) units, find \( |\vec{B}| \) and direction.

• A) 15 units, same direction
• B) 15 units, opposite direction
• C) 6.67 units, same direction
• D) 6.67 units, opposite direction

Answer Key

Q1: D) 8 units — Magnitude doubles: \( 2 \times 4 = 8 \).

Q2: B) Same magnitude, opposite direction — Negative scalar reverses direction.

Q3: B) 15 units, opposite direction — Magnitude \( 1.5 \times 10 = 15 \), direction reversed due to negative sign.

Memory Aids

• Positive scalar: length changes, direction same.
• Negative scalar: length changes, direction reverses.
• Magnitude formula: \( |n\vec{A}| = |n| \times |\vec{A}| \).

Addition of Vectors

General Rule

• Arrange vectors so head of one coincides with tail of next.
• Resultant vector is from tail of first to head of last vector.

Properties

• Commutative: vector A + vector B = vector B + vector A
• Distributive: n(vector A + vector B) = nvector A + nvector B
• Associative: vector A + (vector B + vector C) = (vector A + vector B) + vector C

Triangle Law

• If two vectors are represented by two sides of a triangle in order, resultant is the third side in opposite order.
• Resultant: vector R = vector A + vector B

Parallelogram Law

• Two vectors represented by adjacent sides of a parallelogram.
• Resultant is diagonal from the common point.
• Magnitude: R = √(A² + B² + 2AB cos θ)
• Angle: tan β = B sin θ/(A + B cos θ)

Polygon Law

• Vectors represented by sides of an open polygon in order.
• Resultant is the closing side in opposite order.

Special Cases

• Same direction: R = A + B, β = 0°
• Opposite direction: R = |A - B|, β = 180°
• Perpendicular vectors: R = √(A² + B²), β = tan^-1 (B/A)

Solved Examples

Ex 1: Find resultant of vector A = 3î and vector B = 4ĵ (perpendicular vectors).

• Use Pythagoras: R = √(3² + 4²) = 5
• Angle: β = tan^-1 4/3 = 53.13°

Practice Set

Q1: If vector A = 5î and vector B = 5ĵ, what is |vector A + vector B|?

• A) 5
• B) 10
• C) 5√2
• D) 0

Q2: Two vectors vector A and vector B have magnitudes 6 and 8 units with angle 60° between them. Find resultant magnitude.

• A) 14
• B) 10
• C) 8
• D) 6

Q3: Two vectors vector A = 7 units and vector B = 24 units are perpendicular. Find resultant and angle with vector A.

• A) 25 units, 73.74°
• B) 31 units, 45°
• C) 17 units, 60°
• D) 24 units, 90°

Answer Key

Q1: C; Pythagoras for perpendicular vectors gives 5√2.

Q2: B; Use formula √(6² + 8² + 2 × 6 × 8 × cos 60°) = 10.

Q3: A; Resultant = √(7² + 24²) = 25, angle = tan^-1 (24/7) = 73.74°.

Memory Aids

• Triangle law: two sides form triangle, third side is resultant.
• Parallelogram law: diagonal is resultant.
• Resultant magnitude formula: R = √(A² + B² + 2AB cos θ).

General Rule

• Arrange vectors so head of one coincides with tail of next.
• Resultant vector is from tail of first to head of last vector.

Properties

• Commutative: vector A + vector B = vector B + vector A
• Distributive: n(vector A + vector B) = nvector A + nvector B
• Associative: vector A + (vector B + vector C) = (vector A + vector B) + vector C

Triangle Law

• If two vectors are represented by two sides of a triangle in order, resultant is the third side in opposite order.
• Resultant: vector R = vector A + vector B

Parallelogram Law

• Two vectors represented by adjacent sides of a parallelogram.
• Resultant is diagonal from the common point.
• Magnitude: R = √(A² + B² + 2AB cos θ)
• Angle: tan β = B sin θ/(A + B cos θ)

Polygon Law

• Vectors represented by sides of an open polygon in order.
• Resultant is the closing side in opposite order.

Special Cases

• Same direction: R = A + B, β = 0°
• Opposite direction: R = |A - B|, β = 180°
• Perpendicular vectors: R = √(A² + B²), β = tan^-1 (B/A)

Solved Examples

Ex 1: Find resultant of vector A = 3î and vector B = 4ĵ (perpendicular vectors).

• Use Pythagoras: R = √(3² + 4²) = 5
• Angle: β = tan^-1 4/3 = 53.13°

Practice Set

Q1: If vector A = 5î and vector B = 5ĵ, what is |vector A + vector B|?

• A) 5
• B) 10
• C) 5√2
• D) 0

Q2: Two vectors vector A and vector B have magnitudes 6 and 8 units with angle 60° between them. Find resultant magnitude.

• A) 14
• B) 10
• C) 8
• D) 6

Q3: Two vectors vector A = 7 units and vector B = 24 units are perpendicular. Find resultant and angle with vector A.

• A) 25 units, 73.74°
• B) 31 units, 45°
• C) 17 units, 60°
• D) 24 units, 90°

Answer Key

Q1: C; Pythagoras for perpendicular vectors gives 5√2.

Q2: B; Use formula √(6² + 8² + 2 × 6 × 8 × cos 60°) = 10.

Q3: A; Resultant = √(7² + 24²) = 25, angle = tan^-1 (24/7) = 73.74°.

Memory Aids

• Triangle law: two sides form triangle, third side is resultant.
• Parallelogram law: diagonal is resultant.
• Resultant magnitude formula: R = √(A² + B² + 2AB cos θ).

General Rule

• Arrange vectors so head of one coincides with tail of next.
• Resultant vector is from tail of first to head of last vector.

Properties

• Commutative: vector A + vector B = vector B + vector A
• Distributive: n(vector A + vector B) = nvector A + nvector B
• Associative: vector A + (vector B + vector C) = (vector A + vector B) + vector C

Triangle Law

• If two vectors are represented by two sides of a triangle in order, resultant is the third side in opposite order.
• Resultant: vector R = vector A + vector B

Parallelogram Law

• Two vectors represented by adjacent sides of a parallelogram.
• Resultant is diagonal from the common point.
• Magnitude: R = √(A² + B² + 2AB cos θ)
• Angle: tan β = B sin θ/(A + B cos θ)

Polygon Law

• Vectors represented by sides of an open polygon in order.
• Resultant is the closing side in opposite order.

Special Cases

• Same direction: R = A + B, β = 0°
• Opposite direction: R = |A - B|, β = 180°
• Perpendicular vectors: R = √(A² + B²), β = tan^-1 (B/A)

Solved Examples

Ex 1: Find resultant of vector A = 3î and vector B = 4ĵ (perpendicular vectors).

• Use Pythagoras: R = √(3² + 4²) = 5
• Angle: β = tan^-1 4/3 = 53.13°

Practice Set

Q1: If vector A = 5î and vector B = 5ĵ, what is |vector A + vector B|?

• A) 5
• B) 10
• C) 5√2
• D) 0

Q2: Two vectors vector A and vector B have magnitudes 6 and 8 units with angle 60° between them. Find resultant magnitude.

• A) 14
• B) 10
• C) 8
• D) 6

Q3: Two vectors vector A = 7 units and vector B = 24 units are perpendicular. Find resultant and angle with vector A.

• A) 25 units, 73.74°
• B) 31 units, 45°
• C) 17 units, 60°
• D) 24 units, 90°

Answer Key

Q1: C; Pythagoras for perpendicular vectors gives 5√2.

Q2: B; Use formula √(6² + 8² + 2 × 6 × 8 × cos 60°) = 10.

Q3: A; Resultant = √(7² + 24²) = 25, angle = tan^-1 (24/7) = 73.74°.

Memory Aids

• Triangle law: two sides form triangle, third side is resultant.
• Parallelogram law: diagonal is resultant.
• Resultant magnitude formula: R = √(A² + B² + 2AB cos θ).

General Rule

• Arrange vectors so head of one coincides with tail of next.
• Resultant vector is from tail of first to head of last vector.

Properties

• Commutative: \( \vec{A} + \vec{B} = \vec{B} + \vec{A} \)
• Distributive: \( n(\vec{A} + \vec{B}) = n\vec{A} + n\vec{B} \)
• Associative: \( \vec{A} + (\vec{B} + \vec{C}) = (\vec{A} + \vec{B}) + \vec{C} \)

Triangle Law

• If two vectors are represented by two sides of a triangle in order, resultant is the third side in opposite order.
• Resultant: \( \vec{R} = \vec{A} + \vec{B} \)

Parallelogram Law

• Two vectors represented by adjacent sides of a parallelogram.
• Resultant is diagonal from the common point.
• Magnitude: \( R = \sqrt{A^2 + B^2 + 2AB \cos \theta} \)
• Angle: \( \tan \beta = \frac{B \sin \theta}{A + B \cos \theta} \)

Polygon Law

• Vectors represented by sides of an open polygon in order.
• Resultant is the closing side in opposite order.

Special Cases

• Same direction: \( R = A + B, \beta = 0^\circ \)
• Opposite direction: \( R = |A - B|, \beta = 180^\circ \)
• Perpendicular vectors: \( R = \sqrt{A^2 + B^2}, \beta = \tan^{-1} \left( \frac{B}{A} \right) \)

Solved Examples

Ex 1: Find resultant of \( \vec{A} = 3\hat{i} \) and \( \vec{B} = 4\hat{j} \) (perpendicular vectors).

• Use Pythagoras: \( R = \sqrt{3^2 + 4^2} = 5 \)
• Angle: \( \beta = \tan^{-1} \frac{4}{3} = 53.13^\circ \)

Practice Set

Q1: If \( \vec{A} = 5\hat{i} \) and \( \vec{B} = 5\hat{j} \), what is \( |\vec{A} + \vec{B}| \)?

• A) 5
• B) 10
• C) \( 5\sqrt{2} \)
• D) 0

Q2: Two vectors \( \vec{A} \) and \( \vec{B} \) have magnitudes 6 and 8 units with angle 60° between them. Find resultant magnitude.

• A) 14
• B) 10
• C) 8
• D) 6

Q3: Two vectors \( \vec{A} = 7 \) units and \( \vec{B} = 24 \) units are perpendicular. Find resultant and angle with \( \vec{A} \).

• A) 25 units, 73.74°
• B) 31 units, 45°
• C) 17 units, 60°
• D) 24 units, 90°

Answer Key

Q1: C; Pythagoras for perpendicular vectors gives \( 5\sqrt{2} \).

Q2: B; Use formula \( \sqrt{6^2 + 8^2 + 2 \times 6 \times 8 \times \cos 60^\circ} = 10 \).

Q3: A; Resultant \( = \sqrt{7^2 + 24^2} = 25 \), angle \( = \tan^{-1} (24/7) = 73.74^\circ \).

Memory Aids

• Triangle law: two sides form triangle, third side is resultant.
• Parallelogram law: diagonal is resultant.
• Resultant magnitude formula: \( R = \sqrt{A^2 + B^2 + 2AB \cos \theta} \).

Subtraction of Vectors

Definition

• vector A - vector B = vector A + (-vector B)
• Angle between vector A and -vector B is 180° - θ if angle between vector A and vector B is θ.
• Magnitude: |vector A - vector B| = √(A² + B² - 2AB cos θ)

Properties

• Not commutative: vector A - vector B ≠ vector B - vector A
• Not associative: vector A - (vector B - vector C) ≠ (vector A - vector B) - vector C

Solved Examples

Ex 1: Find magnitude of vector A - vector B if A = 5, B = 3, and angle θ = 60°.

• Use formula: √(5² + 3² - 2 × 5 × 3 × cos 60°) = √(25 + 9 - 15) = √19 ≈ 4.36

Practice Set

Q1: If vector A = 4î and vector B = 3î, what is |vector A - vector B|?

• A) 7
• B) 1
• C) 5
• D) 0

Q2: Two vectors vector A and vector B have magnitudes 8 and 6 units with angle 90° between them. Find magnitude of vector A - vector B.

• A) 10
• B) √100
• C) √(64 + 36)
• D) √(64 + 36 - 2 × 8 × 6 × 0)

Q3: If angle between vector A and vector B is 120°, find magnitude of vector A - vector B for A = 10, B = 5.

• A) 15
• B) 10
• C) 12.99
• D) 5

Answer Key

Q1: B; |4 - 3| = 1.

Q2: D; √(64 + 36 - 0) = 10.

Q3: C; √(10² + 5² - 2 × 10 × 5 × cos 120°) = 12.99.

Memory Aids

• Subtraction as addition of negative vector.
• Magnitude formula uses minus sign in cosine term.
• Not commutative or associative.

Definition

• vector A - vector B = vector A + (-vector B)
• Angle between vector A and -vector B is 180° - θ if angle between vector A and vector B is θ.
• Magnitude: |vector A - vector B| = √(A² + B² - 2AB cos θ)

Properties

• Not commutative: vector A - vector B ≠ vector B - vector A
• Not associative: vector A - (vector B - vector C) ≠ (vector A - vector B) - vector C

Solved Examples

Ex 1: Find magnitude of vector A - vector B if A = 5, B = 3, and angle θ = 60°.

• Use formula: √(5² + 3² - 2 × 5 × 3 × cos 60°) = √(25 + 9 - 15) = √19 ≈ 4.36

Practice Set

Q1: If vector A = 4î and vector B = 3î, what is |vector A - vector B|?

• A) 7
• B) 1
• C) 5
• D) 0

Q2: Two vectors vector A and vector B have magnitudes 8 and 6 units with angle 90° between them. Find magnitude of vector A - vector B.

• A) 10
• B) √100
• C) √(64 + 36)
• D) √(64 + 36 - 2 × 8 × 6 × 0)

Q3: If angle between vector A and vector B is 120°, find magnitude of vector A - vector B for A = 10, B = 5.

• A) 15
• B) 10
• C) 12.99
• D) 5

Answer Key

Q1: B; |4 - 3| = 1.

Q2: D; √(64 + 36 - 0) = 10.

Q3: C; √(10² + 5² - 2 × 10 × 5 × cos 120°) = 12.99.

Memory Aids

• Subtraction as addition of negative vector.
• Magnitude formula uses minus sign in cosine term.
• Not commutative or associative.

Definition

• vector A - vector B = vector A + (-vector B)
• Angle between vector A and -vector B is 180° - θ if angle between vector A and vector B is θ.
• Magnitude: |vector A - vector B| = √(A² + B² - 2AB cos θ)

Properties

• Not commutative: vector A - vector B ≠ vector B - vector A
• Not associative: vector A - (vector B - vector C) ≠ (vector A - vector B) - vector C

Solved Examples

Ex 1: Find magnitude of vector A - vector B if A = 5, B = 3, and angle θ = 60°.

• Use formula: √(5² + 3² - 2 × 5 × 3 × cos 60°) = √(25 + 9 - 15) = √19 ≈ 4.36

Practice Set

Q1: If vector A = 4î and vector B = 3î, what is |vector A - vector B|?

• A) 7
• B) 1
• C) 5
• D) 0

Q2: Two vectors vector A and vector B have magnitudes 8 and 6 units with angle 90° between them. Find magnitude of vector A - vector B.

• A) 10
• B) √100
• C) √(64 + 36)
• D) √(64 + 36 - 2 × 8 × 6 × 0)

Q3: If angle between vector A and vector B is 120°, find magnitude of vector A - vector B for A = 10, B = 5.

• A) 15
• B) 10
• C) 12.99
• D) 5

Answer Key

Q1: B; |4 - 3| = 1.

Q2: D; √(64 + 36 - 0) = 10.

Q3: C; √(10² + 5² - 2 × 10 × 5 × cos 120°) = 12.99.

Memory Aids

• Subtraction as addition of negative vector.
• Magnitude formula uses minus sign in cosine term.
• Not commutative or associative.

Definition

• \( \vec{A} - \vec{B} = \vec{A} + (-\vec{B}) \)
• Angle between \( \vec{A} \) and \( -\vec{B} \) is \( 180^\circ - \theta \) if angle between \( \vec{A} \) and \( \vec{B} \) is \( \theta \).
• Magnitude: \( |\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB \cos \theta} \)

Properties

• Not commutative: \( \vec{A} - \vec{B} \neq \vec{B} - \vec{A} \)
• Not associative: \( \vec{A} - (\vec{B} - \vec{C}) \neq (\vec{A} - \vec{B}) - \vec{C} \)

Solved Examples

Ex 1: Find magnitude of \( \vec{A} - \vec{B} \) if \( A = 5 \), \( B = 3 \), and angle \( \theta = 60^\circ \).

• Use formula: \( \sqrt{5^2 + 3^2 - 2 \times 5 \times 3 \times \cos 60^\circ} = \sqrt{25 + 9 - 15} = \sqrt{19} \approx 4.36 \)

Practice Set

Q1: If \( \vec{A} = 4\hat{i} \) and \( \vec{B} = 3\hat{i} \), what is \( |\vec{A} - \vec{B}| \)?

• A) 7
• B) 1
• C) 5
• D) 0

Q2: Two vectors \( \vec{A} \) and \( \vec{B} \) have magnitudes 8 and 6 units with angle 90° between them. Find magnitude of \( \vec{A} - \vec{B} \).

• A) 10
• B) \( \sqrt{100} \)
• C) \( \sqrt{64 + 36} \)
• D) \( \sqrt{64 + 36 - 2 \times 8 \times 6 \times 0} \)

Q3: If angle between \( \vec{A} \) and \( \vec{B} \) is 120°, find magnitude of \( \vec{A} - \vec{B} \) for \( A = 10 \), \( B = 5 \).

• A) 15
• B) 10
• C) 12.99
• D) 5

Answer Key

Q1: B; \( |4 - 3| = 1 \).

Q2: D; \( \sqrt{64 + 36 - 0} = 10 \).

Q3: C; \( \sqrt{10^2 + 5^2 - 2 \times 10 \times 5 \times \cos 120^\circ} = 12.99 \).

Memory Aids

• Subtraction as addition of negative vector.
• Magnitude formula uses minus sign in cosine term.
• Not commutative or associative.

Resolution of Vectors

Concept and Definition

• Resolution of a vector: Splitting a vector into two or more component vectors.
• Component vectors: Resultant vectors after splitting.
• Rectangular components: Components at right angles (perpendicular) to each other.

Solved Examples

• Ex 1: Vector vector A is split into two perpendicular components A_x and A_y.
• Use A_x = A cos θ, A_y = A sin θ where θ is angle with x-axis.

Practice Set

• Q1: What is the process of splitting a vector into components called?
  • A. Addition
  • B. Resolution
  • C. Multiplication
  • D. Integration
• Q2: If vector A makes an angle 30° with x-axis and magnitude 10 units, what is A_x?
  • A. 5
  • B. 8.66
  • C. 10
  • D. 0
• Q3: Which of the following is true for rectangular components?
  • A. Components are parallel
  • B. Components are perpendicular
  • C. Components are equal
  • D. Components are zero

Answer Key

• Q1: B - Resolution is the process of splitting a vector into components.
• Q2: B - A_x = 10 cos 30° = 8.66
• Q3: B - Rectangular components are perpendicular to each other.

Memory Aids

• Remember: A_x = A cos θ, A_y = A sin θ
• Rectangular components form right angles.
• Resolution = splitting vector into perpendicular parts.

Concept and Definition

• Resolution of a vector: Splitting a vector into two or more component vectors.
• Component vectors: Resultant vectors after splitting.
• Rectangular components: Components at right angles (perpendicular) to each other.

Solved Examples

• Ex 1: Vector \\(\vec{A}\\) is split into two perpendicular components \\(A_x\\) and \\(A_y\\).
• Use \\(A_x = A \cos \theta\\), \\(A_y = A \sin \theta\\) where \\(\theta\\) is angle with x-axis.

Practice Set

• Q1: What is the process of splitting a vector into components called?
  • A. Addition
  • B. Resolution
  • C. Multiplication
  • D. Integration
• Q2: If \\(\vec{A}\\) makes an angle \\(30^\circ\\) with x-axis and magnitude \\(10\\,\mathrm{units}\\), what is \\(A_x\\)?
  • A. 5
  • B. 8.66
  • C. 10
  • D. 0
• Q3: Which of the following is true for rectangular components?
  • A. Components are parallel
  • B. Components are perpendicular
  • C. Components are equal
  • D. Components are zero

Answer Key

• Q1: B - Resolution is the process of splitting a vector into components.
• Q2: B - \\(A_x = 10 \cos 30^\circ = 8.66\\)
• Q3: B - Rectangular components are perpendicular to each other.

Memory Aids

• Remember: \\(A_x = A \cos \theta\\), \\(A_y = A \sin \theta\\)
• Rectangular components form right angles.
• Resolution = splitting vector into perpendicular parts.

Concept and Definition

• Resolution of a vector: Splitting a vector into two or more component vectors.
• Component vectors: Resultant vectors after splitting.
• Rectangular components: Components at right angles (perpendicular) to each other.

Solved Examples

• Ex 1: Vector vector A is split into two perpendicular components A_x and A_y.
• Use A_x = A cos θ, A_y = A sin θ where θ is angle with x-axis.

Practice Set

• Q1: What is the process of splitting a vector into components called?
  • A. Addition
  • B. Resolution
  • C. Multiplication
  • D. Integration
• Q2: If vector A makes an angle 30° with x-axis and magnitude 10 units, what is A_x?
  • A. 5
  • B. 8.66
  • C. 10
  • D. 0
• Q3: Which of the following is true for rectangular components?
  • A. Components are parallel
  • B. Components are perpendicular
  • C. Components are equal
  • D. Components are zero

Answer Key

• Q1: B - Resolution is the process of splitting a vector into components.
• Q2: B - A_x = 10 cos 30° = 8.66
• Q3: B - Rectangular components are perpendicular to each other.

Memory Aids

• Remember: A_x = A cos θ, A_y = A sin θ
• Rectangular components form right angles.
• Resolution = splitting vector into perpendicular parts.

Concept and Definition

• Resolution of a vector: Splitting a vector into two or more component vectors.
• Component vectors: Resultant vectors after splitting.
• Rectangular components: Components at right angles (perpendicular) to each other.

Solved Examples

• Ex 1: Vector \(\vec{A}\) is split into two perpendicular components \(A_x\) and \(A_y\).
• Use \(A_x = A \cos \theta\), \(A_y = A \sin \theta\) where \(\theta\) is angle with x-axis.

Practice Set

• Q1: What is the process of splitting a vector into components called?
  • A. Addition
  • B. Resolution
  • C. Multiplication
  • D. Integration
• Q2: If \(\vec{A}\) makes an angle \(30^\circ\) with x-axis and magnitude \(10\,\mathrm{units}\), what is \(A_x\)?
  • A. 5
  • B. 8.66
  • C. 10
  • D. 0
• Q3: Which of the following is true for rectangular components?
  • A. Components are parallel
  • B. Components are perpendicular
  • C. Components are equal
  • D. Components are zero

Answer Key

• Q1: B - Resolution is the process of splitting a vector into components.
• Q2: B - \(A_x = 10 \cos 30^\circ = 8.66\)
• Q3: B - Rectangular components are perpendicular to each other.

Memory Aids

• Remember: \(A_x = A \cos \theta\), \(A_y = A \sin \theta\)
• Rectangular components form right angles.
• Resolution = splitting vector into perpendicular parts.

Rectangular Components in a Plane

Key Formulas and Concepts

• Vector vector A makes angle θ with x-axis.
• Components: vector A = A_x î + A_y ĵ
• A_x = A cos θ, A_y = A sin θ
• Magnitude: A = √(A_x² + A_y²)
• Direction: tan θ = A_y/A_x

Solved Examples

• Ex 1: Find A_x and A_y if A=5 units and θ=60°.
• A_x = 5 cos 60° = 2.5, A_y = 5 sin 60° = 4.33

Practice Set

• Q1: If A=10 units and θ=45°, what is A_y?
  • A. 7.07
  • B. 5
  • C. 10
  • D. 0
• Q2: The magnitude of vector A is 13 units, A_x=5. Find A_y.
  • A. 12
  • B. 8
  • C. 10
  • D. 7
• Q3: If tan θ = 1, what is the angle θ?
  • A. 30°
  • B. 45°
  • C. 60°
  • D. 90°

Answer Key

• Q1: A - A_y = 10 sin 45° = 7.07
• Q2: B - A_y = √(13² - 5²) = 12 (Check: actually √(169-25)=√144=12)
• Q3: B - tan 45° = 1

Memory Aids

• Use Pythagoras: A² = A_x² + A_y²
• tan θ = A_y/A_x for direction
• Cosine for x-component, sine for y-component

Key Formulas and Concepts

• Vector \\(\vec{A}\\) makes angle \\(\theta\\) with x-axis.
• Components: \\(\vec{A} = A_x \hat{i} + A_y \hat{j}\\)
• \\(A_x = A \cos \theta\\), \\(A_y = A \sin \theta\\)
• Magnitude: \\(A = \sqrt{A_x^2 + A_y^2}\\)
• Direction: \\(\tan \theta = \frac{A_y}{A_x}\\)

Solved Examples

• Ex 1: Find \\(A_x\\) and \\(A_y\\) if \\(A=5\\) units and \\(\theta=60^\circ\\).
• \\(A_x = 5 \cos 60^\circ = 2.5\\), \\(A_y = 5 \sin 60^\circ = 4.33\\)

Practice Set

• Q1: If \\(A=10\\) units and \\(\theta=45^\circ\\), what is \\(A_y\\)?
  • A. 7.07
  • B. 5
  • C. 10
  • D. 0
• Q2: The magnitude of \\(\vec{A}\\) is 13 units, \\(A_x=5\\). Find \\(A_y\\).
  • A. 12
  • B. 8
  • C. 10
  • D. 7
• Q3: If \\(\tan \theta = 1\\), what is the angle \\(\theta\\)?
  • A. 30^\circ
  • B. 45^\circ
  • C. 60^\circ
  • D. 90^\circ

Answer Key

• Q1: A - \\(A_y = 10 \sin 45^\circ = 7.07\\)
• Q2: B - \\(A_y = \sqrt{13^2 - 5^2} = 12\\) (Check: actually \\(\sqrt{169-25}=\sqrt{144}=12\\))
• Q3: B - \\(\tan 45^\circ = 1\\)

Memory Aids

• Use Pythagoras: \\(A^2 = A_x^2 + A_y^2\\)
• \\(\tan \theta = \frac{A_y}{A_x}\\) for direction
• Cosine for x-component, sine for y-component

Key Formulas and Concepts

• Vector vector A makes angle θ with x-axis.
• Components: vector A = A_x î + A_y ĵ
• A_x = A cos θ, A_y = A sin θ
• Magnitude: A = √(A_x² + A_y²)
• Direction: tan θ = A_y/A_x

Solved Examples

• Ex 1: Find A_x and A_y if A=5 units and θ=60°.
• A_x = 5 cos 60° = 2.5, A_y = 5 sin 60° = 4.33

Practice Set

• Q1: If A=10 units and θ=45°, what is A_y?
  • A. 7.07
  • B. 5
  • C. 10
  • D. 0
• Q2: The magnitude of vector A is 13 units, A_x=5. Find A_y.
  • A. 12
  • B. 8
  • C. 10
  • D. 7
• Q3: If tan θ = 1, what is the angle θ?
  • A. 30°
  • B. 45°
  • C. 60°
  • D. 90°

Answer Key

• Q1: A - A_y = 10 sin 45° = 7.07
• Q2: B - A_y = √(13² - 5²) = 12 (Check: actually √(169-25)=√144=12)
• Q3: B - tan 45° = 1

Memory Aids

• Use Pythagoras: A² = A_x² + A_y²
• tan θ = A_y/A_x for direction
• Cosine for x-component, sine for y-component

Key Formulas and Concepts

• Vector \(\vec{A}\) makes angle \(\theta\) with x-axis.
• Components: \(\vec{A} = A_x \hat{i} + A_y \hat{j}\)
• \(A_x = A \cos \theta\), \(A_y = A \sin \theta\)
• Magnitude: \(A = \sqrt{A_x^2 + A_y^2}\)
• Direction: \(\tan \theta = \frac{A_y}{A_x}\)

Solved Examples

• Ex 1: Find \(A_x\) and \(A_y\) if \(A=5\) units and \(\theta=60^\circ\).
• \(A_x = 5 \cos 60^\circ = 2.5\), \(A_y = 5 \sin 60^\circ = 4.33\)

Practice Set

• Q1: If \(A=10\) units and \(\theta=45^\circ\), what is \(A_y\)?
  • A. 7.07
  • B. 5
  • C. 10
  • D. 0
• Q2: The magnitude of \(\vec{A}\) is 13 units, \(A_x=5\). Find \(A_y\).
  • A. 12
  • B. 8
  • C. 10
  • D. 7
• Q3: If \(\tan \theta = 1\), what is the angle \(\theta\)?
  • A. 30^\circ
  • B. 45^\circ
  • C. 60^\circ
  • D. 90^\circ

Answer Key

• Q1: A - \(A_y = 10 \sin 45^\circ = 7.07\)
• Q2: B - \(A_y = \sqrt{13^2 - 5^2} = 12\) (Check: actually \(\sqrt{169-25}=\sqrt{144}=12\))
• Q3: B - \(\tan 45^\circ = 1\)

Memory Aids

• Use Pythagoras: \(A^2 = A_x^2 + A_y^2\)
• \(\tan \theta = \frac{A_y}{A_x}\) for direction
• Cosine for x-component, sine for y-component

Rectangular Components in Space

Key Formulas and Concepts

• Vector vector A makes angles α, β, γ with x, y, z axes.
• Components: vector A = A_x î + A_y ĵ + A_z k̂
• A_x = A cos α, A_y = A cos β, A_z = A cos γ
• Magnitude: A = √(A_x² + A_y² + A_z²)
• Direction cosines: cos² α + cos² β + cos² γ = 1

Solved Examples

• Ex 1: Given A=10, α=60°, β=45°, find A_x and A_y.
• A_x = 10 cos 60° = 5, A_y = 10 cos 45° = 7.07

Practice Set

• Q1: If A=5 and cos α = 0.6, find A_x.
  • A. 3
  • B. 5
  • C. 0.6
  • D. 8.3
• Q2: Which equation represents direction cosines?
  • A. cos α + cos β + cos γ = 1
  • B. cos² α + cos² β + cos² γ = 1
  • C. sin² α + sin² β + sin² γ = 1
  • D. tan α + tan β + tan γ = 1
• Q3: Find magnitude A if A_x=3, A_y=4, A_z=12.
  • A. 13
  • B. 14
  • C. 15
  • D. 16

Answer Key

• Q1: A - A_x = 5 × 0.6 = 3
• Q2: B - Direction cosines satisfy cos² α + cos² β + cos² γ = 1
• Q3: C - A = √(3² + 4² + 12²) = 13

Memory Aids

• Direction cosines squared sum to 1.
• Use cosine of angles for components.
• Magnitude from 3D Pythagoras.

Key Formulas and Concepts

• Vector \(\vec{A}\) makes angles \(\alpha, \beta, \gamma\) with x, y, z axes.
• Components: \(\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}\)
• \(A_x = A \cos \alpha\), \(A_y = A \cos \beta\), \(A_z = A \cos \gamma\)
• Magnitude: \(A = \sqrt{A_x^2 + A_y^2 + A_z^2}\)
• Direction cosines: \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\)

Solved Examples

• Ex 1: Given \(A=10\), \(\alpha=60^\circ\), \(\beta=45^\circ\), find \(A_x\) and \(A_y\).
• \(A_x = 10 \cos 60^\circ = 5\), \(A_y = 10 \cos 45^\circ = 7.07\)

Practice Set

• Q1: If \(A=5\) and \(\cos \alpha = 0.6\), find \(A_x\).
  • A. 3
  • B. 5
  • C. 0.6
  • D. 8.3
• Q2: Which equation represents direction cosines?
  • A. \(\cos \alpha + \cos \beta + \cos \gamma = 1\)
  • B. \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\)
  • C. \(\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 1\)
  • D. \(\tan \alpha + \tan \beta + \tan \gamma = 1\)
• Q3: Find magnitude \(A\) if \(A_x=3\), \(A_y=4\), \(A_z=12\).
  • A. 13
  • B. 14
  • C. 15
  • D. 16

Answer Key

• Q1: A - \(A_x = 5 \times 0.6 = 3\)
• Q2: B - Direction cosines satisfy \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\)
• Q3: C - \(A = \sqrt{3^2 + 4^2 + 12^2} = 13\)

Memory Aids

• Direction cosines squared sum to 1.
• Use cosine of angles for components.
• Magnitude from 3D Pythagoras.

Vector Addition Analytical Method

Component-wise Addition

• Vectors vector A and vector B in x-y plane: vector A = A_x î + A_y ĵ, vector B = B_x î + B_y ĵ
• Resultant vector vector R = vector A + vector B = (A_x + B_x) î + (A_y + B_y) ĵ = R_x î + R_y ĵ
• Each component of vector R is sum of corresponding components of vector A and vector B
• In 3D, vector A = A_x î + A_y ĵ + A_z k̂, vector B = B_x î + B_y ĵ + B_z k̂
• Resultant in 3D: vector R = (A_x + B_x) î + (A_y + B_y) ĵ + (A_z + B_z) k̂ = R_x î + R_y ĵ + R_z k̂
• Method extends to any number of vectors for addition or subtraction

Solved Examples

• Ex 1: Given vector A = 3î + 4ĵ, vector B = 1î + 2ĵ, find vector R = vector A + vector B.
  • Calculate components: R_x = 3 + 1 = 4, R_y = 4 + 2 = 6
  • Result: vector R = 4î + 6ĵ
• Ex 2: For 3D vectors vector A = 2î + 3ĵ + 4k̂, vector B = 1î + 0ĵ + 5k̂, find vector R = vector A + vector B.
  • Calculate components: R_x = 2 + 1 = 3, R_y = 3 + 0 = 3, R_z = 4 + 5 = 9
  • Result: vector R = 3î + 3ĵ + 9k̂

Practice Set

• Q1: If vector A = 5î + 2ĵ and vector B = 3î + 4ĵ, what is R_y of vector R = vector A + vector B?
  • A. 6
  • B. 8
  • C. 2
  • D. 4
• Q2: For vectors vector A = 1î + 2ĵ + 3k̂ and vector B = 4î + 0ĵ + 1k̂, find R_z.
  • A. 4
  • B. 3
  • C. 1
  • D. 2
• Q3: Given vector A = 2î + 3ĵ, vector B = -1î + 4ĵ, and vector C = 3î - 2ĵ, find R_x of vector R = vector A + vector B + vector C.
  • A. 4
  • B. 3
  • C. 2
  • D. 5

Answer Key

• Q1: B. 8 — Sum of y-components: 2 + 4 = 6 (Correction: 2 + 4 = 6, so correct answer is A. 6)
• Q2: D. 4 — Sum of z-components: 3 + 1 = 4
• Q3: A. 4 — Sum of x-components: 2 + (-1) + 3 = 4

Memory Aids

• Resultant components = sum of individual components
• Use R_x = ∑ A_x, R_y = ∑ A_y, R_z = ∑ A_z
• Extend method to any number of vectors

Component-wise Addition

• Vectors vector A and vector B in x-y plane: vector A = A_x î + A_y ĵ, vector B = B_x î + B_y ĵ
• Resultant vector vector R = vector A + vector B = (A_x + B_x) î + (A_y + B_y) ĵ = R_x î + R_y ĵ
• Each component of vector R is sum of corresponding components of vector A and vector B
• In 3D, vector A = A_x î + A_y ĵ + A_z k̂, vector B = B_x î + B_y ĵ + B_z k̂
• Resultant in 3D: vector R = (A_x + B_x) î + (A_y + B_y) ĵ + (A_z + B_z) k̂ = R_x î + R_y ĵ + R_z k̂
• Method extends to any number of vectors for addition or subtraction

Solved Examples

• Ex 1: Given vector A = 3î + 4ĵ, vector B = 1î + 2ĵ, find vector R = vector A + vector B.
  • Calculate components: R_x = 3 + 1 = 4, R_y = 4 + 2 = 6
  • Result: vector R = 4î + 6ĵ
• Ex 2: For 3D vectors vector A = 2î + 3ĵ + 4k̂, vector B = 1î + 0ĵ + 5k̂, find vector R = vector A + vector B.
  • Calculate components: R_x = 2 + 1 = 3, R_y = 3 + 0 = 3, R_z = 4 + 5 = 9
  • Result: vector R = 3î + 3ĵ + 9k̂

Practice Set

• Q1: If vector A = 5î + 2ĵ and vector B = 3î + 4ĵ, what is R_y of vector R = vector A + vector B?
  • A. 6
  • B. 8
  • C. 2
  • D. 4
• Q2: For vectors vector A = 1î + 2ĵ + 3k̂ and vector B = 4î + 0ĵ + 1k̂, find R_z.
  • A. 4
  • B. 3
  • C. 1
  • D. 2
• Q3: Given vector A = 2î + 3ĵ, vector B = -1î + 4ĵ, and vector C = 3î - 2ĵ, find R_x of vector R = vector A + vector B + vector C.
  • A. 4
  • B. 3
  • C. 2
  • D. 5

Answer Key

• Q1: B. 8 — Sum of y-components: 2 + 4 = 6 (Correction: 2 + 4 = 6, so correct answer is A. 6)
• Q2: D. 4 — Sum of z-components: 3 + 1 = 4
• Q3: A. 4 — Sum of x-components: 2 + (-1) + 3 = 4

Memory Aids

• Resultant components = sum of individual components
• Use R_x = ∑ A_x, R_y = ∑ A_y, R_z = ∑ A_z
• Extend method to any number of vectors

Component-wise Addition

• Vectors vector A and vector B in x-y plane: vector A = A_x î + A_y ĵ, vector B = B_x î + B_y ĵ
• Resultant vector vector R = vector A + vector B = (A_x + B_x) î + (A_y + B_y) ĵ = R_x î + R_y ĵ
• Each component of vector R is sum of corresponding components of vector A and vector B
• In 3D, vector A = A_x î + A_y ĵ + A_z k̂, vector B = B_x î + B_y ĵ + B_z k̂
• Resultant in 3D: vector R = (A_x + B_x) î + (A_y + B_y) ĵ + (A_z + B_z) k̂ = R_x î + R_y ĵ + R_z k̂
• Method extends to any number of vectors for addition or subtraction

Solved Examples

• Ex 1: Given vector A = 3î + 4ĵ, vector B = 1î + 2ĵ, find vector R = vector A + vector B.
  • Calculate components: R_x = 3 + 1 = 4, R_y = 4 + 2 = 6
  • Result: vector R = 4î + 6ĵ
• Ex 2: For 3D vectors vector A = 2î + 3ĵ + 4k̂, vector B = 1î + 0ĵ + 5k̂, find vector R = vector A + vector B.
  • Calculate components: R_x = 2 + 1 = 3, R_y = 3 + 0 = 3, R_z = 4 + 5 = 9
  • Result: vector R = 3î + 3ĵ + 9k̂

Practice Set

• Q1: If vector A = 5î + 2ĵ and vector B = 3î + 4ĵ, what is R_y of vector R = vector A + vector B?
  • A. 6
  • B. 8
  • C. 2
  • D. 4
• Q2: For vectors vector A = 1î + 2ĵ + 3k̂ and vector B = 4î + 0ĵ + 1k̂, find R_z.
  • A. 4
  • B. 3
  • C. 1
  • D. 2
• Q3: Given vector A = 2î + 3ĵ, vector B = -1î + 4ĵ, and vector C = 3î - 2ĵ, find R_x of vector R = vector A + vector B + vector C.
  • A. 4
  • B. 3
  • C. 2
  • D. 5

Answer Key

• Q1: B. 8 — Sum of y-components: 2 + 4 = 6 (Correction: 2 + 4 = 6, so correct answer is A. 6)
• Q2: D. 4 — Sum of z-components: 3 + 1 = 4
• Q3: A. 4 — Sum of x-components: 2 + (-1) + 3 = 4

Memory Aids

• Resultant components = sum of individual components
• Use R_x = ∑ A_x, R_y = ∑ A_y, R_z = ∑ A_z
• Extend method to any number of vectors

Component-wise Addition

• Vectors \( \vec{A} \) and \( \vec{B} \) in x-y plane: \( \vec{A} = A_x \hat{i} + A_y \hat{j} \), \( \vec{B} = B_x \hat{i} + B_y \hat{j} \)
• Resultant vector \( \vec{R} = \vec{A} + \vec{B} = (A_x + B_x) \hat{i} + (A_y + B_y) \hat{j} = R_x \hat{i} + R_y \hat{j} \)
• Each component of \( \vec{R} \) is sum of corresponding components of \( \vec{A} \) and \( \vec{B} \)
• In 3D, \( \vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k} \), \( \vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k} \)
• Resultant in 3D: \( \vec{R} = (A_x + B_x) \hat{i} + (A_y + B_y) \hat{j} + (A_z + B_z) \hat{k} = R_x \hat{i} + R_y \hat{j} + R_z \hat{k} \)
• Method extends to any number of vectors for addition or subtraction

Solved Examples

• Ex 1: Given \( \vec{A} = 3\hat{i} + 4\hat{j} \), \( \vec{B} = 1\hat{i} + 2\hat{j} \), find \( \vec{R} = \vec{A} + \vec{B} \).
  • Calculate components: \( R_x = 3 + 1 = 4 \), \( R_y = 4 + 2 = 6 \)
  • Result: \( \vec{R} = 4\hat{i} + 6\hat{j} \)
• Ex 2: For 3D vectors \( \vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k} \), \( \vec{B} = 1\hat{i} + 0\hat{j} + 5\hat{k} \), find \( \vec{R} = \vec{A} + \vec{B} \).
  • Calculate components: \( R_x = 2 + 1 = 3 \), \( R_y = 3 + 0 = 3 \), \( R_z = 4 + 5 = 9 \)
  • Result: \( \vec{R} = 3\hat{i} + 3\hat{j} + 9\hat{k} \)

Practice Set

• Q1: If \( \vec{A} = 5\hat{i} + 2\hat{j} \) and \( \vec{B} = 3\hat{i} + 4\hat{j} \), what is \( R_y \) of \( \vec{R} = \vec{A} + \vec{B} \)?
  • A. 6
  • B. 8
  • C. 2
  • D. 4
• Q2: For vectors \( \vec{A} = 1\hat{i} + 2\hat{j} + 3\hat{k} \) and \( \vec{B} = 4\hat{i} + 0\hat{j} + 1\hat{k} \), find \( R_z \).
  • A. 4
  • B. 3
  • C. 1
  • D. 2
• Q3: Given \( \vec{A} = 2\hat{i} + 3\hat{j} \), \( \vec{B} = -1\hat{i} + 4\hat{j} \), and \( \vec{C} = 3\hat{i} - 2\hat{j} \), find \( R_x \) of \( \vec{R} = \vec{A} + \vec{B} + \vec{C} \).
  • A. 4
  • B. 3
  • C. 2
  • D. 5

Answer Key

• Q1: B. 8 — Sum of y-components: 2 + 4 = 6 (Correction: 2 + 4 = 6, so correct answer is A. 6)
• Q2: D. 4 — Sum of z-components: 3 + 1 = 4
• Q3: A. 4 — Sum of x-components: 2 + (-1) + 3 = 4

Memory Aids

• Resultant components = sum of individual components
• Use \( R_x = \sum A_x \), \( R_y = \sum A_y \), \( R_z = \sum A_z \)
• Extend method to any number of vectors

Position Vector

Concepts

• Position vector vector r = xî + yĵ locates a particle in the xy-plane from origin O.
• Components: x along x-axis, y along y-axis.
• Represents the particle's position as a vector from origin to point P.

Solved Examples

• Ex 1: Find position vector for x=3 m, y=4 m.
  • vector r = 3î + 4ĵ m

Practice Set

• Q1: Position vector of point P(2,5) is?
  • A) 2î + 5ĵ
  • B) 5î + 2ĵ
  • C) 2î - 5ĵ
  • D) -2î + 5ĵ
• Q2: If vector r = 4î + 3ĵ, magnitude is?
  • A) 5
  • B) 7
  • C) 1
  • D) 12
• Q3: For vector r = 6î - 8ĵ, direction angle θ with x-axis is?
  • A) 53.13°
  • B) 126.87°
  • C) 36.87°
  • D) 143.13°

Answer Key

• Q1: A) 2î + 5ĵ - Direct components from coordinates.
• Q2: A) 5 - Magnitude = √(4² + 3²) = 5.
• Q3: B) 126.87° - θ = tan^-1(-8/6) in 2nd quadrant.

Memory Aids

• Position vector = xî + yĵ.
• Magnitude = √(x² + y²).
• Direction θ = tan^-1(y/x).

Concepts

• Position vector \\(\vec{r} = x\hat{i} + y\hat{j}\\) locates a particle in the xy-plane from origin \\(O\\).
• Components: \\(x\\) along x-axis, \\(y\\) along y-axis.
• Represents the particle's position as a vector from origin to point \\(P\\).

Solved Examples

• Ex 1: Find position vector for \\(x=3\\,m\\), \\(y=4\\,m\\).
  • \\(\vec{r} = 3\hat{i} + 4\hat{j}\\,m\\)

Practice Set

• Q1: Position vector of point \\(P(2,5)\\) is?
  • A) \\(2\hat{i} + 5\hat{j}\\)
  • B) \\(5\hat{i} + 2\hat{j}\\)
  • C) \\(2\hat{i} - 5\hat{j}\\)
  • D) \\(-2\hat{i} + 5\hat{j}\\)
• Q2: If \\(\vec{r} = 4\hat{i} + 3\hat{j}\\), magnitude is?
  • A) 5
  • B) 7
  • C) 1
  • D) 12
• Q3: For \\(\vec{r} = 6\hat{i} - 8\hat{j}\\), direction angle \\(\theta\\) with x-axis is?
  • A) \\(53.13^\circ\\)
  • B) \\(126.87^\circ\\)
  • C) \\(36.87^\circ\\)
  • D) \\(143.13^\circ\\)

Answer Key

• Q1: A) \\(2\hat{i} + 5\hat{j}\\) - Direct components from coordinates.
• Q2: A) 5 - Magnitude \\(= \\sqrt{4^2 + 3^2} = 5\\).
• Q3: B) \\(126.87^\circ\\) - \\(\theta = \\tan^{-1}(-8/6)\\) in 2nd quadrant.

Memory Aids

• Position vector = \\(x\hat{i} + y\hat{j}\\).
• Magnitude = \\(\\sqrt{x^2 + y^2}\\).
• Direction \\(\theta = \\tan^{-1}(y/x)\\).

Concepts

• Position vector vector r = xî + yĵ locates a particle in the xy-plane from origin O.
• Components: x along x-axis, y along y-axis.
• Represents the particle's position as a vector from origin to point P.

Solved Examples

• Ex 1: Find position vector for x=3 m, y=4 m.
  • vector r = 3î + 4ĵ m

Practice Set

• Q1: Position vector of point P(2,5) is?
  • A) 2î + 5ĵ
  • B) 5î + 2ĵ
  • C) 2î - 5ĵ
  • D) -2î + 5ĵ
• Q2: If vector r = 4î + 3ĵ, magnitude is?
  • A) 5
  • B) 7
  • C) 1
  • D) 12
• Q3: For vector r = 6î - 8ĵ, direction angle θ with x-axis is?
  • A) 53.13°
  • B) 126.87°
  • C) 36.87°
  • D) 143.13°

Answer Key

• Q1: A) 2î + 5ĵ - Direct components from coordinates.
• Q2: A) 5 - Magnitude = √(4² + 3²) = 5.
• Q3: B) 126.87° - θ = tan^-1(-8/6) in 2nd quadrant.

Memory Aids

• Position vector = xî + yĵ.
• Magnitude = √(x² + y²).
• Direction θ = tan^-1(y/x).

Concepts

• Position vector \(\vec{r} = x\hat{i} + y\hat{j}\) locates a particle in the xy-plane from origin \(O\).
• Components: \(x\) along x-axis, \(y\) along y-axis.
• Represents the particle's position as a vector from origin to point \(P\).

Solved Examples

• Ex 1: Find position vector for \(x=3\,m\), \(y=4\,m\).
  • \(\vec{r} = 3\hat{i} + 4\hat{j}\,m\)

Practice Set

• Q1: Position vector of point \(P(2,5)\) is?
  • A) \(2\hat{i} + 5\hat{j}\)
  • B) \(5\hat{i} + 2\hat{j}\)
  • C) \(2\hat{i} - 5\hat{j}\)
  • D) \(-2\hat{i} + 5\hat{j}\)
• Q2: If \(\vec{r} = 4\hat{i} + 3\hat{j}\), magnitude is?
  • A) 5
  • B) 7
  • C) 1
  • D) 12
• Q3: For \(\vec{r} = 6\hat{i} - 8\hat{j}\), direction angle \(\theta\) with x-axis is?
  • A) \(53.13^\circ\)
  • B) \(126.87^\circ\)
  • C) \(36.87^\circ\)
  • D) \(143.13^\circ\)

Answer Key

• Q1: A) \(2\hat{i} + 5\hat{j}\) - Direct components from coordinates.
• Q2: A) 5 - Magnitude \(= \sqrt{4^2 + 3^2} = 5\).
• Q3: B) \(126.87^\circ\) - \(\theta = \tan^{-1}(-8/6)\) in 2nd quadrant.

Memory Aids

• Position vector = \(x\hat{i} + y\hat{j}\).
• Magnitude = \(\sqrt{x^2 + y^2}\).
• Direction \(\theta = \tan^{-1}(y/x)\).

Displacement Vector

Concepts

• Displacement Δvector r = vector r' - vector r = îΔx + ĵΔy.
• Directed from initial point P to final point P'.
• Represents change in position vector.

Solved Examples

• Ex 1: Particle moves from (2,3) to (5,7).
  • Δvector r = (5-2)î + (7-3)ĵ = 3î + 4ĵ

Practice Set

• Q1: Displacement from (1,2) to (4,6) is?
  • A) 3î + 4ĵ
  • B) 4î + 6ĵ
  • C) 1î + 2ĵ
  • D) -3î - 4ĵ
• Q2: If Δvector r = 5î - 12ĵ, magnitude is?
  • A) 13
  • B) 7
  • C) 17
  • D) 10
• Q3: Displacement vector direction for Δx = -3, Δy = 4 is?
  • A) 126.87°
  • B) 53.13°
  • C) 36.87°
  • D) 143.13°

Answer Key

• Q1: A) 3î + 4ĵ - Difference of coordinates.
• Q2: A) 13 - Magnitude = √(5² + (-12)²) = 13.
• Q3: A) 126.87° - tan^-1(4/-3) in 2nd quadrant.

Memory Aids

• Displacement = final position - initial position.
• Component form: Δvector r = îΔx + ĵΔy.
• Direction θ = tan^-1(Δy / Δx).

Concepts

• Displacement \\(\Delta \vec{r} = \vec{r'} - \vec{r} = \hat{i}\Delta x + \hat{j}\Delta y\\).
• Directed from initial point \\(P\\) to final point \\(P'\\).
• Represents change in position vector.

Solved Examples

• Ex 1: Particle moves from \\((2,3)\\) to \\((5,7)\\).
  • \\(\Delta \vec{r} = (5-2)\hat{i} + (7-3)\hat{j} = 3\hat{i} + 4\hat{j}\\)

Practice Set

• Q1: Displacement from \\((1,2)\\) to \\((4,6)\\) is?
  • A) \\(3\hat{i} + 4\hat{j}\\)
  • B) \\(4\hat{i} + 6\hat{j}\\)
  • C) \\(1\hat{i} + 2\hat{j}\\)
  • D) \\(-3\hat{i} - 4\hat{j}\\)
• Q2: If \\(\Delta \vec{r} = 5\hat{i} - 12\hat{j}\\), magnitude is?
  • A) 13
  • B) 7
  • C) 17
  • D) 10
• Q3: Displacement vector direction for \\(\Delta x = -3\\), \\(\Delta y = 4\\) is?
  • A) \\(126.87^\circ\\)
  • B) \\(53.13^\circ\\)
  • C) \\(36.87^\circ\\)
  • D) \\(143.13^\circ\\)

Answer Key

• Q1: A) \\(3\hat{i} + 4\hat{j}\\) - Difference of coordinates.
• Q2: A) 13 - Magnitude \\(= \\sqrt{5^2 + (-12)^2} = 13\\).
• Q3: A) \\(126.87^\circ\\) - \\(\tan^{-1}(4/-3)\\) in 2nd quadrant.

Memory Aids

• Displacement = final position - initial position.
• Component form: \\(\Delta \vec{r} = \hat{i}\Delta x + \hat{j}\Delta y\\).
• Direction \\(\theta = \\tan^{-1}(\\Delta y / \\Delta x)\\).

Concepts

• Displacement Δvector r = vector r' - vector r = îΔx + ĵΔy.
• Directed from initial point P to final point P'.
• Represents change in position vector.

Solved Examples

• Ex 1: Particle moves from (2,3) to (5,7).
  • Δvector r = (5-2)î + (7-3)ĵ = 3î + 4ĵ

Practice Set

• Q1: Displacement from (1,2) to (4,6) is?
  • A) 3î + 4ĵ
  • B) 4î + 6ĵ
  • C) 1î + 2ĵ
  • D) -3î - 4ĵ
• Q2: If Δvector r = 5î - 12ĵ, magnitude is?
  • A) 13
  • B) 7
  • C) 17
  • D) 10
• Q3: Displacement vector direction for Δx = -3, Δy = 4 is?
  • A) 126.87°
  • B) 53.13°
  • C) 36.87°
  • D) 143.13°

Answer Key

• Q1: A) 3î + 4ĵ - Difference of coordinates.
• Q2: A) 13 - Magnitude = √(5² + (-12)²) = 13.
• Q3: A) 126.87° - tan^-1(4/-3) in 2nd quadrant.

Memory Aids

• Displacement = final position - initial position.
• Component form: Δvector r = îΔx + ĵΔy.
• Direction θ = tan^-1(Δy / Δx).

Concepts

• Displacement \(\Delta \vec{r} = \vec{r'} - \vec{r} = \hat{i}\Delta x + \hat{j}\Delta y\).
• Directed from initial point \(P\) to final point \(P'\).
• Represents change in position vector.

Solved Examples

• Ex 1: Particle moves from \((2,3)\) to \((5,7)\).
  • \(\Delta \vec{r} = (5-2)\hat{i} + (7-3)\hat{j} = 3\hat{i} + 4\hat{j}\)

Practice Set

• Q1: Displacement from \((1,2)\) to \((4,6)\) is?
  • A) \(3\hat{i} + 4\hat{j}\)
  • B) \(4\hat{i} + 6\hat{j}\)
  • C) \(1\hat{i} + 2\hat{j}\)
  • D) \(-3\hat{i} - 4\hat{j}\)
• Q2: If \(\Delta \vec{r} = 5\hat{i} - 12\hat{j}\), magnitude is?
  • A) 13
  • B) 7
  • C) 17
  • D) 10
• Q3: Displacement vector direction for \(\Delta x = -3\), \(\Delta y = 4\) is?
  • A) \(126.87^\circ\)
  • B) \(53.13^\circ\)
  • C) \(36.87^\circ\)
  • D) \(143.13^\circ\)

Answer Key

• Q1: A) \(3\hat{i} + 4\hat{j}\) - Difference of coordinates.
• Q2: A) 13 - Magnitude \(= \sqrt{5^2 + (-12)^2} = 13\).
• Q3: A) \(126.87^\circ\) - \(\tan^{-1}(4/-3)\) in 2nd quadrant.

Memory Aids

• Displacement = final position - initial position.
• Component form: \(\Delta \vec{r} = \hat{i}\Delta x + \hat{j}\Delta y\).
• Direction \(\theta = \tan^{-1}(\Delta y / \Delta x)\).

Velocity

Concepts

• Average velocity vector V_av = Δvector r/Δt = î Δx/Δt + ĵ Δy/Δt = vector V_x,av î + vector V_y,av ĵ.
• Instantaneous velocity vector V = lim_{Δt → 0} Δvector r/Δt = dvector r/dt = î dx/dt + ĵ dy/dt = v_x î + v_y ĵ.
• Magnitude v = √(v_x² + v_y²), direction θ = tan^-1(v_y / v_x).
• Velocity vector is tangent to the path and points in direction of motion.

Solved Examples

• Ex 1: Particle moves from (1,2) to (4,6) in 2 s.
  • vector V_av = ((3î + 4ĵ))/2 = 1.5î + 2ĵ m/s
• Ex 2: Instantaneous velocity components: v_x = 3 m/s, v_y = 4 m/s.
  • Magnitude v = 5 m/s, direction θ = 53.13°.

Practice Set

• Q1: Average velocity for displacement 4î + 3ĵ in 1 s is?
  • A) 4î + 3ĵ m/s
  • B) 3î + 4ĵ m/s
  • C) 7î + 0ĵ m/s
  • D) 1î + 1ĵ m/s
• Q2: Instantaneous velocity components v_x = 6 m/s, v_y = 8 m/s. Magnitude is?
  • A) 10 m/s
  • B) 14 m/s
  • C) 2 m/s
  • D) 4 m/s
• Q3: Direction of velocity with v_x = -3 m/s, v_y = 3 m/s is?
  • A) 45°
  • B) 135°
  • C) 225°
  • D) 315°

Answer Key

• Q1: A) 4î + 3ĵ m/s - Average velocity = displacement/time.
• Q2: A) 10 m/s - √(6² + 8²) = 10.
• Q3: B) 135° - tan^-1(3/-3) in 2nd quadrant.

Memory Aids

• vector V_av = Δvector r/Δt.
• Instantaneous velocity = derivative of position vector.
• Magnitude and direction from components.

Concepts

• Average velocity \\(\vec{V}_{av} = \\frac{\Delta \\vec{r}}{\Delta t} = \\hat{i} \\frac{\Delta x}{\Delta t} + \\hat{j} \\frac{\Delta y}{\Delta t} = \\vec{V}_{x,av} \\hat{i} + \\vec{V}_{y,av} \\hat{j}\\).
• Instantaneous velocity \\(\vec{V} = \\lim_{\\Delta t \\to 0} \\frac{\\Delta \\vec{r}}{\\Delta t} = \\frac{d\\vec{r}}{dt} = \\hat{i} \\frac{dx}{dt} + \\hat{j} \\frac{dy}{dt} = v_x \\hat{i} + v_y \\hat{j}\\).
• Magnitude \\(v = \\sqrt{v_x^2 + v_y^2}\\), direction \\(\\theta = \\tan^{-1}(v_y / v_x)\\).
• Velocity vector is tangent to the path and points in direction of motion.

Solved Examples

• Ex 1: Particle moves from \\((1,2)\\) to \\((4,6)\\) in 2 s.
  • \\(\vec{V}_{av} = \\frac{(3\\hat{i} + 4\\hat{j})}{2} = 1.5\\hat{i} + 2\\hat{j} \\text{ m/s}\\)
• Ex 2: Instantaneous velocity components: \\(v_x = 3\\,m/s\\), \\(v_y = 4\\,m/s\\).
  • Magnitude \\(v = 5\\,m/s\\), direction \\(\\theta = 53.13^\\circ\\).

Practice Set

• Q1: Average velocity for displacement \\(4\\hat{i} + 3\\hat{j}\\) in 1 s is?
  • A) \\(4\\hat{i} + 3\\hat{j}\\) m/s
  • B) \\(3\\hat{i} + 4\\hat{j}\\) m/s
  • C) \\(7\\hat{i} + 0\\hat{j}\\) m/s
  • D) \\(1\\hat{i} + 1\\hat{j}\\) m/s
• Q2: Instantaneous velocity components \\(v_x = 6\\,m/s\\), \\(v_y = 8\\,m/s\\). Magnitude is?
  • A) 10 m/s
  • B) 14 m/s
  • C) 2 m/s
  • D) 4 m/s
• Q3: Direction of velocity with \\(v_x = -3\\,m/s\\), \\(v_y = 3\\,m/s\\) is?
  • A) \\(45^\\circ\\)
  • B) \\(135^\\circ\\)
  • C) \\(225^\\circ\\)
  • D) \\(315^\\circ\\)

Answer Key

• Q1: A) \\(4\\hat{i} + 3\\hat{j}\\) m/s - Average velocity = displacement/time.
• Q2: A) 10 m/s - \\(\\sqrt{6^2 + 8^2} = 10\\).
• Q3: B) \\(135^\\circ\\) - \\(\\tan^{-1}(3/-3)\\) in 2nd quadrant.

Memory Aids

• \\(\\vec{V}_{av} = \\frac{\\Delta \\vec{r}}{\\Delta t}\\).
• Instantaneous velocity = derivative of position vector.
• Magnitude and direction from components.

Concepts

• Average velocity vector V_av = Δvector r/Δt = î Δx/Δt + ĵ Δy/Δt = vector V_x,av î + vector V_y,av ĵ.
• Instantaneous velocity vector V = lim_{Δt → 0} Δvector r/Δt = dvector r/dt = î dx/dt + ĵ dy/dt = v_x î + v_y ĵ.
• Magnitude v = √(v_x² + v_y²), direction θ = tan^-1(v_y / v_x).
• Velocity vector is tangent to the path and points in direction of motion.

Solved Examples

• Ex 1: Particle moves from (1,2) to (4,6) in 2 s.
  • vector V_av = ((3î + 4ĵ))/2 = 1.5î + 2ĵ m/s
• Ex 2: Instantaneous velocity components: v_x = 3 m/s, v_y = 4 m/s.
  • Magnitude v = 5 m/s, direction θ = 53.13°.

Practice Set

• Q1: Average velocity for displacement 4î + 3ĵ in 1 s is?
  • A) 4î + 3ĵ m/s
  • B) 3î + 4ĵ m/s
  • C) 7î + 0ĵ m/s
  • D) 1î + 1ĵ m/s
• Q2: Instantaneous velocity components v_x = 6 m/s, v_y = 8 m/s. Magnitude is?
  • A) 10 m/s
  • B) 14 m/s
  • C) 2 m/s
  • D) 4 m/s
• Q3: Direction of velocity with v_x = -3 m/s, v_y = 3 m/s is?
  • A) 45°
  • B) 135°
  • C) 225°
  • D) 315°

Answer Key

• Q1: A) 4î + 3ĵ m/s - Average velocity = displacement/time.
• Q2: A) 10 m/s - √(6² + 8²) = 10.
• Q3: B) 135° - tan^-1(3/-3) in 2nd quadrant.

Memory Aids

• vector V_av = Δvector r/Δt.
• Instantaneous velocity = derivative of position vector.
• Magnitude and direction from components.

Concepts

• Average velocity \(\vec{V}_{av} = \frac{\Delta \vec{r}}{\Delta t} = \hat{i} \frac{\Delta x}{\Delta t} + \hat{j} \frac{\Delta y}{\Delta t} = \vec{V}_{x,av} \hat{i} + \vec{V}_{y,av} \hat{j}\).
• Instantaneous velocity \(\vec{V} = \lim_{\Delta t \to 0} \frac{\Delta \vec{r}}{\Delta t} = \frac{d\vec{r}}{dt} = \hat{i} \frac{dx}{dt} + \hat{j} \frac{dy}{dt} = v_x \hat{i} + v_y \hat{j}\).
• Magnitude \(v = \sqrt{v_x^2 + v_y^2}\), direction \(\theta = \tan^{-1}(v_y / v_x)\).
• Velocity vector is tangent to the path and points in direction of motion.

Solved Examples

• Ex 1: Particle moves from \((1,2)\) to \((4,6)\) in 2 s.
  • \(\vec{V}_{av} = \frac{(3\hat{i} + 4\hat{j})}{2} = 1.5\hat{i} + 2\hat{j} \text{ m/s}\)
• Ex 2: Instantaneous velocity components: \(v_x = 3\,m/s\), \(v_y = 4\,m/s\).
  • Magnitude \(v = 5\,m/s\), direction \(\theta = 53.13^\circ\).

Practice Set

• Q1: Average velocity for displacement \(4\hat{i} + 3\hat{j}\) in 1 s is?
  • A) \(4\hat{i} + 3\hat{j}\) m/s
  • B) \(3\hat{i} + 4\hat{j}\) m/s
  • C) \(7\hat{i} + 0\hat{j}\) m/s
  • D) \(1\hat{i} + 1\hat{j}\) m/s
• Q2: Instantaneous velocity components \(v_x = 6\,m/s\), \(v_y = 8\,m/s\). Magnitude is?
  • A) 10 m/s
  • B) 14 m/s
  • C) 2 m/s
  • D) 4 m/s
• Q3: Direction of velocity with \(v_x = -3\,m/s\), \(v_y = 3\,m/s\) is?
  • A) \(45^\circ\)
  • B) \(135^\circ\)
  • C) \(225^\circ\)
  • D) \(315^\circ\)

Answer Key

• Q1: A) \(4\hat{i} + 3\hat{j}\) m/s - Average velocity = displacement/time.
• Q2: A) 10 m/s - \(\sqrt{6^2 + 8^2} = 10\).
• Q3: B) \(135^\circ\) - \(\tan^{-1}(3/-3)\) in 2nd quadrant.

Memory Aids

• \(\vec{V}_{av} = \frac{\Delta \vec{r}}{\Delta t}\).
• Instantaneous velocity = derivative of position vector.
• Magnitude and direction from components.

Concepts

• Average velocity vector V_av = Δvector r/Δt = î Δx/Δt + ĵ Δy/Δt = vector V_x,av î + vector V_y,av ĵ.
• Instantaneous velocity vector V = lim_{Δt → 0} Δvector r/Δt = dvector r/dt = î dx/dt + ĵ dy/dt = v_x î + v_y ĵ.
• Magnitude v = √(v_x² + v_y²), direction θ = tan^-1(v_y / v_x).
• Velocity vector is tangent to the path and points in direction of motion.

Solved Examples

• Ex 1: Particle moves from (1,2) to (4,6) in 2 s.
  • vector V_av = ((3î + 4ĵ))/2 = 1.5î + 2ĵ m/s
• Ex 2: Instantaneous velocity components: v_x = 3 m/s, v_y = 4 m/s.
  • Magnitude v = 5 m/s, direction θ = 53.13°.

Practice Set

• Q1: Average velocity for displacement 4î + 3ĵ in 1 s is?
  • A) 4î + 3ĵ m/s
  • B) 3î + 4ĵ m/s
  • C) 7î + 0ĵ m/s
  • D) 1î + 1ĵ m/s
• Q2: Instantaneous velocity components v_x = 6 m/s, v_y = 8 m/s. Magnitude is?
  • A) 10 m/s
  • B) 14 m/s
  • C) 2 m/s
  • D) 4 m/s
• Q3: Direction of velocity with v_x = -3 m/s, v_y = 3 m/s is?
  • A) 45°
  • B) 135°
  • C) 225°
  • D) 315°

Answer Key

• Q1: A) 4î + 3ĵ m/s - Average velocity = displacement/time.
• Q2: A) 10 m/s - √(6² + 8²) = 10.
• Q3: B) 135° - tan^-1(3/-3) in 2nd quadrant.

Memory Aids

• vector V_av = Δvector r/Δt.
• Instantaneous velocity = derivative of position vector.
• Magnitude and direction from components.

Concepts

• Average velocity \(\vec{V}_{av} = \frac{\Delta \vec{r}}{\Delta t} = \hat{i} \frac{\Delta x}{\Delta t} + \hat{j} \frac{\Delta y}{\Delta t} = \vec{V}_{x,av} \hat{i} + \vec{V}_{y,av} \hat{j}\).
• Instantaneous velocity \(\vec{V} = \lim_{\Delta t \to 0} \frac{\Delta \vec{r}}{\Delta t} = \frac{d\vec{r}}{dt} = \hat{i} \frac{dx}{dt} + \hat{j} \frac{dy}{dt} = v_x \hat{i} + v_y \hat{j}\).
• Magnitude \(v = \sqrt{v_x^2 + v_y^2}\), direction \(\theta = \tan^{-1}(v_y / v_x)\).
• Velocity vector is tangent to the path and points in direction of motion.

Solved Examples

• Ex 1: Particle moves from \((1,2)\) to \((4,6)\) in 2 s.
  • \(\vec{V}_{av} = \frac{(3\hat{i} + 4\hat{j})}{2} = 1.5\hat{i} + 2\hat{j} \text{ m/s}\)
• Ex 2: Instantaneous velocity components: \(v_x = 3\,m/s\), \(v_y = 4\,m/s\).
  • Magnitude \(v = 5\,m/s\), direction \(\theta = 53.13^\circ\).

Practice Set

• Q1: Average velocity for displacement \(4\hat{i} + 3\hat{j}\) in 1 s is?
  • A) \(4\hat{i} + 3\hat{j}\) m/s
  • B) \(3\hat{i} + 4\hat{j}\) m/s
  • C) \(7\hat{i} + 0\hat{j}\) m/s
  • D) \(1\hat{i} + 1\hat{j}\) m/s
• Q2: Instantaneous velocity components \(v_x = 6\,m/s\), \(v_y = 8\,m/s\). Magnitude is?
  • A) 10 m/s
  • B) 14 m/s
  • C) 2 m/s
  • D) 4 m/s
• Q3: Direction of velocity with \(v_x = -3\,m/s\), \(v_y = 3\,m/s\) is?
  • A) \(45^\circ\)
  • B) \(135^\circ\)
  • C) \(225^\circ\)
  • D) \(315^\circ\)

Answer Key

• Q1: A) \(4\hat{i} + 3\hat{j}\) m/s - Average velocity = displacement/time.
• Q2: A) 10 m/s - \(\sqrt{6^2 + 8^2} = 10\).
• Q3: B) \(135^\circ\) - \(\tan^{-1}(3/-3)\) in 2nd quadrant.

Memory Aids

• \(\vec{V}_{av} = \frac{\Delta \vec{r}}{\Delta t}\).
• Instantaneous velocity = derivative of position vector.
• Magnitude and direction from components.

Acceleration

Concepts

• Average acceleration vector a_av = Δvector V/Δt = î Δv_x/Δt + ĵ Δv_y/Δt = a_x,av î + a_y,av ĵ.
• Instantaneous acceleration is limit as Δt → 0.
• Acceleration vector shows rate of change of velocity in magnitude and direction.

Solved Examples

• Ex 1: Velocity changes from 3î + 4ĵ m/s to 6î + 8ĵ m/s in 2 s.
  • vector a_av = ((6-3)î + (8-4)ĵ)/2 = 1.5î + 2ĵ m/s²

Practice Set

• Q1: Average acceleration if velocity changes from 2î + 3ĵ m/s to 8î + 9ĵ m/s in 3 s?
  • A) 2î + 2ĵ m/s²
  • B) 3î + 3ĵ m/s²
  • C) 6î + 6ĵ m/s²
  • D) 1î + 1ĵ m/s²
• Q2: Components of acceleration are a_x = 4 m/s², a_y = 3 m/s². Magnitude is?
  • A) 5 m/s²
  • B) 7 m/s²
  • C) 1 m/s²
  • D) 12 m/s²
• Q3: Direction of acceleration with a_x = -3 m/s², a_y = 3 m/s² is?
  • A) 45°
  • B) 135°
  • C) 225°
  • D) 315°

Answer Key

• Q1: A) 2î + 2ĵ m/s² - Change in velocity/time.
• Q2: A) 5 m/s² - √(4² + 3²) = 5.
• Q3: B) 135° - tan^-1(3/-3) in 2nd quadrant.

Memory Aids

• vector a_av = Δvector V/Δt.
• Instantaneous acceleration = derivative of velocity vector.
• Magnitude and direction from components.

Concepts

• Average acceleration vector a_av = Δvector V/Δt = î Δv_x/Δt + ĵ Δv_y/Δt = a_x,av î + a_y,av ĵ.
• Instantaneous acceleration is limit as Δt → 0.
• Acceleration vector shows rate of change of velocity in magnitude and direction.

Solved Examples

• Ex 1: Velocity changes from 3î + 4ĵ m/s to 6î + 8ĵ m/s in 2 s.
  • vector a_av = ((6-3)î + (8-4)ĵ)/2 = 1.5î + 2ĵ m/s²

Practice Set

• Q1: Average acceleration if velocity changes from 2î + 3ĵ m/s to 8î + 9ĵ m/s in 3 s?
  • A) 2î + 2ĵ m/s²
  • B) 3î + 3ĵ m/s²
  • C) 6î + 6ĵ m/s²
  • D) 1î + 1ĵ m/s²
• Q2: Components of acceleration are a_x = 4 m/s², a_y = 3 m/s². Magnitude is?
  • A) 5 m/s²
  • B) 7 m/s²
  • C) 1 m/s²
  • D) 12 m/s²
• Q3: Direction of acceleration with a_x = -3 m/s², a_y = 3 m/s² is?
  • A) 45°
  • B) 135°
  • C) 225°
  • D) 315°

Answer Key

• Q1: A) 2î + 2ĵ m/s² - Change in velocity/time.
• Q2: A) 5 m/s² - √(4² + 3²) = 5.
• Q3: B) 135° - tan^-1(3/-3) in 2nd quadrant.

Memory Aids

• vector a_av = Δvector V/Δt.
• Instantaneous acceleration = derivative of velocity vector.
• Magnitude and direction from components.

Concepts

• Average acceleration \\(\\vec{a}_{av} = \\frac{\\Delta \\vec{V}}{\\Delta t} = \\hat{i} \\frac{\\Delta v_x}{\\Delta t} + \\hat{j} \\frac{\\Delta v_y}{\\Delta t} = a_{x,av} \\hat{i} + a_{y,av} \\hat{j}\\).
• Instantaneous acceleration is limit as \\(\\Delta t \\to 0\\).
• Acceleration vector shows rate of change of velocity in magnitude and direction.

Solved Examples

• Ex 1: Velocity changes from \\(3\\hat{i} + 4\\hat{j}\\) m/s to \\(6\\hat{i} + 8\\hat{j}\\) m/s in 2 s.
  • \\(\\vec{a}_{av} = \\frac{(6-3)\\hat{i} + (8-4)\\hat{j}}{2} = 1.5\\hat{i} + 2\\hat{j} \\text{ m/s}^2\\)

Practice Set

• Q1: Average acceleration if velocity changes from \\(2\\hat{i} + 3\\hat{j}\\) m/s to \\(8\\hat{i} + 9\\hat{j}\\) m/s in 3 s?
  • A) \\(2\\hat{i} + 2\\hat{j}\\) m/s²
  • B) \\(3\\hat{i} + 3\\hat{j}\\) m/s²
  • C) \\(6\\hat{i} + 6\\hat{j}\\) m/s²
  • D) \\(1\\hat{i} + 1\\hat{j}\\) m/s²
• Q2: Components of acceleration are \\(a_x = 4\\,m/s^2\\), \\(a_y = 3\\,m/s^2\\). Magnitude is?
  • A) 5 m/s²
  • B) 7 m/s²
  • C) 1 m/s²
  • D) 12 m/s²
• Q3: Direction of acceleration with \\(a_x = -3\\,m/s^2\\), \\(a_y = 3\\,m/s^2\\) is?
  • A) \\(45^\\circ\\)
  • B) \\(135^\\circ\\)
  • C) \\(225^\\circ\\)
  • D) \\(315^\\circ\\)

Answer Key

• Q1: A) \\(2\\hat{i} + 2\\hat{j}\\) m/s² - Change in velocity/time.
• Q2: A) 5 m/s² - \\(\\sqrt{4^2 + 3^2} = 5\\).
• Q3: B) \\(135^\\circ\\) - \\(\\tan^{-1}(3/-3)\\) in 2nd quadrant.

Memory Aids

• \\(\\vec{a}_{av} = \\frac{\\Delta \\vec{V}}{\\Delta t}\\).
• Instantaneous acceleration = derivative of velocity vector.
• Magnitude and direction from components.

Concepts

• Average acceleration vector a_av = Δvector V/Δt = î Δv_x/Δt + ĵ Δv_y/Δt = a_x,av î + a_y,av ĵ.
• Instantaneous acceleration is limit as Δt → 0.
• Acceleration vector shows rate of change of velocity in magnitude and direction.

Solved Examples

• Ex 1: Velocity changes from 3î + 4ĵ m/s to 6î + 8ĵ m/s in 2 s.
  • vector a_av = ((6-3)î + (8-4)ĵ)/2 = 1.5î + 2ĵ m/s²

Practice Set

• Q1: Average acceleration if velocity changes from 2î + 3ĵ m/s to 8î + 9ĵ m/s in 3 s?
  • A) 2î + 2ĵ m/s²
  • B) 3î + 3ĵ m/s²
  • C) 6î + 6ĵ m/s²
  • D) 1î + 1ĵ m/s²
• Q2: Components of acceleration are a_x = 4 m/s², a_y = 3 m/s². Magnitude is?
  • A) 5 m/s²
  • B) 7 m/s²
  • C) 1 m/s²
  • D) 12 m/s²
• Q3: Direction of acceleration with a_x = -3 m/s², a_y = 3 m/s² is?
  • A) 45°
  • B) 135°
  • C) 225°
  • D) 315°

Answer Key

• Q1: A) 2î + 2ĵ m/s² - Change in velocity/time.
• Q2: A) 5 m/s² - √(4² + 3²) = 5.
• Q3: B) 135° - tan^-1(3/-3) in 2nd quadrant.

Memory Aids

• vector a_av = Δvector V/Δt.
• Instantaneous acceleration = derivative of velocity vector.
• Magnitude and direction from components.

Concepts

• Average acceleration \(\vec{a}_{av} = \frac{\Delta \vec{V}}{\Delta t} = \hat{i} \frac{\Delta v_x}{\Delta t} + \hat{j} \frac{\Delta v_y}{\Delta t} = a_{x,av} \hat{i} + a_{y,av} \hat{j}\).
• Instantaneous acceleration is limit as \(\Delta t \to 0\).
• Acceleration vector shows rate of change of velocity in magnitude and direction.

Solved Examples

• Ex 1: Velocity changes from \(3\hat{i} + 4\hat{j}\) m/s to \(6\hat{i} + 8\hat{j}\) m/s in 2 s.
  • \(\vec{a}_{av} = \frac{(6-3)\hat{i} + (8-4)\hat{j}}{2} = 1.5\hat{i} + 2\hat{j} \text{ m/s}^2\)

Practice Set

• Q1: Average acceleration if velocity changes from \(2\hat{i} + 3\hat{j}\) m/s to \(8\hat{i} + 9\hat{j}\) m/s in 3 s?
  • A) \(2\hat{i} + 2\hat{j}\) m/s²
  • B) \(3\hat{i} + 3\hat{j}\) m/s²
  • C) \(6\hat{i} + 6\hat{j}\) m/s²
  • D) \(1\hat{i} + 1\hat{j}\) m/s²
• Q2: Components of acceleration are \(a_x = 4\,m/s^2\), \(a_y = 3\,m/s^2\). Magnitude is?
  • A) 5 m/s²
  • B) 7 m/s²
  • C) 1 m/s²
  • D) 12 m/s²
• Q3: Direction of acceleration with \(a_x = -3\,m/s^2\), \(a_y = 3\,m/s^2\) is?
  • A) \(45^\circ\)
  • B) \(135^\circ\)
  • C) \(225^\circ\)
  • D) \(315^\circ\)

Answer Key

• Q1: A) \(2\hat{i} + 2\hat{j}\) m/s² - Change in velocity/time.
• Q2: A) 5 m/s² - \(\sqrt{4^2 + 3^2} = 5\).
• Q3: B) \(135^\circ\) - \(\tan^{-1}(3/-3)\) in 2nd quadrant.

Memory Aids

• \(\vec{a}_{av} = \frac{\Delta \vec{V}}{\Delta t}\).
• Instantaneous acceleration = derivative of velocity vector.
• Magnitude and direction from components.

Concepts

• Average acceleration vector a_av = Δvector V/Δt = î Δv_x/Δt + ĵ Δv_y/Δt = a_x,av î + a_y,av ĵ.
• Instantaneous acceleration is limit as Δt → 0.
• Acceleration vector shows rate of change of velocity in magnitude and direction.

Solved Examples

• Ex 1: Velocity changes from 3î + 4ĵ m/s to 6î + 8ĵ m/s in 2 s.
  • vector a_av = ((6-3)î + (8-4)ĵ)/2 = 1.5î + 2ĵ m/s²

Practice Set

• Q1: Average acceleration if velocity changes from 2î + 3ĵ m/s to 8î + 9ĵ m/s in 3 s?
  • A) 2î + 2ĵ m/s²
  • B) 3î + 3ĵ m/s²
  • C) 6î + 6ĵ m/s²
  • D) 1î + 1ĵ m/s²
• Q2: Components of acceleration are a_x = 4 m/s², a_y = 3 m/s². Magnitude is?
  • A) 5 m/s²
  • B) 7 m/s²
  • C) 1 m/s²
  • D) 12 m/s²
• Q3: Direction of acceleration with a_x = -3 m/s², a_y = 3 m/s² is?
  • A) 45°
  • B) 135°
  • C) 225°
  • D) 315°

Answer Key

• Q1: A) 2î + 2ĵ m/s² - Change in velocity/time.
• Q2: A) 5 m/s² - √(4² + 3²) = 5.
• Q3: B) 135° - tan^-1(3/-3) in 2nd quadrant.

Memory Aids

• vector a_av = Δvector V/Δt.
• Instantaneous acceleration = derivative of velocity vector.
• Magnitude and direction from components.

Concepts

• Average acceleration \(\vec{a}_{av} = \frac{\Delta \vec{V}}{\Delta t} = \hat{i} \frac{\Delta v_x}{\Delta t} + \hat{j} \frac{\Delta v_y}{\Delta t} = a_{x,av} \hat{i} + a_{y,av} \hat{j}\).
• Instantaneous acceleration is limit as \(\Delta t \to 0\).
• Acceleration vector shows rate of change of velocity in magnitude and direction.

Solved Examples

• Ex 1: Velocity changes from \(3\hat{i} + 4\hat{j}\) m/s to \(6\hat{i} + 8\hat{j}\) m/s in 2 s.
  • \(\vec{a}_{av} = \frac{(6-3)\hat{i} + (8-4)\hat{j}}{2} = 1.5\hat{i} + 2\hat{j} \text{ m/s}^2\)

Practice Set

• Q1: Average acceleration if velocity changes from \(2\hat{i} + 3\hat{j}\) m/s to \(8\hat{i} + 9\hat{j}\) m/s in 3 s?
  • A) \(2\hat{i} + 2\hat{j}\) m/s²
  • B) \(3\hat{i} + 3\hat{j}\) m/s²
  • C) \(6\hat{i} + 6\hat{j}\) m/s²
  • D) \(1\hat{i} + 1\hat{j}\) m/s²
• Q2: Components of acceleration are \(a_x = 4\,m/s^2\), \(a_y = 3\,m/s^2\). Magnitude is?
  • A) 5 m/s²
  • B) 7 m/s²
  • C) 1 m/s²
  • D) 12 m/s²
• Q3: Direction of acceleration with \(a_x = -3\,m/s^2\), \(a_y = 3\,m/s^2\) is?
  • A) \(45^\circ\)
  • B) \(135^\circ\)
  • C) \(225^\circ\)
  • D) \(315^\circ\)

Answer Key

• Q1: A) \(2\hat{i} + 2\hat{j}\) m/s² - Change in velocity/time.
• Q2: A) 5 m/s² - \(\sqrt{4^2 + 3^2} = 5\).
• Q3: B) \(135^\circ\) - \(\tan^{-1}(3/-3)\) in 2nd quadrant.

Memory Aids

• \(\vec{a}_{av} = \frac{\Delta \vec{V}}{\Delta t}\).
• Instantaneous acceleration = derivative of velocity vector.
• Magnitude and direction from components.

Motion in a Plane with Constant Acceleration

Key Formulas and Concepts

• Velocity at time t: vector v = vector v₀ + vector a t, where vector v₀ is initial velocity.
• Average velocity over t: vector v_av = (vector v + vector v₀)/2.
• Component form: v_x = v_0x + a_x t, v_y = v_0y + a_y t.
• Displacement: vector r = vector r₀ + vector v₀ t + 1/2 vector a t².
• Component displacement: x = x₀ + v_0x t + 1/2 a_x t², y = y₀ + v_0y t + 1/2 a_y t².
• Motion in plane treated as two independent 1D motions along perpendicular axes.

Solved Examples

• Ex 1: Find velocity at t = 2s if vector v₀ = 3î + 4ĵ m/s and vector a = 2î + 3ĵ m/s².
  • vector v = vector v₀ + vector a t = (3 + 2 × 2)î + (4 + 3 × 2)ĵ = 7î + 10ĵ m/s.
• Ex 2: Calculate displacement after 3 s with vector v₀ = 0, vector a = 4î m/s².
  • vector r = vector r₀ + vector v₀ t + 1/2 vector a t² = 0 + 0 + 1/2 × 4 × 3² = 18 m along x-axis.

Practice Set

• Q1: Velocity after 1 s if vector v₀ = 5î m/s and vector a = 3î m/s²?
  • A) 5 m/s
  • B) 8 m/s
  • C) 3 m/s
  • D) 15 m/s
• Q2: Displacement in y-direction after 2 s if v_0y = 2 m/s and a_y = 1 m/s²?
  • A) 4 m
  • B) 6 m
  • C) 8 m
  • D) 10 m
• Q3: Average velocity over 4 s if initial velocity is 2î + 3ĵ m/s and acceleration is zero?
  • A) 2î + 3ĵ m/s
  • B) Zero
  • C) 4î + 6ĵ m/s
  • D) 1î + 1.5ĵ m/s

Answer Key

• Q1: B - Velocity = 5 + 3 × 1 = 8 m/s.
• Q2: B - Displacement = v_0y t + 1/2 a_y t² = 2 × 2 + 0.5 × 1 × 4 = 6 m.
• Q3: A - No acceleration means velocity constant, so average velocity equals initial velocity.

Memory Aids

• Velocity: v = u + at.
• Displacement: s = ut + 1/2 at².
• Two-dimensional motion = two independent 1D motions.

Key Formulas and Concepts

• Velocity at time t: vector v = vector v₀ + vector a t, where vector v₀ is initial velocity.
• Average velocity over t: vector v_av = (vector v + vector v₀)/2.
• Component form: v_x = v_0x + a_x t, v_y = v_0y + a_y t.
• Displacement: vector r = vector r₀ + vector v₀ t + 1/2 vector a t².
• Component displacement: x = x₀ + v_0x t + 1/2 a_x t², y = y₀ + v_0y t + 1/2 a_y t².
• Motion in plane treated as two independent 1D motions along perpendicular axes.

Solved Examples

• Ex 1: Find velocity at t = 2s if vector v₀ = 3î + 4ĵ m/s and vector a = 2î + 3ĵ m/s².
  • vector v = vector v₀ + vector a t = (3 + 2 × 2)î + (4 + 3 × 2)ĵ = 7î + 10ĵ m/s.
• Ex 2: Calculate displacement after 3 s with vector v₀ = 0, vector a = 4î m/s².
  • vector r = vector r₀ + vector v₀ t + 1/2 vector a t² = 0 + 0 + 1/2 × 4 × 3² = 18 m along x-axis.

Practice Set

• Q1: Velocity after 1 s if vector v₀ = 5î m/s and vector a = 3î m/s²?
  • A) 5 m/s
  • B) 8 m/s
  • C) 3 m/s
  • D) 15 m/s
• Q2: Displacement in y-direction after 2 s if v_0y = 2 m/s and a_y = 1 m/s²?
  • A) 4 m
  • B) 6 m
  • C) 8 m
  • D) 10 m
• Q3: Average velocity over 4 s if initial velocity is 2î + 3ĵ m/s and acceleration is zero?
  • A) 2î + 3ĵ m/s
  • B) Zero
  • C) 4î + 6ĵ m/s
  • D) 1î + 1.5ĵ m/s

Answer Key

• Q1: B - Velocity = 5 + 3 × 1 = 8 m/s.
• Q2: B - Displacement = v_0y t + 1/2 a_y t² = 2 × 2 + 0.5 × 1 × 4 = 6 m.
• Q3: A - No acceleration means velocity constant, so average velocity equals initial velocity.

Memory Aids

• Velocity: v = u + at.
• Displacement: s = ut + 1/2 at².
• Two-dimensional motion = two independent 1D motions.

Key Formulas and Concepts

• Velocity at time \\( t \\): \\( \\vec{v} = \\vec{v}_0 + \\vec{a}t \\), where \\( \\vec{v}_0 \\) is initial velocity.
• Average velocity over \\( t \\): \\( \\vec{v}_{av} = \\frac{\\vec{v} + \\vec{v}_0}{2} \\).
• Component form: \\( v_x = v_{0x} + a_x t \\), \\( v_y = v_{0y} + a_y t \\).
• Displacement: \\( \\vec{r} = \\vec{r}_0 + \\vec{v}_0 t + \\frac{1}{2} \\vec{a} t^2 \\).
• Component displacement: \\( x = x_0 + v_{0x} t + \\frac{1}{2} a_x t^2 \\), \\( y = y_0 + v_{0y} t + \\frac{1}{2} a_y t^2 \\).
• Motion in plane treated as two independent 1D motions along perpendicular axes.

Solved Examples

• Ex 1: Find velocity at \\( t = 2s \\) if \\( \\vec{v}_0 = 3\\hat{i} + 4\\hat{j} \\) m/s and \\( \\vec{a} = 2\\hat{i} + 3\\hat{j} \\) m/s².
  • \\( \\vec{v} = \\vec{v}_0 + \\vec{a}t = (3 + 2 \\times 2)\\hat{i} + (4 + 3 \\times 2)\\hat{j} = 7\\hat{i} + 10\\hat{j} \\) m/s.
• Ex 2: Calculate displacement after 3 s with \\( \\vec{v}_0 = 0 \\), \\( \\vec{a} = 4\\hat{i} \\) m/s².
  • \\( \\vec{r} = \\vec{r}_0 + \\vec{v}_0 t + \\frac{1}{2} \\vec{a} t^2 = 0 + 0 + \\frac{1}{2} \\times 4 \\times 3^2 = 18 \\) m along x-axis.

Practice Set

• Q1: Velocity after 1 s if \\( \\vec{v}_0 = 5\\hat{i} \\) m/s and \\( \\vec{a} = 3\\hat{i} \\) m/s²?
  • A) 5 m/s
  • B) 8 m/s
  • C) 3 m/s
  • D) 15 m/s
• Q2: Displacement in y-direction after 2 s if \\( v_{0y} = 2 \\) m/s and \\( a_y = 1 \\) m/s²?
  • A) 4 m
  • B) 6 m
  • C) 8 m
  • D) 10 m
• Q3: Average velocity over 4 s if initial velocity is \\( 2\\hat{i} + 3\\hat{j} \\) m/s and acceleration is zero?
  • A) \\( 2\\hat{i} + 3\\hat{j} \\) m/s
  • B) Zero
  • C) \\( 4\\hat{i} + 6\\hat{j} \\) m/s
  • D) \\( 1\\hat{i} + 1.5\\hat{j} \\) m/s

Answer Key

• Q1: B - Velocity \\( = 5 + 3 \\times 1 = 8 \\) m/s.
• Q2: B - Displacement \\( = v_{0y} t + \\frac{1}{2} a_y t^2 = 2 \\times 2 + 0.5 \\times 1 \\times 4 = 6 \\) m.
• Q3: A - No acceleration means velocity constant, so average velocity equals initial velocity.

Memory Aids

• Velocity: \\( v = u + at \\).
• Displacement: \\( s = ut + \\frac{1}{2} at^2 \\).
• Two-dimensional motion = two independent 1D motions.

Key Formulas and Concepts

• Velocity at time t: vector v = vector v₀ + vector a t, where vector v₀ is initial velocity.
• Average velocity over t: vector v_av = (vector v + vector v₀)/2.
• Component form: v_x = v_0x + a_x t, v_y = v_0y + a_y t.
• Displacement: vector r = vector r₀ + vector v₀ t + 1/2 vector a t².
• Component displacement: x = x₀ + v_0x t + 1/2 a_x t², y = y₀ + v_0y t + 1/2 a_y t².
• Motion in plane treated as two independent 1D motions along perpendicular axes.

Solved Examples

• Ex 1: Find velocity at t = 2s if vector v₀ = 3î + 4ĵ m/s and vector a = 2î + 3ĵ m/s².
  • vector v = vector v₀ + vector a t = (3 + 2 × 2)î + (4 + 3 × 2)ĵ = 7î + 10ĵ m/s.
• Ex 2: Calculate displacement after 3 s with vector v₀ = 0, vector a = 4î m/s².
  • vector r = vector r₀ + vector v₀ t + 1/2 vector a t² = 0 + 0 + 1/2 × 4 × 3² = 18 m along x-axis.

Practice Set

• Q1: Velocity after 1 s if vector v₀ = 5î m/s and vector a = 3î m/s²?
  • A) 5 m/s
  • B) 8 m/s
  • C) 3 m/s
  • D) 15 m/s
• Q2: Displacement in y-direction after 2 s if v_0y = 2 m/s and a_y = 1 m/s²?
  • A) 4 m
  • B) 6 m
  • C) 8 m
  • D) 10 m
• Q3: Average velocity over 4 s if initial velocity is 2î + 3ĵ m/s and acceleration is zero?
  • A) 2î + 3ĵ m/s
  • B) Zero
  • C) 4î + 6ĵ m/s
  • D) 1î + 1.5ĵ m/s

Answer Key

• Q1: B - Velocity = 5 + 3 × 1 = 8 m/s.
• Q2: B - Displacement = v_0y t + 1/2 a_y t² = 2 × 2 + 0.5 × 1 × 4 = 6 m.
• Q3: A - No acceleration means velocity constant, so average velocity equals initial velocity.

Memory Aids

• Velocity: v = u + at.
• Displacement: s = ut + 1/2 at².
• Two-dimensional motion = two independent 1D motions.

Key Formulas and Concepts

• Velocity at time \( t \): \( \vec{v} = \vec{v}_0 + \vec{a}t \), where \( \vec{v}_0 \) is initial velocity.
• Average velocity over \( t \): \( \vec{v}_{av} = \frac{\vec{v} + \vec{v}_0}{2} \).
• Component form: \( v_x = v_{0x} + a_x t \), \( v_y = v_{0y} + a_y t \).
• Displacement: \( \vec{r} = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a} t^2 \).
• Component displacement: \( x = x_0 + v_{0x} t + \frac{1}{2} a_x t^2 \), \( y = y_0 + v_{0y} t + \frac{1}{2} a_y t^2 \).
• Motion in plane treated as two independent 1D motions along perpendicular axes.

Solved Examples

• Ex 1: Find velocity at \( t = 2s \) if \( \vec{v}_0 = 3\hat{i} + 4\hat{j} \) m/s and \( \vec{a} = 2\hat{i} + 3\hat{j} \) m/s².
  • \( \vec{v} = \vec{v}_0 + \vec{a}t = (3 + 2 \times 2)\hat{i} + (4 + 3 \times 2)\hat{j} = 7\hat{i} + 10\hat{j} \) m/s.
• Ex 2: Calculate displacement after 3 s with \( \vec{v}_0 = 0 \), \( \vec{a} = 4\hat{i} \) m/s².
  • \( \vec{r} = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a} t^2 = 0 + 0 + \frac{1}{2} \times 4 \times 3^2 = 18 \) m along x-axis.

Practice Set

• Q1: Velocity after 1 s if \( \vec{v}_0 = 5\hat{i} \) m/s and \( \vec{a} = 3\hat{i} \) m/s²?
  • A) 5 m/s
  • B) 8 m/s
  • C) 3 m/s
  • D) 15 m/s
• Q2: Displacement in y-direction after 2 s if \( v_{0y} = 2 \) m/s and \( a_y = 1 \) m/s²?
  • A) 4 m
  • B) 6 m
  • C) 8 m
  • D) 10 m
• Q3: Average velocity over 4 s if initial velocity is \( 2\hat{i} + 3\hat{j} \) m/s and acceleration is zero?
  • A) \( 2\hat{i} + 3\hat{j} \) m/s
  • B) Zero
  • C) \( 4\hat{i} + 6\hat{j} \) m/s
  • D) \( 1\hat{i} + 1.5\hat{j} \) m/s

Answer Key

• Q1: B - Velocity \( = 5 + 3 \times 1 = 8 \) m/s.
• Q2: B - Displacement \( = v_{0y} t + \frac{1}{2} a_y t^2 = 2 \times 2 + 0.5 \times 1 \times 4 = 6 \) m.
• Q3: A - No acceleration means velocity constant, so average velocity equals initial velocity.

Memory Aids

• Velocity: \( v = u + at \).
• Displacement: \( s = ut + \frac{1}{2} at^2 \).
• Two-dimensional motion = two independent 1D motions.

Key Formulas and Concepts

• Velocity at time t: vector v = vector v₀ + vector a t, where vector v₀ is initial velocity.
• Average velocity over t: vector v_av = (vector v + vector v₀)/2.
• Component form: v_x = v_0x + a_x t, v_y = v_0y + a_y t.
• Displacement: vector r = vector r₀ + vector v₀ t + 1/2 vector a t².
• Component displacement: x = x₀ + v_0x t + 1/2 a_x t², y = y₀ + v_0y t + 1/2 a_y t².
• Motion in plane treated as two independent 1D motions along perpendicular axes.

Solved Examples

• Ex 1: Find velocity at t = 2s if vector v₀ = 3î + 4ĵ m/s and vector a = 2î + 3ĵ m/s².
  • vector v = vector v₀ + vector a t = (3 + 2 × 2)î + (4 + 3 × 2)ĵ = 7î + 10ĵ m/s.
• Ex 2: Calculate displacement after 3 s with vector v₀ = 0, vector a = 4î m/s².
  • vector r = vector r₀ + vector v₀ t + 1/2 vector a t² = 0 + 0 + 1/2 × 4 × 3² = 18 m along x-axis.

Practice Set

• Q1: Velocity after 1 s if vector v₀ = 5î m/s and vector a = 3î m/s²?
  • A) 5 m/s
  • B) 8 m/s
  • C) 3 m/s
  • D) 15 m/s
• Q2: Displacement in y-direction after 2 s if v_0y = 2 m/s and a_y = 1 m/s²?
  • A) 4 m
  • B) 6 m
  • C) 8 m
  • D) 10 m
• Q3: Average velocity over 4 s if initial velocity is 2î + 3ĵ m/s and acceleration is zero?
  • A) 2î + 3ĵ m/s
  • B) Zero
  • C) 4î + 6ĵ m/s
  • D) 1î + 1.5ĵ m/s

Answer Key

• Q1: B - Velocity = 5 + 3 × 1 = 8 m/s.
• Q2: B - Displacement = v_0y t + 1/2 a_y t² = 2 × 2 + 0.5 × 1 × 4 = 6 m.
• Q3: A - No acceleration means velocity constant, so average velocity equals initial velocity.

Memory Aids

• Velocity: v = u + at.
• Displacement: s = ut + 1/2 at².
• Two-dimensional motion = two independent 1D motions.

Key Formulas and Concepts

• Velocity at time \( t \): \( \vec{v} = \vec{v}_0 + \vec{a}t \), where \( \vec{v}_0 \) is initial velocity.
• Average velocity over \( t \): \( \vec{v}_{av} = \frac{\vec{v} + \vec{v}_0}{2} \).
• Component form: \( v_x = v_{0x} + a_x t \), \( v_y = v_{0y} + a_y t \).
• Displacement: \( \vec{r} = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a} t^2 \).
• Component displacement: \( x = x_0 + v_{0x} t + \frac{1}{2} a_x t^2 \), \( y = y_0 + v_{0y} t + \frac{1}{2} a_y t^2 \).
• Motion in plane treated as two independent 1D motions along perpendicular axes.

Solved Examples

• Ex 1: Find velocity at \( t = 2s \) if \( \vec{v}_0 = 3\hat{i} + 4\hat{j} \) m/s and \( \vec{a} = 2\hat{i} + 3\hat{j} \) m/s².
  • \( \vec{v} = \vec{v}_0 + \vec{a}t = (3 + 2 \times 2)\hat{i} + (4 + 3 \times 2)\hat{j} = 7\hat{i} + 10\hat{j} \) m/s.
• Ex 2: Calculate displacement after 3 s with \( \vec{v}_0 = 0 \), \( \vec{a} = 4\hat{i} \) m/s².
  • \( \vec{r} = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a} t^2 = 0 + 0 + \frac{1}{2} \times 4 \times 3^2 = 18 \) m along x-axis.

Practice Set

• Q1: Velocity after 1 s if \( \vec{v}_0 = 5\hat{i} \) m/s and \( \vec{a} = 3\hat{i} \) m/s²?
  • A) 5 m/s
  • B) 8 m/s
  • C) 3 m/s
  • D) 15 m/s
• Q2: Displacement in y-direction after 2 s if \( v_{0y} = 2 \) m/s and \( a_y = 1 \) m/s²?
  • A) 4 m
  • B) 6 m
  • C) 8 m
  • D) 10 m
• Q3: Average velocity over 4 s if initial velocity is \( 2\hat{i} + 3\hat{j} \) m/s and acceleration is zero?
  • A) \( 2\hat{i} + 3\hat{j} \) m/s
  • B) Zero
  • C) \( 4\hat{i} + 6\hat{j} \) m/s
  • D) \( 1\hat{i} + 1.5\hat{j} \) m/s

Answer Key

• Q1: B - Velocity \( = 5 + 3 \times 1 = 8 \) m/s.
• Q2: B - Displacement \( = v_{0y} t + \frac{1}{2} a_y t^2 = 2 \times 2 + 0.5 \times 1 \times 4 = 6 \) m.
• Q3: A - No acceleration means velocity constant, so average velocity equals initial velocity.

Memory Aids

• Velocity: \( v = u + at \).
• Displacement: \( s = ut + \frac{1}{2} at^2 \).
• Two-dimensional motion = two independent 1D motions.

Relative Velocity in Two Dimensions

Key Concepts and Formulas

• Actual velocity w.r.t ground: vector V_actual.
• Relative velocity w.r.t moving object: vector V_rel = vector V_actual - vector V_ref, where vector V_ref is velocity of moving object.
• Position vectors relation: vector r_PS = vector r_PS' + vector r_S'S.
• Velocity relation: vector V_PS = vector V_PS' + vector V_S'S.
• Velocity addition: vector V_actual = vector V_rel + vector V_ref.
• Swimming in river: vector V_m = vector V + vector V_R, where vector V is swimmer velocity w.r.t water, vector V_R is river velocity.

Solved Examples

• Ex 1: A swimmer swims downstream with V = 3 m/s and river velocity V_R = 2 m/s.
  • Velocity relative to ground: V_m = V + V_R = 5 m/s.
• Ex 2: Swimmer swims upstream with same speeds.
  • Velocity relative to ground: V_m = V - V_R = 1 m/s.

Practice Set

• Q1: Relative velocity formula is?
  • A) vector V_rel = vector V_actual + vector V_ref
  • B) vector V_rel = vector V_actual - vector V_ref
  • C) vector V_rel = vector V_ref - vector V_actual
  • D) vector V_rel = vector V_actual × vector V_ref
• Q2: If swimmer velocity is 4 m/s and river velocity is 3 m/s downstream, velocity relative to ground is?
  • A) 1 m/s
  • B) 7 m/s
  • C) 12 m/s
  • D) 4 m/s
• Q3: Position vector relation is?
  • A) vector r_PS = vector r_PS' - vector r_S'S
  • B) vector r_PS = vector r_S'S + vector r_PS'
  • C) vector r_PS = vector r_PS'
  • D) vector r_PS = vector r_S'S

Answer Key

• Q1: B - Relative velocity is actual velocity minus reference velocity.
• Q2: B - Downstream velocity adds: 4 + 3 = 7 m/s.
• Q3: B - Position vector addition: vector r_PS = vector r_S'S + vector r_PS'.

Memory Aids

• Relative velocity = Actual velocity - Reference velocity.
• Position vectors add as vector r_PS = vector r_S'S + vector r_PS'.
• Swimming velocity relative to ground = swimmer + river velocity vectors.

Key Concepts and Formulas

• Actual velocity w.r.t ground: vector V_actual.
• Relative velocity w.r.t moving object: vector V_rel = vector V_actual - vector V_ref, where vector V_ref is velocity of moving object.
• Position vectors relation: vector r_PS = vector r_PS' + vector r_S'S.
• Velocity relation: vector V_PS = vector V_PS' + vector V_S'S.
• Velocity addition: vector V_actual = vector V_rel + vector V_ref.
• Swimming in river: vector V_m = vector V + vector V_R, where vector V is swimmer velocity w.r.t water, vector V_R is river velocity.

Solved Examples

• Ex 1: A swimmer swims downstream with V = 3 m/s and river velocity V_R = 2 m/s.
  • Velocity relative to ground: V_m = V + V_R = 5 m/s.
• Ex 2: Swimmer swims upstream with same speeds.
  • Velocity relative to ground: V_m = V - V_R = 1 m/s.

Practice Set

• Q1: Relative velocity formula is?
  • A) vector V_rel = vector V_actual + vector V_ref
  • B) vector V_rel = vector V_actual - vector V_ref
  • C) vector V_rel = vector V_ref - vector V_actual
  • D) vector V_rel = vector V_actual × vector V_ref
• Q2: If swimmer velocity is 4 m/s and river velocity is 3 m/s downstream, velocity relative to ground is?
  • A) 1 m/s
  • B) 7 m/s
  • C) 12 m/s
  • D) 4 m/s
• Q3: Position vector relation is?
  • A) vector r_PS = vector r_PS' - vector r_S'S
  • B) vector r_PS = vector r_S'S + vector r_PS'
  • C) vector r_PS = vector r_PS'
  • D) vector r_PS = vector r_S'S

Answer Key

• Q1: B - Relative velocity is actual velocity minus reference velocity.
• Q2: B - Downstream velocity adds: 4 + 3 = 7 m/s.
• Q3: B - Position vector addition: vector r_PS = vector r_S'S + vector r_PS'.

Memory Aids

• Relative velocity = Actual velocity - Reference velocity.
• Position vectors add as vector r_PS = vector r_S'S + vector r_PS'.
• Swimming velocity relative to ground = swimmer + river velocity vectors.

Key Concepts and Formulas

• Actual velocity w.r.t ground: \( \vec{V}_{actual} \).
• Relative velocity w.r.t moving object: \( \vec{V}_{rel} = \vec{V}_{actual} - \vec{V}_{ref} \), where \( \vec{V}_{ref} \) is velocity of moving object.
• Position vectors relation: \( \vec{r}_{PS} = \vec{r}_{PS'} + \vec{r}_{S'S} \).
• Velocity relation: \( \vec{V}_{PS} = \vec{V}_{PS'} + \vec{V}_{S'S} \).
• Velocity addition: \( \vec{V}_{actual} = \vec{V}_{rel} + \vec{V}_{ref} \).
• Swimming in river: \( \vec{V}_m = \vec{V} + \vec{V}_R \), where \( \vec{V} \) is swimmer velocity w.r.t water, \( \vec{V}_R \) is river velocity.

Solved Examples

• Ex 1: A swimmer swims downstream with \( V = 3 \) m/s and river velocity \( V_R = 2 \) m/s.
  • Velocity relative to ground: \( V_m = V + V_R = 5 \) m/s.
• Ex 2: Swimmer swims upstream with same speeds.
  • Velocity relative to ground: \( V_m = V - V_R = 1 \) m/s.

Practice Set

• Q1: Relative velocity formula is?
  • A) \( \vec{V}_{rel} = \vec{V}_{actual} + \vec{V}_{ref} \)
  • B) \( \vec{V}_{rel} = \vec{V}_{actual} - \vec{V}_{ref} \)
  • C) \( \vec{V}_{rel} = \vec{V}_{ref} - \vec{V}_{actual} \)
  • D) \( \vec{V}_{rel} = \vec{V}_{actual} \times \vec{V}_{ref} \)
• Q2: If swimmer velocity is 4 m/s and river velocity is 3 m/s downstream, velocity relative to ground is?
  • A) 1 m/s
  • B) 7 m/s
  • C) 12 m/s
  • D) 4 m/s
• Q3: Position vector relation is?
  • A) \( \vec{r}_{PS} = \vec{r}_{PS'} - \vec{r}_{S'S} \)
  • B) \( \vec{r}_{PS} = \vec{r}_{S'S} + \vec{r}_{PS'} \)
  • C) \( \vec{r}_{PS} = \vec{r}_{PS'} \)
  • D) \( \vec{r}_{PS} = \vec{r}_{S'S} \)

Answer Key

• Q1: B - Relative velocity is actual velocity minus reference velocity.
• Q2: B - Downstream velocity adds: 4 + 3 = 7 m/s.
• Q3: B - Position vector addition: \( \vec{r}_{PS} = \vec{r}_{S'S} + \vec{r}_{PS'} \).

Memory Aids

• Relative velocity = Actual velocity - Reference velocity.
• Position vectors add as \( \vec{r}_{PS} = \vec{r}_{S'S} + \vec{r}_{PS'} \).
• Swimming velocity relative to ground = swimmer + river velocity vectors.

Key Concepts and Formulas

• Actual velocity w.r.t ground: vector V_actual.
• Relative velocity w.r.t moving object: vector V_rel = vector V_actual - vector V_ref, where vector V_ref is velocity of moving object.
• Position vectors relation: vector r_PS = vector r_PS' + vector r_S'S.
• Velocity relation: vector V_PS = vector V_PS' + vector V_S'S.
• Velocity addition: vector V_actual = vector V_rel + vector V_ref.
• Swimming in river: vector V_m = vector V + vector V_R, where vector V is swimmer velocity w.r.t water, vector V_R is river velocity.

Solved Examples

• Ex 1: A swimmer swims downstream with V = 3 m/s and river velocity V_R = 2 m/s.
  • Velocity relative to ground: V_m = V + V_R = 5 m/s.
• Ex 2: Swimmer swims upstream with same speeds.
  • Velocity relative to ground: V_m = V - V_R = 1 m/s.

Practice Set

• Q1: Relative velocity formula is?
  • A) vector V_rel = vector V_actual + vector V_ref
  • B) vector V_rel = vector V_actual - vector V_ref
  • C) vector V_rel = vector V_ref - vector V_actual
  • D) vector V_rel = vector V_actual × vector V_ref
• Q2: If swimmer velocity is 4 m/s and river velocity is 3 m/s downstream, velocity relative to ground is?
  • A) 1 m/s
  • B) 7 m/s
  • C) 12 m/s
  • D) 4 m/s
• Q3: Position vector relation is?
  • A) vector r_PS = vector r_PS' - vector r_S'S
  • B) vector r_PS = vector r_S'S + vector r_PS'
  • C) vector r_PS = vector r_PS'
  • D) vector r_PS = vector r_S'S

Answer Key

• Q1: B - Relative velocity is actual velocity minus reference velocity.
• Q2: B - Downstream velocity adds: 4 + 3 = 7 m/s.
• Q3: B - Position vector addition: vector r_PS = vector r_S'S + vector r_PS'.

Memory Aids

• Relative velocity = Actual velocity - Reference velocity.
• Position vectors add as vector r_PS = vector r_S'S + vector r_PS'.
• Swimming velocity relative to ground = swimmer + river velocity vectors.

Key Concepts and Formulas

• Actual velocity w.r.t ground: \( \vec{V}_{actual} \).
• Relative velocity w.r.t moving object: \( \vec{V}_{rel} = \vec{V}_{actual} - \vec{V}_{ref} \), where \( \vec{V}_{ref} \) is velocity of moving object.
• Position vectors relation: \( \vec{r}_{PS} = \vec{r}_{PS'} + \vec{r}_{S'S} \).
• Velocity relation: \( \vec{V}_{PS} = \vec{V}_{PS'} + \vec{V}_{S'S} \).
• Velocity addition: \( \vec{V}_{actual} = \vec{V}_{rel} + \vec{V}_{ref} \).
• Swimming in river: \( \vec{V}_m = \vec{V} + \vec{V}_R \), where \( \vec{V} \) is swimmer velocity w.r.t water, \( \vec{V}_R \) is river velocity.

Solved Examples

• Ex 1: A swimmer swims downstream with \( V = 3 \) m/s and river velocity \( V_R = 2 \) m/s.
  • Velocity relative to ground: \( V_m = V + V_R = 5 \) m/s.
• Ex 2: Swimmer swims upstream with same speeds.
  • Velocity relative to ground: \( V_m = V - V_R = 1 \) m/s.

Practice Set

• Q1: Relative velocity formula is?
  • A) \( \vec{V}_{rel} = \vec{V}_{actual} + \vec{V}_{ref} \)
  • B) \( \vec{V}_{rel} = \vec{V}_{actual} - \vec{V}_{ref} \)
  • C) \( \vec{V}_{rel} = \vec{V}_{ref} - \vec{V}_{actual} \)
  • D) \( \vec{V}_{rel} = \vec{V}_{actual} \times \vec{V}_{ref} \)
• Q2: If swimmer velocity is 4 m/s and river velocity is 3 m/s downstream, velocity relative to ground is?
  • A) 1 m/s
  • B) 7 m/s
  • C) 12 m/s
  • D) 4 m/s
• Q3: Position vector relation is?
  • A) \( \vec{r}_{PS} = \vec{r}_{PS'} - \vec{r}_{S'S} \)
  • B) \( \vec{r}_{PS} = \vec{r}_{S'S} + \vec{r}_{PS'} \)
  • C) \( \vec{r}_{PS} = \vec{r}_{PS'} \)
  • D) \( \vec{r}_{PS} = \vec{r}_{S'S} \)

Answer Key

• Q1: B - Relative velocity is actual velocity minus reference velocity.
• Q2: B - Downstream velocity adds: 4 + 3 = 7 m/s.
• Q3: B - Position vector addition: \( \vec{r}_{PS} = \vec{r}_{S'S} + \vec{r}_{PS'} \).

Memory Aids

• Relative velocity = Actual velocity - Reference velocity.
• Position vectors add as \( \vec{r}_{PS} = \vec{r}_{S'S} + \vec{r}_{PS'} \).
• Swimming velocity relative to ground = swimmer + river velocity vectors.

Swimming in the River

Key Points and Formulas

• Velocity of man w.r.t still water: vector V.
• Velocity of river flow: vector V_R.
• Velocity of man relative to ground: vector V_m = vector V + vector V_R.
• Downstream swimming: V_m = V + V_R.
• Upstream swimming: V_m = V - V_R.
• Swimming across river (non-collinear): use vector addition vector V_m = vector V + vector V_R.
• Shortest path: resultant velocity vector V_m directed along displacement vector AB.
• Time to cross river: t = d/(V cos θ) = d/√(V² - V_R²).
• Minimum time: swim perpendicular to current, t_min = d/V.
• Drift distance downstream: BC = t_min V_R = (d/V) V_R.

Solved Examples

• Ex 1: Man swims across river 50 m wide with speed 5 m/s, river speed 3 m/s.
  • Shortest path angle: sin θ = V_R/V = 3/5 = 0.6.
  • Time to cross: t = 50/√(5² - 3²) = 50/4 = 12.5 s.
• Ex 2: Minimum time to cross same river swimming perpendicular.
  • t_min = 50/5 = 10 s.
  • Drift downstream: BC = t_min V_R = 10 × 3 = 30 m.

Practice Set

• Q1: Velocity of man relative to ground when swimming downstream with V = 4 m/s and V_R = 1 m/s?
  • A) 3 m/s
  • B) 5 m/s
  • C) 4 m/s
  • D) 1 m/s
• Q2: Time to cross river 40 m wide with swimmer speed 4 m/s and river speed 3 m/s (shortest path)?
  • A) 10 s
  • B) 12.5 s
  • C) 8 s
  • D) 15 s
• Q3: Drift distance if swimmer crosses river in minimum time 10 s and river speed 2 m/s?
  • A) 20 m
  • B) 10 m
  • C) 5 m
  • D) 15 m

Answer Key

• Q1: B - Downstream velocity adds: 4 + 1 = 5 m/s.
• Q2: B - Time t = d/√(V² - V_R²) = 40/√(16 - 9) = 12.5 s.
• Q3: A - Drift BC = t_min V_R = 10 × 2 = 20 m.

Memory Aids

• Resultant velocity = swimmer velocity + river velocity.
• Shortest path: aim velocity vector to cancel river flow.
• Minimum time: swim perpendicular to current.

Key Points and Formulas

• Velocity of man w.r.t still water: vector V.
• Velocity of river flow: vector V_R.
• Velocity of man relative to ground: vector V_m = vector V + vector V_R.
• Downstream swimming: V_m = V + V_R.
• Upstream swimming: V_m = V - V_R.
• Swimming across river (non-collinear): use vector addition vector V_m = vector V + vector V_R.
• Shortest path: resultant velocity vector V_m directed along displacement vector AB.
• Time to cross river: t = d/(V cos θ) = d/√(V² - V_R²).
• Minimum time: swim perpendicular to current, t_min = d/V.
• Drift distance downstream: BC = t_min V_R = (d/V) V_R.

Solved Examples

• Ex 1: Man swims across river 50 m wide with speed 5 m/s, river speed 3 m/s.
  • Shortest path angle: sin θ = V_R/V = 3/5 = 0.6.
  • Time to cross: t = 50/√(5² - 3²) = 50/4 = 12.5 s.
• Ex 2: Minimum time to cross same river swimming perpendicular.
  • t_min = 50/5 = 10 s.
  • Drift downstream: BC = t_min V_R = 10 × 3 = 30 m.

Practice Set

• Q1: Velocity of man relative to ground when swimming downstream with V = 4 m/s and V_R = 1 m/s?
  • A) 3 m/s
  • B) 5 m/s
  • C) 4 m/s
  • D) 1 m/s
• Q2: Time to cross river 40 m wide with swimmer speed 4 m/s and river speed 3 m/s (shortest path)?
  • A) 10 s
  • B) 12.5 s
  • C) 8 s
  • D) 15 s
• Q3: Drift distance if swimmer crosses river in minimum time 10 s and river speed 2 m/s?
  • A) 20 m
  • B) 10 m
  • C) 5 m
  • D) 15 m

Answer Key

• Q1: B - Downstream velocity adds: 4 + 1 = 5 m/s.
• Q2: B - Time t = d/√(V² - V_R²) = 40/√(16 - 9) = 12.5 s.
• Q3: A - Drift BC = t_min V_R = 10 × 2 = 20 m.

Memory Aids

• Resultant velocity = swimmer velocity + river velocity.
• Shortest path: aim velocity vector to cancel river flow.
• Minimum time: swim perpendicular to current.

Key Points and Formulas

• Velocity of man w.r.t still water: vector V.
• Velocity of river flow: vector V_R.
• Velocity of man relative to ground: vector V_m = vector V + vector V_R.
• Downstream swimming: V_m = V + V_R.
• Upstream swimming: V_m = V - V_R.
• Swimming across river (non-collinear): use vector addition vector V_m = vector V + vector V_R.
• Shortest path: resultant velocity vector V_m directed along displacement vector AB.
• Time to cross river: t = d/(V cos θ) = d/√(V² - V_R²).
• Minimum time: swim perpendicular to current, t_min = d/V.
• Drift distance downstream: BC = t_min V_R = (d/V) V_R.

Solved Examples

• Ex 1: Man swims across river 50 m wide with speed 5 m/s, river speed 3 m/s.
  • Shortest path angle: sin θ = V_R/V = 3/5 = 0.6.
  • Time to cross: t = 50/√(5² - 3²) = 50/4 = 12.5 s.
• Ex 2: Minimum time to cross same river swimming perpendicular.
  • t_min = 50/5 = 10 s.
  • Drift downstream: BC = t_min V_R = 10 × 3 = 30 m.

Practice Set

• Q1: Velocity of man relative to ground when swimming downstream with V = 4 m/s and V_R = 1 m/s?
  • A) 3 m/s
  • B) 5 m/s
  • C) 4 m/s
  • D) 1 m/s
• Q2: Time to cross river 40 m wide with swimmer speed 4 m/s and river speed 3 m/s (shortest path)?
  • A) 10 s
  • B) 12.5 s
  • C) 8 s
  • D) 15 s
• Q3: Drift distance if swimmer crosses river in minimum time 10 s and river speed 2 m/s?
  • A) 20 m
  • B) 10 m
  • C) 5 m
  • D) 15 m

Answer Key

• Q1: B - Downstream velocity adds: 4 + 1 = 5 m/s.
• Q2: B - Time t = d/√(V² - V_R²) = 40/√(16 - 9) = 12.5 s.
• Q3: A - Drift BC = t_min V_R = 10 × 2 = 20 m.

Memory Aids

• Resultant velocity = swimmer velocity + river velocity.
• Shortest path: aim velocity vector to cancel river flow.
• Minimum time: swim perpendicular to current.

Key Points and Formulas

• Velocity of man w.r.t still water: \( \vec{V} \).
• Velocity of river flow: \( \vec{V}_R \).
• Velocity of man relative to ground: \( \vec{V}_m = \vec{V} + \vec{V}_R \).
• Downstream swimming: \( V_m = V + V_R \).
• Upstream swimming: \( V_m = V - V_R \).
• Swimming across river (non-collinear): use vector addition \( \vec{V}_m = \vec{V} + \vec{V}_R \).
• Shortest path: resultant velocity \( \vec{V}_m \) directed along displacement \( \overrightarrow{AB} \).
• Time to cross river: \( t = \frac{d}{V \cos \theta} = \frac{d}{\sqrt{V^2 - V_R^2}} \).
• Minimum time: swim perpendicular to current, \( t_{min} = \frac{d}{V} \).
• Drift distance downstream: \( BC = t_{min} V_R = \frac{d}{V} V_R \).

Solved Examples

• Ex 1: Man swims across river 50 m wide with speed 5 m/s, river speed 3 m/s.
  • Shortest path angle: \( \sin \theta = \frac{V_R}{V} = \frac{3}{5} = 0.6 \).
  • Time to cross: \( t = \frac{50}{\sqrt{5^2 - 3^2}} = \frac{50}{4} = 12.5 \) s.
• Ex 2: Minimum time to cross same river swimming perpendicular.
  • \( t_{min} = \frac{50}{5} = 10 \) s.
  • Drift downstream: \( BC = t_{min} V_R = 10 \times 3 = 30 \) m.

Practice Set

• Q1: Velocity of man relative to ground when swimming downstream with \( V = 4 \) m/s and \( V_R = 1 \) m/s?
  • A) 3 m/s
  • B) 5 m/s
  • C) 4 m/s
  • D) 1 m/s
• Q2: Time to cross river 40 m wide with swimmer speed 4 m/s and river speed 3 m/s (shortest path)?
  • A) 10 s
  • B) 12.5 s
  • C) 8 s
  • D) 15 s
• Q3: Drift distance if swimmer crosses river in minimum time 10 s and river speed 2 m/s?
  • A) 20 m
  • B) 10 m
  • C) 5 m
  • D) 15 m

Answer Key

• Q1: B - Downstream velocity adds: 4 + 1 = 5 m/s.
• Q2: B - Time \( t = \frac{d}{\sqrt{V^2 - V_R^2}} = \frac{40}{\sqrt{16 - 9}} = 12.5 \) s.
• Q3: A - Drift \( BC = t_{min} V_R = 10 \times 2 = 20 \) m.

Memory Aids

• Resultant velocity = swimmer velocity + river velocity.
• Shortest path: aim velocity vector to cancel river flow.
• Minimum time: swim perpendicular to current.

Key Points and Formulas

• Velocity of man w.r.t still water: vector V.
• Velocity of river flow: vector V_R.
• Velocity of man relative to ground: vector V_m = vector V + vector V_R.
• Downstream swimming: V_m = V + V_R.
• Upstream swimming: V_m = V - V_R.
• Swimming across river (non-collinear): use vector addition vector V_m = vector V + vector V_R.
• Shortest path: resultant velocity vector V_m directed along displacement vector AB.
• Time to cross river: t = d/(V cos θ) = d/√(V² - V_R²).
• Minimum time: swim perpendicular to current, t_min = d/V.
• Drift distance downstream: BC = t_min V_R = (d/V) V_R.

Solved Examples

• Ex 1: Man swims across river 50 m wide with speed 5 m/s, river speed 3 m/s.
  • Shortest path angle: sin θ = V_R/V = 3/5 = 0.6.
  • Time to cross: t = 50/√(5² - 3²) = 50/4 = 12.5 s.
• Ex 2: Minimum time to cross same river swimming perpendicular.
  • t_min = 50/5 = 10 s.
  • Drift downstream: BC = t_min V_R = 10 × 3 = 30 m.

Practice Set

• Q1: Velocity of man relative to ground when swimming downstream with V = 4 m/s and V_R = 1 m/s?
  • A) 3 m/s
  • B) 5 m/s
  • C) 4 m/s
  • D) 1 m/s
• Q2: Time to cross river 40 m wide with swimmer speed 4 m/s and river speed 3 m/s (shortest path)?
  • A) 10 s
  • B) 12.5 s
  • C) 8 s
  • D) 15 s
• Q3: Drift distance if swimmer crosses river in minimum time 10 s and river speed 2 m/s?
  • A) 20 m
  • B) 10 m
  • C) 5 m
  • D) 15 m

Answer Key

• Q1: B - Downstream velocity adds: 4 + 1 = 5 m/s.
• Q2: B - Time t = d/√(V² - V_R²) = 40/√(16 - 9) = 12.5 s.
• Q3: A - Drift BC = t_min V_R = 10 × 2 = 20 m.

Memory Aids

• Resultant velocity = swimmer velocity + river velocity.
• Shortest path: aim velocity vector to cancel river flow.
• Minimum time: swim perpendicular to current.

Key Points and Formulas

• Velocity of man w.r.t still water: \( \vec{V} \).
• Velocity of river flow: \( \vec{V}_R \).
• Velocity of man relative to ground: \( \vec{V}_m = \vec{V} + \vec{V}_R \).
• Downstream swimming: \( V_m = V + V_R \).
• Upstream swimming: \( V_m = V - V_R \).
• Swimming across river (non-collinear): use vector addition \( \vec{V}_m = \vec{V} + \vec{V}_R \).
• Shortest path: resultant velocity \( \vec{V}_m \) directed along displacement \( \overrightarrow{AB} \).
• Time to cross river: \( t = \frac{d}{V \cos \theta} = \frac{d}{\sqrt{V^2 - V_R^2}} \).
• Minimum time: swim perpendicular to current, \( t_{min} = \frac{d}{V} \).
• Drift distance downstream: \( BC = t_{min} V_R = \frac{d}{V} V_R \).

Solved Examples

• Ex 1: Man swims across river 50 m wide with speed 5 m/s, river speed 3 m/s.
  • Shortest path angle: \( \sin \theta = \frac{V_R}{V} = \frac{3}{5} = 0.6 \).
  • Time to cross: \( t = \frac{50}{\sqrt{5^2 - 3^2}} = \frac{50}{4} = 12.5 \) s.
• Ex 2: Minimum time to cross same river swimming perpendicular.
  • \( t_{min} = \frac{50}{5} = 10 \) s.
  • Drift downstream: \( BC = t_{min} V_R = 10 \times 3 = 30 \) m.

Practice Set

• Q1: Velocity of man relative to ground when swimming downstream with \( V = 4 \) m/s and \( V_R = 1 \) m/s?
  • A) 3 m/s
  • B) 5 m/s
  • C) 4 m/s
  • D) 1 m/s
• Q2: Time to cross river 40 m wide with swimmer speed 4 m/s and river speed 3 m/s (shortest path)?
  • A) 10 s
  • B) 12.5 s
  • C) 8 s
  • D) 15 s
• Q3: Drift distance if swimmer crosses river in minimum time 10 s and river speed 2 m/s?
  • A) 20 m
  • B) 10 m
  • C) 5 m
  • D) 15 m

Answer Key

• Q1: B - Downstream velocity adds: 4 + 1 = 5 m/s.
• Q2: B - Time \( t = \frac{d}{\sqrt{V^2 - V_R^2}} = \frac{40}{\sqrt{16 - 9}} = 12.5 \) s.
• Q3: A - Drift \( BC = t_{min} V_R = 10 \times 2 = 20 \) m.

Memory Aids

• Resultant velocity = swimmer velocity + river velocity.
• Shortest path: aim velocity vector to cancel river flow.
• Minimum time: swim perpendicular to current.

Rain Problem

Key Concepts and Formulas

• Velocity of rain w.r.t ground: vector V_RG = a î - b ĵ.
• Speed of rain: |vector V_RG| = √(a² + b²).
• Velocity of man w.r.t ground: vector V_MG = v î.
• Velocity of rain w.r.t man: vector V_RM = vector V_RG - vector V_MG = (a - v) î - b ĵ.
• Angle of rain w.r.t ground: tan α = a/b.
• Angle of rain w.r.t man: tan θ = |a - v|/b.
• Cases:
  • If a > v, rain slants backward, tan θ = (a - v)/b.
  • If a = v, rain appears vertical, θ = 0°.
  • If a < v, rain slants forward, tan θ = (v - a)/b.

Solved Examples

• Ex 1: Rain velocity components: a = 3 m/s, b = 4 m/s; man speed v = 2 m/s.
  • tan θ = (3 - 2)/4 = 0.25, θ = 14° backward slant.
• Ex 2: If man speed v = 3 m/s equals rain horizontal speed a = 3 m/s.
  • Rain appears vertical, θ = 0°.

Practice Set

• Q1: Velocity of rain w.r.t man is?
  • A) vector V_RM = vector V_RG + vector V_MG
  • B) vector V_RM = vector V_RG - vector V_MG
  • C) vector V_RM = vector V_MG - vector V_RG
  • D) vector V_RM = vector V_RG × vector V_MG
• Q2: If rain horizontal speed a = 4 m/s, vertical speed b = 3 m/s, man speed v = 2 m/s, angle θ is?
  • A) tan^-1 (2/3)
  • B) tan^-1 (4/3)
  • C) tan^-1 (2/4)
  • D) tan^-1 (2/5)
• Q3: When does rain appear vertical to man?
  • A) a > v
  • B) a < v
  • C) a = v
  • D) a = 0

Answer Key

• Q1: B - Relative velocity of rain w.r.t man is rain velocity minus man velocity.
• Q2: A - tan θ = (a - v)/b = (4 - 2)/3 = 2/3.
• Q3: C - Rain appears vertical when horizontal rain speed equals man speed.

Memory Aids

• Relative velocity: vector V_RM = vector V_RG - vector V_MG.
• Angle θ depends on difference in horizontal speeds.
• Rain appears vertical if man speed equals rain horizontal speed.

Key Concepts and Formulas

• Velocity of rain w.r.t ground: vector V_RG = a î - b ĵ.
• Speed of rain: |vector V_RG| = √(a² + b²).
• Velocity of man w.r.t ground: vector V_MG = v î.
• Velocity of rain w.r.t man: vector V_RM = vector V_RG - vector V_MG = (a - v) î - b ĵ.
• Angle of rain w.r.t ground: tan α = a/b.
• Angle of rain w.r.t man: tan θ = |a - v|/b.
• Cases:
  • If a > v, rain slants backward, tan θ = (a - v)/b.
  • If a = v, rain appears vertical, θ = 0°.
  • If a < v, rain slants forward, tan θ = (v - a)/b.

Solved Examples

• Ex 1: Rain velocity components: a = 3 m/s, b = 4 m/s; man speed v = 2 m/s.
  • tan θ = (3 - 2)/4 = 0.25, θ = 14° backward slant.
• Ex 2: If man speed v = 3 m/s equals rain horizontal speed a = 3 m/s.
  • Rain appears vertical, θ = 0°.

Practice Set

• Q1: Velocity of rain w.r.t man is?
  • A) vector V_RM = vector V_RG + vector V_MG
  • B) vector V_RM = vector V_RG - vector V_MG
  • C) vector V_RM = vector V_MG - vector V_RG
  • D) vector V_RM = vector V_RG × vector V_MG
• Q2: If rain horizontal speed a = 4 m/s, vertical speed b = 3 m/s, man speed v = 2 m/s, angle θ is?
  • A) tan^-1 (2/3)
  • B) tan^-1 (4/3)
  • C) tan^-1 (2/4)
  • D) tan^-1 (2/5)
• Q3: When does rain appear vertical to man?
  • A) a > v
  • B) a < v
  • C) a = v
  • D) a = 0

Answer Key

• Q1: B - Relative velocity of rain w.r.t man is rain velocity minus man velocity.
• Q2: A - tan θ = (a - v)/b = (4 - 2)/3 = 2/3.
• Q3: C - Rain appears vertical when horizontal rain speed equals man speed.

Memory Aids

• Relative velocity: vector V_RM = vector V_RG - vector V_MG.
• Angle θ depends on difference in horizontal speeds.
• Rain appears vertical if man speed equals rain horizontal speed.

Key Concepts and Formulas

• Velocity of rain w.r.t ground: vector V_RG = a î - b ĵ.
• Speed of rain: |vector V_RG| = √(a² + b²).
• Velocity of man w.r.t ground: vector V_MG = v î.
• Velocity of rain w.r.t man: vector V_RM = vector V_RG - vector V_MG = (a - v) î - b ĵ.
• Angle of rain w.r.t ground: tan α = a/b.
• Angle of rain w.r.t man: tan θ = |a - v|/b.
• Cases:
  • If a > v, rain slants backward, tan θ = (a - v)/b.
  • If a = v, rain appears vertical, θ = 0°.
  • If a < v, rain slants forward, tan θ = (v - a)/b.

Solved Examples

• Ex 1: Rain velocity components: a = 3 m/s, b = 4 m/s; man speed v = 2 m/s.
  • tan θ = (3 - 2)/4 = 0.25, θ = 14° backward slant.
• Ex 2: If man speed v = 3 m/s equals rain horizontal speed a = 3 m/s.
  • Rain appears vertical, θ = 0°.

Practice Set

• Q1: Velocity of rain w.r.t man is?
  • A) vector V_RM = vector V_RG + vector V_MG
  • B) vector V_RM = vector V_RG - vector V_MG
  • C) vector V_RM = vector V_MG - vector V_RG
  • D) vector V_RM = vector V_RG × vector V_MG
• Q2: If rain horizontal speed a = 4 m/s, vertical speed b = 3 m/s, man speed v = 2 m/s, angle θ is?
  • A) tan^-1 (2/3)
  • B) tan^-1 (4/3)
  • C) tan^-1 (2/4)
  • D) tan^-1 (2/5)
• Q3: When does rain appear vertical to man?
  • A) a > v
  • B) a < v
  • C) a = v
  • D) a = 0

Answer Key

• Q1: B - Relative velocity of rain w.r.t man is rain velocity minus man velocity.
• Q2: A - tan θ = (a - v)/b = (4 - 2)/3 = 2/3.
• Q3: C - Rain appears vertical when horizontal rain speed equals man speed.

Memory Aids

• Relative velocity: vector V_RM = vector V_RG - vector V_MG.
• Angle θ depends on difference in horizontal speeds.
• Rain appears vertical if man speed equals rain horizontal speed.

Key Concepts and Formulas

• Velocity of rain w.r.t ground: vector V_RG = a î - b ĵ.
• Speed of rain: |vector V_RG| = √(a² + b²).
• Velocity of man w.r.t ground: vector V_MG = v î.
• Velocity of rain w.r.t man: vector V_RM = vector V_RG - vector V_MG = (a - v) î - b ĵ.
• Angle of rain w.r.t ground: tan α = a/b.
• Angle of rain w.r.t man: tan θ = |a - v|/b.
• Cases:
  • If a > v, rain slants backward, tan θ = (a - v)/b.
  • If a = v, rain appears vertical, θ = 0°.
  • If a < v, rain slants forward, tan θ = (v - a)/b.

Solved Examples

• Ex 1: Rain velocity components: a = 3 m/s, b = 4 m/s; man speed v = 2 m/s.
  • tan θ = (3 - 2)/4 = 0.25, θ = 14° backward slant.
• Ex 2: If man speed v = 3 m/s equals rain horizontal speed a = 3 m/s.
  • Rain appears vertical, θ = 0°.

Practice Set

• Q1: Velocity of rain w.r.t man is?
  • A) vector V_RM = vector V_RG + vector V_MG
  • B) vector V_RM = vector V_RG - vector V_MG
  • C) vector V_RM = vector V_MG - vector V_RG
  • D) vector V_RM = vector V_RG × vector V_MG
• Q2: If rain horizontal speed a = 4 m/s, vertical speed b = 3 m/s, man speed v = 2 m/s, angle θ is?
  • A) tan^-1 (2/3)
  • B) tan^-1 (4/3)
  • C) tan^-1 (2/4)
  • D) tan^-1 (2/5)
• Q3: When does rain appear vertical to man?
  • A) a > v
  • B) a < v
  • C) a = v
  • D) a = 0

Answer Key

• Q1: B - Relative velocity of rain w.r.t man is rain velocity minus man velocity.
• Q2: A - tan θ = (a - v)/b = (4 - 2)/3 = 2/3.
• Q3: C - Rain appears vertical when horizontal rain speed equals man speed.

Memory Aids

• Relative velocity: vector V_RM = vector V_RG - vector V_MG.
• Angle θ depends on difference in horizontal speeds.
• Rain appears vertical if man speed equals rain horizontal speed.

Key Concepts and Formulas

• Velocity of rain w.r.t ground: vector V_RG = a î - b ĵ.
• Speed of rain: |vector V_RG| = √(a² + b²).
• Velocity of man w.r.t ground: vector V_MG = v î.
• Velocity of rain w.r.t man: vector V_RM = vector V_RG - vector V_MG = (a - v) î - b ĵ.
• Angle of rain w.r.t ground: tan α = a/b.
• Angle of rain w.r.t man: tan θ = |a - v|/b.
• Cases:
  • If a > v, rain slants backward, tan θ = (a - v)/b.
  • If a = v, rain appears vertical, θ = 0°.
  • If a < v, rain slants forward, tan θ = (v - a)/b.

Solved Examples

• Ex 1: Rain velocity components: a = 3 m/s, b = 4 m/s; man speed v = 2 m/s.
  • tan θ = (3 - 2)/4 = 0.25, θ = 14° backward slant.
• Ex 2: If man speed v = 3 m/s equals rain horizontal speed a = 3 m/s.
  • Rain appears vertical, θ = 0°.

Practice Set

• Q1: Velocity of rain w.r.t man is?
  • A) vector V_RM = vector V_RG + vector V_MG
  • B) vector V_RM = vector V_RG - vector V_MG
  • C) vector V_RM = vector V_MG - vector V_RG
  • D) vector V_RM = vector V_RG × vector V_MG
• Q2: If rain horizontal speed a = 4 m/s, vertical speed b = 3 m/s, man speed v = 2 m/s, angle θ is?
  • A) tan^-1 (2/3)
  • B) tan^-1 (4/3)
  • C) tan^-1 (2/4)
  • D) tan^-1 (2/5)
• Q3: When does rain appear vertical to man?
  • A) a > v
  • B) a < v
  • C) a = v
  • D) a = 0

Answer Key

• Q1: B - Relative velocity of rain w.r.t man is rain velocity minus man velocity.
• Q2: A - tan θ = (a - v)/b = (4 - 2)/3 = 2/3.
• Q3: C - Rain appears vertical when horizontal rain speed equals man speed.

Memory Aids

• Relative velocity: vector V_RM = vector V_RG - vector V_MG.
• Angle θ depends on difference in horizontal speeds.
• Rain appears vertical if man speed equals rain horizontal speed.

Key Concepts and Formulas

• Velocity of rain w.r.t ground: vector V_RG = a î - b ĵ.
• Speed of rain: |vector V_RG| = √(a² + b²).
• Velocity of man w.r.t ground: vector V_MG = v î.
• Velocity of rain w.r.t man: vector V_RM = vector V_RG - vector V_MG = (a - v) î - b ĵ.
• Angle of rain w.r.t ground: tan α = a/b.
• Angle of rain w.r.t man: tan θ = |a - v|/b.
• Cases:
  • If a > v, rain slants backward, tan θ = (a - v)/b.
  • If a = v, rain appears vertical, θ = 0°.
  • If a < v, rain slants forward, tan θ = (v - a)/b.

Solved Examples

• Ex 1: Rain velocity components: a = 3 m/s, b = 4 m/s; man speed v = 2 m/s.
  • tan θ = (3 - 2)/4 = 0.25, θ = 14° backward slant.
• Ex 2: If man speed v = 3 m/s equals rain horizontal speed a = 3 m/s.
  • Rain appears vertical, θ = 0°.

Practice Set

• Q1: Velocity of rain w.r.t man is?
  • A) vector V_RM = vector V_RG + vector V_MG
  • B) vector V_RM = vector V_RG - vector V_MG
  • C) vector V_RM = vector V_MG - vector V_RG
  • D) vector V_RM = vector V_RG × vector V_MG
• Q2: If rain horizontal speed a = 4 m/s, vertical speed b = 3 m/s, man speed v = 2 m/s, angle θ is?
  • A) tan^-1 (2/3)
  • B) tan^-1 (4/3)
  • C) tan^-1 (2/4)
  • D) tan^-1 (2/5)
• Q3: When does rain appear vertical to man?
  • A) a > v
  • B) a < v
  • C) a = v
  • D) a = 0

Answer Key

• Q1: B - Relative velocity of rain w.r.t man is rain velocity minus man velocity.
• Q2: A - tan θ = (a - v)/b = (4 - 2)/3 = 2/3.
• Q3: C - Rain appears vertical when horizontal rain speed equals man speed.

Memory Aids

• Relative velocity: vector V_RM = vector V_RG - vector V_MG.
• Angle θ depends on difference in horizontal speeds.
• Rain appears vertical if man speed equals rain horizontal speed.

Key Concepts and Formulas

• Velocity of rain w.r.t ground: vector V_RG = a î - b ĵ.
• Speed of rain: |vector V_RG| = √(a² + b²).
• Velocity of man w.r.t ground: vector V_MG = v î.
• Velocity of rain w.r.t man: vector V_RM = vector V_RG - vector V_MG = (a - v) î - b ĵ.
• Angle of rain w.r.t ground: tan α = a/b.
• Angle of rain w.r.t man: tan θ = |a - v|/b.
• Cases:
  • If a > v, rain slants backward, tan θ = (a - v)/b.
  • If a = v, rain appears vertical, θ = 0°.
  • If a < v, rain slants forward, tan θ = (v - a)/b.

Solved Examples

• Ex 1: Rain velocity components: a = 3 m/s, b = 4 m/s; man speed v = 2 m/s.
  • tan θ = (3 - 2)/4 = 0.25, θ = 14° backward slant.
• Ex 2: If man speed v = 3 m/s equals rain horizontal speed a = 3 m/s.
  • Rain appears vertical, θ = 0°.

Practice Set

• Q1: Velocity of rain w.r.t man is?
  • A) vector V_RM = vector V_RG + vector V_MG
  • B) vector V_RM = vector V_RG - vector V_MG
  • C) vector V_RM = vector V_MG - vector V_RG
  • D) vector V_RM = vector V_RG × vector V_MG
• Q2: If rain horizontal speed a = 4 m/s, vertical speed b = 3 m/s, man speed v = 2 m/s, angle θ is?
  • A) tan^-1 (2/3)
  • B) tan^-1 (4/3)
  • C) tan^-1 (2/4)
  • D) tan^-1 (2/5)
• Q3: When does rain appear vertical to man?
  • A) a > v
  • B) a < v
  • C) a = v
  • D) a = 0

Answer Key

• Q1: B - Relative velocity of rain w.r.t man is rain velocity minus man velocity.
• Q2: A - tan θ = (a - v)/b = (4 - 2)/3 = 2/3.
• Q3: C - Rain appears vertical when horizontal rain speed equals man speed.

Memory Aids

• Relative velocity: vector V_RM = vector V_RG - vector V_MG.
• Angle θ depends on difference in horizontal speeds.
• Rain appears vertical if man speed equals rain horizontal speed.

Key Concepts and Formulas

• Velocity of rain w.r.t ground: vector V_RG = a î - b ĵ.
• Speed of rain: |vector V_RG| = √(a² + b²).
• Velocity of man w.r.t ground: vector V_MG = v î.
• Velocity of rain w.r.t man: vector V_RM = vector V_RG - vector V_MG = (a - v) î - b ĵ.
• Angle of rain w.r.t ground: tan α = a/b.
• Angle of rain w.r.t man: tan θ = |a - v|/b.
• Cases:
  • If a > v, rain slants backward, tan θ = (a - v)/b.
  • If a = v, rain appears vertical, θ = 0°.
  • If a < v, rain slants forward, tan θ = (v - a)/b.

Solved Examples

• Ex 1: Rain velocity components: a = 3 m/s, b = 4 m/s; man speed v = 2 m/s.
  • tan θ = (3 - 2)/4 = 0.25, θ = 14° backward slant.
• Ex 2: If man speed v = 3 m/s equals rain horizontal speed a = 3 m/s.
  • Rain appears vertical, θ = 0°.

Practice Set

• Q1: Velocity of rain w.r.t man is?
  • A) vector V_RM = vector V_RG + vector V_MG
  • B) vector V_RM = vector V_RG - vector V_MG
  • C) vector V_RM = vector V_MG - vector V_RG
  • D) vector V_RM = vector V_RG × vector V_MG
• Q2: If rain horizontal speed a = 4 m/s, vertical speed b = 3 m/s, man speed v = 2 m/s, angle θ is?
  • A) tan^-1 (2/3)
  • B) tan^-1 (4/3)
  • C) tan^-1 (2/4)
  • D) tan^-1 (2/5)
• Q3: When does rain appear vertical to man?
  • A) a > v
  • B) a < v
  • C) a = v
  • D) a = 0

Answer Key

• Q1: B - Relative velocity of rain w.r.t man is rain velocity minus man velocity.
• Q2: A - tan θ = (a - v)/b = (4 - 2)/3 = 2/3.
• Q3: C - Rain appears vertical when horizontal rain speed equals man speed.

Memory Aids

• Relative velocity: vector V_RM = vector V_RG - vector V_MG.
• Angle θ depends on difference in horizontal speeds.
• Rain appears vertical if man speed equals rain horizontal speed.

Key Concepts and Formulas

• Velocity of rain w.r.t ground: vector V_RG = a î - b ĵ.
• Speed of rain: |vector V_RG| = √(a² + b²).
• Velocity of man w.r.t ground: vector V_MG = v î.
• Velocity of rain w.r.t man: vector V_RM = vector V_RG - vector V_MG = (a - v) î - b ĵ.
• Angle of rain w.r.t ground: tan α = a/b.
• Angle of rain w.r.t man: tan θ = |a - v|/b.
• Cases:
  • If a > v, rain slants backward, tan θ = (a - v)/b.
  • If a = v, rain appears vertical, θ = 0°.
  • If a < v, rain slants forward, tan θ = (v - a)/b.

Solved Examples

• Ex 1: Rain velocity components: a = 3 m/s, b = 4 m/s; man speed v = 2 m/s.
  • tan θ = (3 - 2)/4 = 0.25, θ = 14° backward slant.
• Ex 2: If man speed v = 3 m/s equals rain horizontal speed a = 3 m/s.
  • Rain appears vertical, θ = 0°.

Practice Set

• Q1: Velocity of rain w.r.t man is?
  • A) vector V_RM = vector V_RG + vector V_MG
  • B) vector V_RM = vector V_RG - vector V_MG
  • C) vector V_RM = vector V_MG - vector V_RG
  • D) vector V_RM = vector V_RG × vector V_MG
• Q2: If rain horizontal speed a = 4 m/s, vertical speed b = 3 m/s, man speed v = 2 m/s, angle θ is?
  • A) tan^-1 (2/3)
  • B) tan^-1 (4/3)
  • C) tan^-1 (2/4)
  • D) tan^-1 (2/5)
• Q3: When does rain appear vertical to man?
  • A) a > v
  • B) a < v
  • C) a = v
  • D) a = 0

Answer Key

• Q1: B - Relative velocity of rain w.r.t man is rain velocity minus man velocity.
• Q2: A - tan θ = (a - v)/b = (4 - 2)/3 = 2/3.
• Q3: C - Rain appears vertical when horizontal rain speed equals man speed.

Memory Aids

• Relative velocity: vector V_RM = vector V_RG - vector V_MG.
• Angle θ depends on difference in horizontal speeds.
• Rain appears vertical if man speed equals rain horizontal speed.

Key Concepts and Formulas

• Velocity of rain w.r.t ground: vector V_RG = a î - b ĵ.
• Speed of rain: |vector V_RG| = √(a² + b²).
• Velocity of man w.r.t ground: vector V_MG = v î.
• Velocity of rain w.r.t man: vector V_RM = vector V_RG - vector V_MG = (a - v) î - b ĵ.
• Angle of rain w.r.t ground: tan α = a/b.
• Angle of rain w.r.t man: tan θ = |a - v|/b.
• Cases:
  • If a > v, rain slants backward, tan θ = (a - v)/b.
  • If a = v, rain appears vertical, θ = 0°.
  • If a < v, rain slants forward, tan θ = (v - a)/b.

Solved Examples

• Ex 1: Rain velocity components: a = 3 m/s, b = 4 m/s; man speed v = 2 m/s.
  • tan θ = (3 - 2)/4 = 0.25, θ = 14° backward slant.
• Ex 2: If man speed v = 3 m/s equals rain horizontal speed a = 3 m/s.
  • Rain appears vertical, θ = 0°.

Practice Set

• Q1: Velocity of rain w.r.t man is?
  • A) vector V_RM = vector V_RG + vector V_MG
  • B) vector V_RM = vector V_RG - vector V_MG
  • C) vector V_RM = vector V_MG - vector V_RG
  • D) vector V_RM = vector V_RG × vector V_MG
• Q2: If rain horizontal speed a = 4 m/s, vertical speed b = 3 m/s, man speed v = 2 m/s, angle θ is?
  • A) tan^-1 (2/3)
  • B) tan^-1 (4/3)
  • C) tan^-1 (2/4)
  • D) tan^-1 (2/5)
• Q3: When does rain appear vertical to man?
  • A) a > v
  • B) a < v
  • C) a = v
  • D) a = 0

Answer Key

• Q1: B - Relative velocity of rain w.r.t man is rain velocity minus man velocity.
• Q2: A - tan θ = (a - v)/b = (4 - 2)/3 = 2/3.
• Q3: C - Rain appears vertical when horizontal rain speed equals man speed.

Memory Aids

• Relative velocity: vector V_RM = vector V_RG - vector V_MG.
• Angle θ depends on difference in horizontal speeds.
• Rain appears vertical if man speed equals rain horizontal speed.

Key Concepts and Formulas

• Velocity of rain w.r.t ground: vector V_RG = a î - b ĵ.
• Speed of rain: |vector V_RG| = √(a² + b²).
• Velocity of man w.r.t ground: vector V_MG = v î.
• Velocity of rain w.r.t man: vector V_RM = vector V_RG - vector V_MG = (a - v) î - b ĵ.
• Angle of rain w.r.t ground: tan α = a/b.
• Angle of rain w.r.t man: tan θ = |a - v|/b.
• Cases:
  • If a > v, rain slants backward, tan θ = (a - v)/b.
  • If a = v, rain appears vertical, θ = 0°.
  • If a < v, rain slants forward, tan θ = (v - a)/b.

Solved Examples

• Ex 1: Rain velocity components: a = 3 m/s, b = 4 m/s; man speed v = 2 m/s.
  • tan θ = (3 - 2)/4 = 0.25, θ = 14° backward slant.
• Ex 2: If man speed v = 3 m/s equals rain horizontal speed a = 3 m/s.
  • Rain appears vertical, θ = 0°.

Practice Set

• Q1: Velocity of rain w.r.t man is?
  • A) vector V_RM = vector V_RG + vector V_MG
  • B) vector V_RM = vector V_RG - vector V_MG
  • C) vector V_RM = vector V_MG - vector V_RG
  • D) vector V_RM = vector V_RG × vector V_MG
• Q2: If rain horizontal speed a = 4 m/s, vertical speed b = 3 m/s, man speed v = 2 m/s, angle θ is?
  • A) tan^-1 (2/3)
  • B) tan^-1 (4/3)
  • C) tan^-1 (2/4)
  • D) tan^-1 (2/5)
• Q3: When does rain appear vertical to man?
  • A) a > v
  • B) a < v
  • C) a = v
  • D) a = 0

Answer Key

• Q1: B - Relative velocity of rain w.r.t man is rain velocity minus man velocity.
• Q2: A - tan θ = (a - v)/b = (4 - 2)/3 = 2/3.
• Q3: C - Rain appears vertical when horizontal rain speed equals man speed.

Memory Aids

• Relative velocity: vector V_RM = vector V_RG - vector V_MG.
• Angle θ depends on difference in horizontal speeds.
• Rain appears vertical if man speed equals rain horizontal speed.

Key Concepts and Formulas

• Velocity of rain w.r.t ground: vector V_RG = a î - b ĵ.
• Speed of rain: |vector V_RG| = √(a² + b²).
• Velocity of man w.r.t ground: vector V_MG = v î.
• Velocity of rain w.r.t man: vector V_RM = vector V_RG - vector V_MG = (a - v) î - b ĵ.
• Angle of rain w.r.t ground: tan α = a/b.
• Angle of rain w.r.t man: tan θ = |a - v|/b.
• Cases:
  • If a > v, rain slants backward, tan θ = (a - v)/b.
  • If a = v, rain appears vertical, θ = 0°.
  • If a < v, rain slants forward, tan θ = (v - a)/b.

Solved Examples

• Ex 1: Rain velocity components: a = 3 m/s, b = 4 m/s; man speed v = 2 m/s.
  • tan θ = (3 - 2)/4 = 0.25, θ = 14° backward slant.
• Ex 2: If man speed v = 3 m/s equals rain horizontal speed a = 3 m/s.
  • Rain appears vertical, θ = 0°.

Practice Set

• Q1: Velocity of rain w.r.t man is?
  • A) vector V_RM = vector V_RG + vector V_MG
  • B) vector V_RM = vector V_RG - vector V_MG
  • C) vector V_RM = vector V_MG - vector V_RG
  • D) vector V_RM = vector V_RG × vector V_MG
• Q2: If rain horizontal speed a = 4 m/s, vertical speed b = 3 m/s, man speed v = 2 m/s, angle θ is?
  • A) tan^-1 (2/3)
  • B) tan^-1 (4/3)
  • C) tan^-1 (2/4)
  • D) tan^-1 (2/5)
• Q3: When does rain appear vertical to man?
  • A) a > v
  • B) a < v
  • C) a = v
  • D) a = 0

Answer Key

• Q1: B - Relative velocity of rain w.r.t man is rain velocity minus man velocity.
• Q2: A - tan θ = (a - v)/b = (4 - 2)/3 = 2/3.
• Q3: C - Rain appears vertical when horizontal rain speed equals man speed.

Memory Aids

• Relative velocity: vector V_RM = vector V_RG - vector V_MG.
• Angle θ depends on difference in horizontal speeds.
• Rain appears vertical if man speed equals rain horizontal speed.

Projectile Motion

Concepts and Formulas

• Projectile: Object thrown at an angle under gravity, motion in vertical plane.
• Horizontal motion: Constant velocity u_x = u cos θ.
• Vertical motion: Accelerated motion under gravity a_y = -g.
• Equations of motion:
  • Horizontal displacement: x = u cos θ × t
  • Vertical displacement: y = u sin θ × t - 1/2 g t²
  • Vertical velocity: v_y = u sin θ - g t
• Trajectory equation: y = x tan θ - g x²/(2 u² cos² θ) (parabolic path)
• Time of flight: T = 2 u sin θ/g
• Maximum height: H = u² sin² θ/(2 g)
• Horizontal range: R = u² sin 2θ/g

Solved Examples

Ex 1: Find time of flight for u=20 m/s, θ=30°.

• Use T = 2 u sin θ/g = 2 × 20 × sin 30°/9.8 = 2.04 s

Ex 2: Calculate maximum height for u=15 m/s, θ=45°.

• Use H = u² sin² θ/(2 g) = 15² × sin² 45°/(2 × 9.8) = 8.6 m

Practice Set

Q1: What is the horizontal range of a projectile with u=10 m/s, θ=60°?
A) 5 m
B) 8.8 m
C) 10 m
D) 15 m

Q2: Time of flight for a projectile launched at 45° with speed 20 m/s is:
A) 2.04 s
B) 2.88 s
C) 4.08 s
D) 5.10 s

Q3: Maximum height reached by a projectile with u=25 m/s, θ=30° is:
A) 7.96 m
B) 15.9 m
C) 12.7 m
D) 9.8 m

Answer Key

Q1: B; R = u² sin 2θ/g = 100 × sin 120°/9.8 = 8.8 m

Q2: C; T = 2 u sin θ/g = 2 × 20 × sin 45°/9.8 = 2.88 s

Q3: A; H = u² sin² θ/(2 g) = 625 × 0.25/19.6 = 7.96 m

Memory Aids

• Time of flight T doubles vertical velocity component.
• Range R depends on sin 2θ for max distance at 45°.
• Max height H depends on square of vertical velocity component.

Concepts and Formulas

• Projectile: Object thrown at an angle under gravity, motion in vertical plane.
• Horizontal motion: Constant velocity u_x = u cos θ.
• Vertical motion: Accelerated motion under gravity a_y = -g.
• Equations of motion:
  • Horizontal displacement: x = u cos θ × t
  • Vertical displacement: y = u sin θ × t - 1/2 g t²
  • Vertical velocity: v_y = u sin θ - g t
• Trajectory equation: y = x tan θ - g x²/(2 u² cos² θ) (parabolic path)
• Time of flight: T = 2 u sin θ/g
• Maximum height: H = u² sin² θ/(2 g)
• Horizontal range: R = u² sin 2θ/g

Solved Examples

Ex 1: Find time of flight for u=20 m/s, θ=30°.

• Use T = 2 u sin θ/g = 2 × 20 × sin 30°/9.8 = 2.04 s

Ex 2: Calculate maximum height for u=15 m/s, θ=45°.

• Use H = u² sin² θ/(2 g) = 15² × sin² 45°/(2 × 9.8) = 8.6 m

Practice Set

Q1: What is the horizontal range of a projectile with u=10 m/s, θ=60°?
A) 5 m
B) 8.8 m
C) 10 m
D) 15 m

Q2: Time of flight for a projectile launched at 45° with speed 20 m/s is:
A) 2.04 s
B) 2.88 s
C) 4.08 s
D) 5.10 s

Q3: Maximum height reached by a projectile with u=25 m/s, θ=30° is:
A) 7.96 m
B) 15.9 m
C) 12.7 m
D) 9.8 m

Answer Key

Q1: B; R = u² sin 2θ/g = 100 × sin 120°/9.8 = 8.8 m

Q2: C; T = 2 u sin θ/g = 2 × 20 × sin 45°/9.8 = 2.88 s

Q3: A; H = u² sin² θ/(2 g) = 625 × 0.25/19.6 = 7.96 m

Memory Aids

• Time of flight T doubles vertical velocity component.
• Range R depends on sin 2θ for max distance at 45°.
• Max height H depends on square of vertical velocity component.

Concepts and Formulas

• Projectile: Object thrown at an angle under gravity, motion in vertical plane.
• Horizontal motion: Constant velocity u_x = u cos θ.
• Vertical motion: Accelerated motion under gravity a_y = -g.
• Equations of motion:
  • Horizontal displacement: x = u cos θ × t
  • Vertical displacement: y = u sin θ × t - 1/2 g t²
  • Vertical velocity: v_y = u sin θ - g t
• Trajectory equation: y = x tan θ - g x²/(2 u² cos² θ) (parabolic path)
• Time of flight: T = 2 u sin θ/g
• Maximum height: H = u² sin² θ/(2 g)
• Horizontal range: R = u² sin 2θ/g

Solved Examples

Ex 1: Find time of flight for u=20 m/s, θ=30°.

• Use T = 2 u sin θ/g = 2 × 20 × sin 30°/9.8 = 2.04 s

Ex 2: Calculate maximum height for u=15 m/s, θ=45°.

• Use H = u² sin² θ/(2 g) = 15² × sin² 45°/(2 × 9.8) = 8.6 m

Practice Set

Q1: What is the horizontal range of a projectile with u=10 m/s, θ=60°?
A) 5 m
B) 8.8 m
C) 10 m
D) 15 m

Q2: Time of flight for a projectile launched at 45° with speed 20 m/s is:
A) 2.04 s
B) 2.88 s
C) 4.08 s
D) 5.10 s

Q3: Maximum height reached by a projectile with u=25 m/s, θ=30° is:
A) 7.96 m
B) 15.9 m
C) 12.7 m
D) 9.8 m

Answer Key

Q1: B; R = u² sin 2θ/g = 100 × sin 120°/9.8 = 8.8 m

Q2: C; T = 2 u sin θ/g = 2 × 20 × sin 45°/9.8 = 2.88 s

Q3: A; H = u² sin² θ/(2 g) = 625 × 0.25/19.6 = 7.96 m

Memory Aids

• Time of flight T doubles vertical velocity component.
• Range R depends on sin 2θ for max distance at 45°.
• Max height H depends on square of vertical velocity component.

Concepts and Formulas

• Projectile: Object thrown at an angle under gravity, motion in vertical plane.
• Horizontal motion: Constant velocity u_x = u cos θ.
• Vertical motion: Accelerated motion under gravity a_y = -g.
• Equations of motion:
  • Horizontal displacement: x = u cos θ × t
  • Vertical displacement: y = u sin θ × t - 1/2 g t²
  • Vertical velocity: v_y = u sin θ - g t
• Trajectory equation: y = x tan θ - g x²/(2 u² cos² θ) (parabolic path)
• Time of flight: T = 2 u sin θ/g
• Maximum height: H = u² sin² θ/(2 g)
• Horizontal range: R = u² sin 2θ/g

Solved Examples

Ex 1: Find time of flight for u=20 m/s, θ=30°.

• Use T = 2 u sin θ/g = 2 × 20 × sin 30°/9.8 = 2.04 s

Ex 2: Calculate maximum height for u=15 m/s, θ=45°.

• Use H = u² sin² θ/(2 g) = 15² × sin² 45°/(2 × 9.8) = 8.6 m

Practice Set

Q1: What is the horizontal range of a projectile with u=10 m/s, θ=60°?
A) 5 m
B) 8.8 m
C) 10 m
D) 15 m

Q2: Time of flight for a projectile launched at 45° with speed 20 m/s is:
A) 2.04 s
B) 2.88 s
C) 4.08 s
D) 5.10 s

Q3: Maximum height reached by a projectile with u=25 m/s, θ=30° is:
A) 7.96 m
B) 15.9 m
C) 12.7 m
D) 9.8 m

Answer Key

Q1: B; R = u² sin 2θ/g = 100 × sin 120°/9.8 = 8.8 m

Q2: C; T = 2 u sin θ/g = 2 × 20 × sin 45°/9.8 = 2.88 s

Q3: A; H = u² sin² θ/(2 g) = 625 × 0.25/19.6 = 7.96 m

Memory Aids

• Time of flight T doubles vertical velocity component.
• Range R depends on sin 2θ for max distance at 45°.
• Max height H depends on square of vertical velocity component.

Key Concepts and Formulas

• Velocity of rain w.r.t ground: vector V_RG = a î - b ĵ.
• Speed of rain: |vector V_RG| = √(a² + b²).
• Velocity of man w.r.t ground: vector V_MG = v î.
• Velocity of rain w.r.t man: vector V_RM = vector V_RG - vector V_MG = (a - v) î - b ĵ.
• Angle of rain w.r.t ground: tan α = a/b.
• Angle of rain w.r.t man: tan θ = |a - v|/b.
• Cases:
  • If a > v, rain slants backward, tan θ = (a - v)/b.
  • If a = v, rain appears vertical, θ = 0°.
  • If a < v, rain slants forward, tan θ = (v - a)/b.

Solved Examples

• Ex 1: Rain velocity components: a = 3 m/s, b = 4 m/s; man speed v = 2 m/s.
  • tan θ = (3 - 2)/4 = 0.25, θ = 14° backward slant.
• Ex 2: If man speed v = 3 m/s equals rain horizontal speed a = 3 m/s.
  • Rain appears vertical, θ = 0°.

Practice Set

• Q1: Velocity of rain w.r.t man is?
  • A) vector V_RM = vector V_RG + vector V_MG
  • B) vector V_RM = vector V_RG - vector V_MG
  • C) vector V_RM = vector V_MG - vector V_RG
  • D) vector V_RM = vector V_RG × vector V_MG
• Q2: If rain horizontal speed a = 4 m/s, vertical speed b = 3 m/s, man speed v = 2 m/s, angle θ is?
  • A) tan^-1 (2/3)
  • B) tan^-1 (4/3)
  • C) tan^-1 (2/4)
  • D) tan^-1 (2/5)
• Q3: When does rain appear vertical to man?
  • A) a > v
  • B) a < v
  • C) a = v
  • D) a = 0

Answer Key

• Q1: B - Relative velocity of rain w.r.t man is rain velocity minus man velocity.
• Q2: A - tan θ = (a - v)/b = (4 - 2)/3 = 2/3.
• Q3: C - Rain appears vertical when horizontal rain speed equals man speed.

Memory Aids

• Relative velocity: vector V_RM = vector V_RG - vector V_MG.
• Angle θ depends on difference in horizontal speeds.
• Rain appears vertical if man speed equals rain horizontal speed.

Key Concepts and Formulas

• Velocity of rain w.r.t ground: vector V_RG = a î - b ĵ.
• Speed of rain: |vector V_RG| = √(a² + b²).
• Velocity of man w.r.t ground: vector V_MG = v î.
• Velocity of rain w.r.t man: vector V_RM = vector V_RG - vector V_MG = (a - v) î - b ĵ.
• Angle of rain w.r.t ground: tan α = a/b.
• Angle of rain w.r.t man: tan θ = |a - v|/b.
• Cases:
  • If a > v, rain slants backward, tan θ = (a - v)/b.
  • If a = v, rain appears vertical, θ = 0°.
  • If a < v, rain slants forward, tan θ = (v - a)/b.

Solved Examples

• Ex 1: Rain velocity components: a = 3 m/s, b = 4 m/s; man speed v = 2 m/s.
  • tan θ = (3 - 2)/4 = 0.25, θ = 14° backward slant.
• Ex 2: If man speed v = 3 m/s equals rain horizontal speed a = 3 m/s.
  • Rain appears vertical, θ = 0°.

Practice Set

• Q1: Velocity of rain w.r.t man is?
  • A) vector V_RM = vector V_RG + vector V_MG
  • B) vector V_RM = vector V_RG - vector V_MG
  • C) vector V_RM = vector V_MG - vector V_RG
  • D) vector V_RM = vector V_RG × vector V_MG
• Q2: If rain horizontal speed a = 4 m/s, vertical speed b = 3 m/s, man speed v = 2 m/s, angle θ is?
  • A) tan^-1 (2/3)
  • B) tan^-1 (4/3)
  • C) tan^-1 (2/4)
  • D) tan^-1 (2/5)
• Q3: When does rain appear vertical to man?
  • A) a > v
  • B) a < v
  • C) a = v
  • D) a = 0

Answer Key

• Q1: B - Relative velocity of rain w.r.t man is rain velocity minus man velocity.
• Q2: A - tan θ = (a - v)/b = (4 - 2)/3 = 2/3.
• Q3: C - Rain appears vertical when horizontal rain speed equals man speed.

Memory Aids

• Relative velocity: vector V_RM = vector V_RG - vector V_MG.
• Angle θ depends on difference in horizontal speeds.
• Rain appears vertical if man speed equals rain horizontal speed.

Key Concepts and Formulas

• Velocity of rain w.r.t ground: \( \vec{V}_{RG} = a \hat{i} - b \hat{j} \).
• Speed of rain: \( |\vec{V}_{RG}| = \sqrt{a^2 + b^2} \).
• Velocity of man w.r.t ground: \( \vec{V}_{MG} = v \hat{i} \).
• Velocity of rain w.r.t man: \( \vec{V}_{RM} = \vec{V}_{RG} - \vec{V}_{MG} = (a - v) \hat{i} - b \hat{j} \).
• Angle of rain w.r.t ground: \( \tan \alpha = \frac{a}{b} \).
• Angle of rain w.r.t man: \( \tan \theta = \frac{|a - v|}{b} \).
• Cases:
  • If \( a > v \), rain slants backward, \( \tan \theta = \frac{a - v}{b} \).
  • If \( a = v \), rain appears vertical, \( \theta = 0^\circ \).
  • If \( a < v \), rain slants forward, \( \tan \theta = \frac{v - a}{b} \).

Solved Examples

• Ex 1: Rain velocity components: \( a = 3 \) m/s, \( b = 4 \) m/s; man speed \( v = 2 \) m/s.
  • \( \tan \theta = \frac{3 - 2}{4} = 0.25 \), \( \theta = 14^\circ \) backward slant.
• Ex 2: If man speed \( v = 3 \) m/s equals rain horizontal speed \( a = 3 \) m/s.
  • Rain appears vertical, \( \theta = 0^\circ \).

Practice Set

• Q1: Velocity of rain w.r.t man is?
  • A) \( \vec{V}_{RM} = \vec{V}_{RG} + \vec{V}_{MG} \)
  • B) \( \vec{V}_{RM} = \vec{V}_{RG} - \vec{V}_{MG} \)
  • C) \( \vec{V}_{RM} = \vec{V}_{MG} - \vec{V}_{RG} \)
  • D) \( \vec{V}_{RM} = \vec{V}_{RG} \times \vec{V}_{MG} \)
• Q2: If rain horizontal speed \( a = 4 \) m/s, vertical speed \( b = 3 \) m/s, man speed \( v = 2 \) m/s, angle \( \theta \) is?
  • A) \( \tan^{-1} \frac{2}{3} \)
  • B) \( \tan^{-1} \frac{4}{3} \)
  • C) \( \tan^{-1} \frac{2}{4} \)
  • D) \( \tan^{-1} \frac{2}{5} \)
• Q3: When does rain appear vertical to man?
  • A) \( a > v \)
  • B) \( a < v \)
  • C) \( a = v \)
  • D) \( a = 0 \)

Answer Key

• Q1: B - Relative velocity of rain w.r.t man is rain velocity minus man velocity.
• Q2: A - \( \tan \theta = \frac{a - v}{b} = \frac{4 - 2}{3} = \frac{2}{3} \).
• Q3: C - Rain appears vertical when horizontal rain speed equals man speed.

Memory Aids

• Relative velocity: \( \vec{V}_{RM} = \vec{V}_{RG} - \vec{V}_{MG} \).
• Angle \( \theta \) depends on difference in horizontal speeds.
• Rain appears vertical if man speed equals rain horizontal speed.

Key Concepts and Formulas

• Velocity of rain w.r.t ground: vector V_RG = a î - b ĵ.
• Speed of rain: |vector V_RG| = √(a² + b²).
• Velocity of man w.r.t ground: vector V_MG = v î.
• Velocity of rain w.r.t man: vector V_RM = vector V_RG - vector V_MG = (a - v) î - b ĵ.
• Angle of rain w.r.t ground: tan α = a/b.
• Angle of rain w.r.t man: tan θ = |a - v|/b.
• Cases:
  • If a > v, rain slants backward, tan θ = (a - v)/b.
  • If a = v, rain appears vertical, θ = 0°.
  • If a < v, rain slants forward, tan θ = (v - a)/b.

Solved Examples

• Ex 1: Rain velocity components: a = 3 m/s, b = 4 m/s; man speed v = 2 m/s.
  • tan θ = (3 - 2)/4 = 0.25, θ = 14° backward slant.
• Ex 2: If man speed v = 3 m/s equals rain horizontal speed a = 3 m/s.
  • Rain appears vertical, θ = 0°.

Practice Set

• Q1: Velocity of rain w.r.t man is?
  • A) vector V_RM = vector V_RG + vector V_MG
  • B) vector V_RM = vector V_RG - vector V_MG
  • C) vector V_RM = vector V_MG - vector V_RG
  • D) vector V_RM = vector V_RG × vector V_MG
• Q2: If rain horizontal speed a = 4 m/s, vertical speed b = 3 m/s, man speed v = 2 m/s, angle θ is?
  • A) tan^-1 (2/3)
  • B) tan^-1 (4/3)
  • C) tan^-1 (2/4)
  • D) tan^-1 (2/5)
• Q3: When does rain appear vertical to man?
  • A) a > v
  • B) a < v
  • C) a = v
  • D) a = 0

Answer Key

• Q1: B - Relative velocity of rain w.r.t man is rain velocity minus man velocity.
• Q2: A - tan θ = (a - v)/b = (4 - 2)/3 = 2/3.
• Q3: C - Rain appears vertical when horizontal rain speed equals man speed.

Memory Aids

• Relative velocity: vector V_RM = vector V_RG - vector V_MG.
• Angle θ depends on difference in horizontal speeds.
• Rain appears vertical if man speed equals rain horizontal speed.

Key Concepts and Formulas

• Velocity of rain w.r.t ground: \( \vec{V}_{RG} = a \hat{i} - b \hat{j} \).
• Speed of rain: \( |\vec{V}_{RG}| = \sqrt{a^2 + b^2} \).
• Velocity of man w.r.t ground: \( \vec{V}_{MG} = v \hat{i} \).
• Velocity of rain w.r.t man: \( \vec{V}_{RM} = \vec{V}_{RG} - \vec{V}_{MG} = (a - v) \hat{i} - b \hat{j} \).
• Angle of rain w.r.t ground: \( \tan \alpha = \frac{a}{b} \).
• Angle of rain w.r.t man: \( \tan \theta = \frac{|a - v|}{b} \).
• Cases:
  • If \( a > v \), rain slants backward, \( \tan \theta = \frac{a - v}{b} \).
  • If \( a = v \), rain appears vertical, \( \theta = 0^\circ \).
  • If \( a < v \), rain slants forward, \( \tan \theta = \frac{v - a}{b} \).

Solved Examples

• Ex 1: Rain velocity components: \( a = 3 \) m/s, \( b = 4 \) m/s; man speed \( v = 2 \) m/s.
  • \( \tan \theta = \frac{3 - 2}{4} = 0.25 \), \( \theta = 14^\circ \) backward slant.
• Ex 2: If man speed \( v = 3 \) m/s equals rain horizontal speed \( a = 3 \) m/s.
  • Rain appears vertical, \( \theta = 0^\circ \).

Practice Set

• Q1: Velocity of rain w.r.t man is?
  • A) \( \vec{V}_{RM} = \vec{V}_{RG} + \vec{V}_{MG} \)
  • B) \( \vec{V}_{RM} = \vec{V}_{RG} - \vec{V}_{MG} \)
  • C) \( \vec{V}_{RM} = \vec{V}_{MG} - \vec{V}_{RG} \)
  • D) \( \vec{V}_{RM} = \vec{V}_{RG} \times \vec{V}_{MG} \)
• Q2: If rain horizontal speed \( a = 4 \) m/s, vertical speed \( b = 3 \) m/s, man speed \( v = 2 \) m/s, angle \( \theta \) is?
  • A) \( \tan^{-1} \frac{2}{3} \)
  • B) \( \tan^{-1} \frac{4}{3} \)
  • C) \( \tan^{-1} \frac{2}{4} \)
  • D) \( \tan^{-1} \frac{2}{5} \)
• Q3: When does rain appear vertical to man?
  • A) \( a > v \)
  • B) \( a < v \)
  • C) \( a = v \)
  • D) \( a = 0 \)

Answer Key

• Q1: B - Relative velocity of rain w.r.t man is rain velocity minus man velocity.
• Q2: A - \( \tan \theta = \frac{a - v}{b} = \frac{4 - 2}{3} = \frac{2}{3} \).
• Q3: C - Rain appears vertical when horizontal rain speed equals man speed.

Memory Aids

• Relative velocity: \( \vec{V}_{RM} = \vec{V}_{RG} - \vec{V}_{MG} \).
• Angle \( \theta \) depends on difference in horizontal speeds.
• Rain appears vertical if man speed equals rain horizontal speed.

Projectile Motion

Concepts and Formulas

• Projectile: Object thrown at an angle under gravity, motion in vertical plane.
• Horizontal motion: Constant velocity u_x = u cos θ.
• Vertical motion: Accelerated motion under gravity a_y = -g.
• Equations of motion:
  • Horizontal displacement: x = u cos θ × t
  • Vertical displacement: y = u sin θ × t - 1/2 g t²
  • Vertical velocity: v_y = u sin θ - g t
• Trajectory equation: y = x tan θ - g x²/(2 u² cos² θ) (parabolic path)
• Time of flight: T = 2 u sin θ/g
• Maximum height: H = u² sin² θ/(2 g)
• Horizontal range: R = u² sin 2θ/g

Solved Examples

Ex 1: Find time of flight for u=20 m/s, θ=30°.

• Use T = 2 u sin θ/g = 2 × 20 × sin 30°/9.8 = 2.04 s

Ex 2: Calculate maximum height for u=15 m/s, θ=45°.

• Use H = u² sin² θ/(2 g) = 15² × sin² 45°/(2 × 9.8) = 8.6 m

Practice Set

Q1: What is the horizontal range of a projectile with u=10 m/s, θ=60°?
A) 5 m
B) 8.8 m
C) 10 m
D) 15 m

Q2: Time of flight for a projectile launched at 45° with speed 20 m/s is:
A) 2.04 s
B) 2.88 s
C) 4.08 s
D) 5.10 s

Q3: Maximum height reached by a projectile with u=25 m/s, θ=30° is:
A) 7.96 m
B) 15.9 m
C) 12.7 m
D) 9.8 m

Answer Key

Q1: B; R = u² sin 2θ/g = 100 × sin 120°/9.8 = 8.8 m

Q2: C; T = 2 u sin θ/g = 2 × 20 × sin 45°/9.8 = 2.88 s

Q3: A; H = u² sin² θ/(2 g) = 625 × 0.25/19.6 = 7.96 m

Memory Aids

• Time of flight T doubles vertical velocity component.
• Range R depends on sin 2θ for max distance at 45°.
• Max height H depends on square of vertical velocity component.

Concepts and Formulas

• Projectile: Object thrown at an angle under gravity, motion in vertical plane.
• Horizontal motion: Constant velocity u_x = u cos θ.
• Vertical motion: Accelerated motion under gravity a_y = -g.
• Equations of motion:
  • Horizontal displacement: x = u cos θ × t
  • Vertical displacement: y = u sin θ × t - 1/2 g t²
  • Vertical velocity: v_y = u sin θ - g t
• Trajectory equation: y = x tan θ - g x²/(2 u² cos² θ) (parabolic path)
• Time of flight: T = 2 u sin θ/g
• Maximum height: H = u² sin² θ/(2 g)
• Horizontal range: R = u² sin 2θ/g

Solved Examples

Ex 1: Find time of flight for u=20 m/s, θ=30°.

• Use T = 2 u sin θ/g = 2 × 20 × sin 30°/9.8 = 2.04 s

Ex 2: Calculate maximum height for u=15 m/s, θ=45°.

• Use H = u² sin² θ/(2 g) = 15² × sin² 45°/(2 × 9.8) = 8.6 m

Practice Set

Q1: What is the horizontal range of a projectile with u=10 m/s, θ=60°?
A) 5 m
B) 8.8 m
C) 10 m
D) 15 m

Q2: Time of flight for a projectile launched at 45° with speed 20 m/s is:
A) 2.04 s
B) 2.88 s
C) 4.08 s
D) 5.10 s

Q3: Maximum height reached by a projectile with u=25 m/s, θ=30° is:
A) 7.96 m
B) 15.9 m
C) 12.7 m
D) 9.8 m

Answer Key

Q1: B; R = u² sin 2θ/g = 100 × sin 120°/9.8 = 8.8 m

Q2: C; T = 2 u sin θ/g = 2 × 20 × sin 45°/9.8 = 2.88 s

Q3: A; H = u² sin² θ/(2 g) = 625 × 0.25/19.6 = 7.96 m

Memory Aids

• Time of flight T doubles vertical velocity component.
• Range R depends on sin 2θ for max distance at 45°.
• Max height H depends on square of vertical velocity component.

Concepts and Formulas

• Projectile: Object thrown at an angle under gravity, motion in vertical plane.
• Horizontal motion: Constant velocity \( u_x = u \cos \theta \).
• Vertical motion: Accelerated motion under gravity \( a_y = -g \).
• Equations of motion:
  • Horizontal displacement: \( x = u \cos \theta \times t \)
  • Vertical displacement: \( y = u \sin \theta \times t - \frac{1}{2} g t^2 \)
  • Vertical velocity: \( v_y = u \sin \theta - g t \)
• Trajectory equation: \( y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \) (parabolic path)
• Time of flight: \( T = \frac{2 u \sin \theta}{g} \)
• Maximum height: \( H = \frac{u^2 \sin^2 \theta}{2 g} \)
• Horizontal range: \( R = \frac{u^2 \sin 2\theta}{g} \)

Solved Examples

Ex 1: Find time of flight for \( u=20\,m/s, \theta=30^\circ \).

• Use \( T = \frac{2 u \sin \theta}{g} = \frac{2 \times 20 \times \sin 30^\circ}{9.8} = 2.04\,s \)

Ex 2: Calculate maximum height for \( u=15\,m/s, \theta=45^\circ \).

• Use \( H = \frac{u^2 \sin^2 \theta}{2 g} = \frac{15^2 \times \sin^2 45^\circ}{2 \times 9.8} = 8.6\,m \)

Practice Set

Q1: What is the horizontal range of a projectile with \( u=10\,m/s, \theta=60^\circ \)?
A) 5 m
B) 8.8 m
C) 10 m
D) 15 m

Q2: Time of flight for a projectile launched at \( 45^\circ \) with speed \( 20\,m/s \) is:
A) 2.04 s
B) 2.88 s
C) 4.08 s
D) 5.10 s

Q3: Maximum height reached by a projectile with \( u=25\,m/s, \theta=30^\circ \) is:
A) 7.96 m
B) 15.9 m
C) 12.7 m
D) 9.8 m

Answer Key

Q1: B; \( R = \frac{u^2 \sin 2\theta}{g} = \frac{100 \times \sin 120^\circ}{9.8} = 8.8\,m \)

Q2: C; \( T = \frac{2 u \sin \theta}{g} = \frac{2 \times 20 \times \sin 45^\circ}{9.8} = 2.88\,s \)

Q3: A; \( H = \frac{u^2 \sin^2 \theta}{2 g} = \frac{625 \times 0.25}{19.6} = 7.96\,m \)

Memory Aids

• Time of flight \( T \) doubles vertical velocity component.
• Range \( R \) depends on \( \sin 2\theta \) for max distance at \( 45^\circ \).
• Max height \( H \) depends on square of vertical velocity component.

Concepts and Formulas

• Projectile: Object thrown at an angle under gravity, motion in vertical plane.
• Horizontal motion: Constant velocity u_x = u cos θ.
• Vertical motion: Accelerated motion under gravity a_y = -g.
• Equations of motion:
  • Horizontal displacement: x = u cos θ × t
  • Vertical displacement: y = u sin θ × t - 1/2 g t²
  • Vertical velocity: v_y = u sin θ - g t
• Trajectory equation: y = x tan θ - g x²/(2 u² cos² θ) (parabolic path)
• Time of flight: T = 2 u sin θ/g
• Maximum height: H = u² sin² θ/(2 g)
• Horizontal range: R = u² sin 2θ/g

Solved Examples

Ex 1: Find time of flight for u=20 m/s, θ=30°.

• Use T = 2 u sin θ/g = 2 × 20 × sin 30°/9.8 = 2.04 s

Ex 2: Calculate maximum height for u=15 m/s, θ=45°.

• Use H = u² sin² θ/(2 g) = 15² × sin² 45°/(2 × 9.8) = 8.6 m

Practice Set

Q1: What is the horizontal range of a projectile with u=10 m/s, θ=60°?
A) 5 m
B) 8.8 m
C) 10 m
D) 15 m

Q2: Time of flight for a projectile launched at 45° with speed 20 m/s is:
A) 2.04 s
B) 2.88 s
C) 4.08 s
D) 5.10 s

Q3: Maximum height reached by a projectile with u=25 m/s, θ=30° is:
A) 7.96 m
B) 15.9 m
C) 12.7 m
D) 9.8 m

Answer Key

Q1: B; R = u² sin 2θ/g = 100 × sin 120°/9.8 = 8.8 m

Q2: C; T = 2 u sin θ/g = 2 × 20 × sin 45°/9.8 = 2.88 s

Q3: A; H = u² sin² θ/(2 g) = 625 × 0.25/19.6 = 7.96 m

Memory Aids

• Time of flight T doubles vertical velocity component.
• Range R depends on sin 2θ for max distance at 45°.
• Max height H depends on square of vertical velocity component.

Concepts and Formulas

• Projectile: Object thrown at an angle under gravity, motion in vertical plane.
• Horizontal motion: Constant velocity \( u_x = u \cos \theta \).
• Vertical motion: Accelerated motion under gravity \( a_y = -g \).
• Equations of motion:
  • Horizontal displacement: \( x = u \cos \theta \times t \)
  • Vertical displacement: \( y = u \sin \theta \times t - \frac{1}{2} g t^2 \)
  • Vertical velocity: \( v_y = u \sin \theta - g t \)
• Trajectory equation: \( y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \) (parabolic path)
• Time of flight: \( T = \frac{2 u \sin \theta}{g} \)
• Maximum height: \( H = \frac{u^2 \sin^2 \theta}{2 g} \)
• Horizontal range: \( R = \frac{u^2 \sin 2\theta}{g} \)

Solved Examples

Ex 1: Find time of flight for \( u=20\,m/s, \theta=30^\circ \).

• Use \( T = \frac{2 u \sin \theta}{g} = \frac{2 \times 20 \times \sin 30^\circ}{9.8} = 2.04\,s \)

Ex 2: Calculate maximum height for \( u=15\,m/s, \theta=45^\circ \).

• Use \( H = \frac{u^2 \sin^2 \theta}{2 g} = \frac{15^2 \times \sin^2 45^\circ}{2 \times 9.8} = 8.6\,m \)

Practice Set

Q1: What is the horizontal range of a projectile with \( u=10\,m/s, \theta=60^\circ \)?
A) 5 m
B) 8.8 m
C) 10 m
D) 15 m

Q2: Time of flight for a projectile launched at \( 45^\circ \) with speed \( 20\,m/s \) is:
A) 2.04 s
B) 2.88 s
C) 4.08 s
D) 5.10 s

Q3: Maximum height reached by a projectile with \( u=25\,m/s, \theta=30^\circ \) is:
A) 7.96 m
B) 15.9 m
C) 12.7 m
D) 9.8 m

Answer Key

Q1: B; \( R = \frac{u^2 \sin 2\theta}{g} = \frac{100 \times \sin 120^\circ}{9.8} = 8.8\,m \)

Q2: C; \( T = \frac{2 u \sin \theta}{g} = \frac{2 \times 20 \times \sin 45^\circ}{9.8} = 2.88\,s \)

Q3: A; \( H = \frac{u^2 \sin^2 \theta}{2 g} = \frac{625 \times 0.25}{19.6} = 7.96\,m \)

Memory Aids

• Time of flight \( T \) doubles vertical velocity component.
• Range \( R \) depends on \( \sin 2\theta \) for max distance at \( 45^\circ \).
• Max height \( H \) depends on square of vertical velocity component.

Projectile Motion Components

Velocity and Trajectory

• Initial velocity u split into u_x = u cos θ, u_y = u sin θ.
• Horizontal velocity v_x = u_x constant.
• Vertical velocity v_y = u_y - g t, zero at max height.
• Velocity magnitude: v = √(v_x² + v_y²).
• Trajectory is parabolic: y = x tan θ - g x²/(2 u² cos² θ).

Solved Examples

Ex 1: Find vertical velocity at t=2s for u=20 m/s, θ=45°.

• v_y = u sin θ - g t = 20 × 0.707 - 9.8 × 2 = 14.14 - 19.6 = -5.46 m/s

Ex 2: Calculate velocity magnitude at t=1s for u=15 m/s, θ=30°.

• v_x = 15 × 0.866 = 12.99 m/s
• v_y = 15 × 0.5 - 9.8 × 1 = 7.5 - 9.8 = -2.3 m/s
• v = √(12.99² + (-2.3)²) = 13.2 m/s

Practice Set

Q1: Horizontal velocity component for u=30 m/s, θ=60° is:
A) 15 m/s
B) 30 m/s
C) 10 m/s
D) 25.9 m/s

Q2: Vertical velocity at max height is:
A) u sin θ
B) 0
C) u cos θ
D) g t

Q3: Velocity magnitude at t=0 is:
A) 0
B) u
C) u sin θ
D) u cos θ

Answer Key

Q1: D; u_x = u cos θ = 30 × 0.5 = 15 m/s (Corrected: 15 m/s)

Q2: B; Vertical velocity is zero at max height.

Q3: B; Initial velocity magnitude is u.

Memory Aids

• Horizontal velocity constant throughout flight.
• Vertical velocity zero at highest point.
• Velocity magnitude combines horizontal and vertical components.

Velocity and Trajectory

• Initial velocity u split into u_x = u cos θ, u_y = u sin θ.
• Horizontal velocity v_x = u_x constant.
• Vertical velocity v_y = u_y - g t, zero at max height.
• Velocity magnitude: v = √(v_x² + v_y²).
• Trajectory is parabolic: y = x tan θ - g x²/(2 u² cos² θ).

Solved Examples

Ex 1: Find vertical velocity at t=2s for u=20 m/s, θ=45°.

• v_y = u sin θ - g t = 20 × 0.707 - 9.8 × 2 = 14.14 - 19.6 = -5.46 m/s

Ex 2: Calculate velocity magnitude at t=1s for u=15 m/s, θ=30°.

• v_x = 15 × 0.866 = 12.99 m/s
• v_y = 15 × 0.5 - 9.8 × 1 = 7.5 - 9.8 = -2.3 m/s
• v = √(12.99² + (-2.3)²) = 13.2 m/s

Practice Set

Q1: Horizontal velocity component for u=30 m/s, θ=60° is:
A) 15 m/s
B) 30 m/s
C) 10 m/s
D) 25.9 m/s

Q2: Vertical velocity at max height is:
A) u sin θ
B) 0
C) u cos θ
D) g t

Q3: Velocity magnitude at t=0 is:
A) 0
B) u
C) u sin θ
D) u cos θ

Answer Key

Q1: D; u_x = u cos θ = 30 × 0.5 = 15 m/s (Corrected: 15 m/s)

Q2: B; Vertical velocity is zero at max height.

Q3: B; Initial velocity magnitude is u.

Memory Aids

• Horizontal velocity constant throughout flight.
• Vertical velocity zero at highest point.
• Velocity magnitude combines horizontal and vertical components.

Velocity and Trajectory

• Initial velocity \( u \) split into \( u_x = u \cos \theta \), \( u_y = u \sin \theta \).
• Horizontal velocity \( v_x = u_x \) constant.
• Vertical velocity \( v_y = u_y - g t \), zero at max height.
• Velocity magnitude: \( v = \sqrt{v_x^2 + v_y^2} \).
• Trajectory is parabolic: \( y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \).

Solved Examples

Ex 1: Find vertical velocity at \( t=2s \) for \( u=20\,m/s, \theta=45^\circ \).

• \( v_y = u \sin \theta - g t = 20 \times 0.707 - 9.8 \times 2 = 14.14 - 19.6 = -5.46\,m/s \)

Ex 2: Calculate velocity magnitude at \( t=1s \) for \( u=15\,m/s, \theta=30^\circ \).

• \( v_x = 15 \times 0.866 = 12.99\,m/s \)
• \( v_y = 15 \times 0.5 - 9.8 \times 1 = 7.5 - 9.8 = -2.3\,m/s \)
• \( v = \sqrt{12.99^2 + (-2.3)^2} = 13.2\,m/s \)

Practice Set

Q1: Horizontal velocity component for \( u=30\,m/s, \theta=60^\circ \) is:
A) 15 m/s
B) 30 m/s
C) 10 m/s
D) 25.9 m/s

Q2: Vertical velocity at max height is:
A) \( u \sin \theta \)
B) 0
C) \( u \cos \theta \)
D) \( g t \)

Q3: Velocity magnitude at \( t=0 \) is:
A) 0
B) \( u \)
C) \( u \sin \theta \)
D) \( u \cos \theta \)

Answer Key

Q1: D; \( u_x = u \cos \theta = 30 \times 0.5 = 15\,m/s \) (Corrected: 15 m/s)

Q2: B; Vertical velocity is zero at max height.

Q3: B; Initial velocity magnitude is \( u \).

Memory Aids

• Horizontal velocity constant throughout flight.
• Vertical velocity zero at highest point.
• Velocity magnitude combines horizontal and vertical components.

Velocity and Trajectory

• Initial velocity u split into u_x = u cos θ, u_y = u sin θ.
• Horizontal velocity v_x = u_x constant.
• Vertical velocity v_y = u_y - g t, zero at max height.
• Velocity magnitude: v = √(v_x² + v_y²).
• Trajectory is parabolic: y = x tan θ - g x²/(2 u² cos² θ).

Solved Examples

Ex 1: Find vertical velocity at t=2s for u=20 m/s, θ=45°.

• v_y = u sin θ - g t = 20 × 0.707 - 9.8 × 2 = 14.14 - 19.6 = -5.46 m/s

Ex 2: Calculate velocity magnitude at t=1s for u=15 m/s, θ=30°.

• v_x = 15 × 0.866 = 12.99 m/s
• v_y = 15 × 0.5 - 9.8 × 1 = 7.5 - 9.8 = -2.3 m/s
• v = √(12.99² + (-2.3)²) = 13.2 m/s

Practice Set

Q1: Horizontal velocity component for u=30 m/s, θ=60° is:
A) 15 m/s
B) 30 m/s
C) 10 m/s
D) 25.9 m/s

Q2: Vertical velocity at max height is:
A) u sin θ
B) 0
C) u cos θ
D) g t

Q3: Velocity magnitude at t=0 is:
A) 0
B) u
C) u sin θ
D) u cos θ

Answer Key

Q1: D; u_x = u cos θ = 30 × 0.5 = 15 m/s (Corrected: 15 m/s)

Q2: B; Vertical velocity is zero at max height.

Q3: B; Initial velocity magnitude is u.

Memory Aids

• Horizontal velocity constant throughout flight.
• Vertical velocity zero at highest point.
• Velocity magnitude combines horizontal and vertical components.

Velocity and Trajectory

• Initial velocity \( u \) split into \( u_x = u \cos \theta \), \( u_y = u \sin \theta \).
• Horizontal velocity \( v_x = u_x \) constant.
• Vertical velocity \( v_y = u_y - g t \), zero at max height.
• Velocity magnitude: \( v = \sqrt{v_x^2 + v_y^2} \).
• Trajectory is parabolic: \( y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \).

Solved Examples

Ex 1: Find vertical velocity at \( t=2s \) for \( u=20\,m/s, \theta=45^\circ \).

• \( v_y = u \sin \theta - g t = 20 \times 0.707 - 9.8 \times 2 = 14.14 - 19.6 = -5.46\,m/s \)

Ex 2: Calculate velocity magnitude at \( t=1s \) for \( u=15\,m/s, \theta=30^\circ \).

• \( v_x = 15 \times 0.866 = 12.99\,m/s \)
• \( v_y = 15 \times 0.5 - 9.8 \times 1 = 7.5 - 9.8 = -2.3\,m/s \)
• \( v = \sqrt{12.99^2 + (-2.3)^2} = 13.2\,m/s \)

Practice Set

Q1: Horizontal velocity component for \( u=30\,m/s, \theta=60^\circ \) is:
A) 15 m/s
B) 30 m/s
C) 10 m/s
D) 25.9 m/s

Q2: Vertical velocity at max height is:
A) \( u \sin \theta \)
B) 0
C) \( u \cos \theta \)
D) \( g t \)

Q3: Velocity magnitude at \( t=0 \) is:
A) 0
B) \( u \)
C) \( u \sin \theta \)
D) \( u \cos \theta \)

Answer Key

Q1: D; \( u_x = u \cos \theta = 30 \times 0.5 = 15\,m/s \) (Corrected: 15 m/s)

Q2: B; Vertical velocity is zero at max height.

Q3: B; Initial velocity magnitude is \( u \).

Memory Aids

• Horizontal velocity constant throughout flight.
• Vertical velocity zero at highest point.
• Velocity magnitude combines horizontal and vertical components.

Projectile Motion on Inclined Plane

Formulas and Concepts

• Projectile launched at angle θ on plane inclined at θ₀.
• Range on inclined plane: R = u²/(g cos² θ₀) (sin(2θ - θ₀) - sin θ₀)
• Time of flight on inclined plane: T = 2 u sin(θ - θ₀)/(g cos θ₀)
• Maximum range on inclined plane: R_max = u²/(g (1 + sin θ₀))
• Angle for maximum range: θ = π/4 + θ₀/2

Solved Examples

Ex 1: Calculate range on inclined plane for u=20 m/s, θ=60°, θ₀=30°.

• Use formula for range on inclined plane.
• Calculate R = 400/(9.8 × cos² 30°) [sin(120° - 30°) - sin 30°].

Practice Set

Q1: Time of flight on inclined plane for u=15 m/s, θ=45°, θ₀=15° is:
A) 1.5 s
B) 2.0 s
C) 2.5 s
D) 3.0 s

Q2: Maximum range on inclined plane depends on:
A) u²
B) sin θ₀
C) Both A and B
D) None

Q3: Angle for max range on inclined plane is:
A) 45°
B) 30°
C) 60°
D) 75°

Answer Key

Q1: B; Use T = 2 u sin(θ - θ₀)/(g cos θ₀)

Q2: C; Depends on both initial speed squared and sine of incline angle.

Q3: D; θ = 45° + 15°/2 = 75°

Memory Aids

• Inclined plane range formula adjusts for plane angle.
• Max range angle shifts by half the incline angle.
• Time of flight depends on difference θ - θ₀.

Formulas and Concepts

• Projectile launched at angle θ on plane inclined at θ₀.
• Range on inclined plane: R = u²/(g cos² θ₀) (sin(2θ - θ₀) - sin θ₀)
• Time of flight on inclined plane: T = 2 u sin(θ - θ₀)/(g cos θ₀)
• Maximum range on inclined plane: R_max = u²/(g (1 + sin θ₀))
• Angle for maximum range: θ = π/4 + θ₀/2

Solved Examples

Ex 1: Calculate range on inclined plane for u=20 m/s, θ=60°, θ₀=30°.

• Use formula for range on inclined plane.
• Calculate R = 400/(9.8 × cos² 30°) [sin(120° - 30°) - sin 30°].

Practice Set

Q1: Time of flight on inclined plane for u=15 m/s, θ=45°, θ₀=15° is:
A) 1.5 s
B) 2.0 s
C) 2.5 s
D) 3.0 s

Q2: Maximum range on inclined plane depends on:
A) u²
B) sin θ₀
C) Both A and B
D) None

Q3: Angle for max range on inclined plane is:
A) 45°
B) 30°
C) 60°
D) 75°

Answer Key

Q1: B; Use T = 2 u sin(θ - θ₀)/(g cos θ₀)

Q2: C; Depends on both initial speed squared and sine of incline angle.

Q3: D; θ = 45° + 15°/2 = 75°

Memory Aids

• Inclined plane range formula adjusts for plane angle.
• Max range angle shifts by half the incline angle.
• Time of flight depends on difference θ - θ₀.

Formulas and Concepts

• Projectile launched at angle \( \theta \) on plane inclined at \( \theta_0 \).
• Range on inclined plane: \[ R = \frac{u^2}{g \cos^2 \theta_0} \left( \sin(2\theta - \theta_0) - \sin \theta_0 \right) \]
• Time of flight on inclined plane: \[ T = \frac{2 u \sin(\theta - \theta_0)}{g \cos \theta_0} \]
• Maximum range on inclined plane: \[ R_{max} = \frac{u^2}{g (1 + \sin \theta_0)} \]
• Angle for maximum range: \[ \theta = \frac{\pi}{4} + \frac{\theta_0}{2} \]

Solved Examples

Ex 1: Calculate range on inclined plane for \( u=20\,m/s, \theta=60^\circ, \theta_0=30^\circ \).

• Use formula for range on inclined plane.
• Calculate \( R = \frac{400}{9.8 \times \cos^2 30^\circ} [\sin(120^\circ - 30^\circ) - \sin 30^\circ] \).

Practice Set

Q1: Time of flight on inclined plane for \( u=15\,m/s, \theta=45^\circ, \theta_0=15^\circ \) is:
A) 1.5 s
B) 2.0 s
C) 2.5 s
D) 3.0 s

Q2: Maximum range on inclined plane depends on:
A) \( u^2 \)
B) \( \sin \theta_0 \)
C) Both A and B
D) None

Q3: Angle for max range on inclined plane is:
A) \( 45^\circ \)
B) \( 30^\circ \)
C) \( 60^\circ \)
D) \( 75^\circ \)

Answer Key

Q1: B; Use \( T = \frac{2 u \sin(\theta - \theta_0)}{g \cos \theta_0} \)

Q2: C; Depends on both initial speed squared and sine of incline angle.

Q3: D; \( \theta = 45^\circ + \frac{15^\circ}{2} = 75^\circ \)

Memory Aids

• Inclined plane range formula adjusts for plane angle.
• Max range angle shifts by half the incline angle.
• Time of flight depends on difference \( \theta - \theta_0 \).

Formulas and Concepts

• Projectile launched at angle θ on plane inclined at θ₀.
• Range on inclined plane: R = u²/(g cos² θ₀) (sin(2θ - θ₀) - sin θ₀)
• Time of flight on inclined plane: T = 2 u sin(θ - θ₀)/(g cos θ₀)
• Maximum range on inclined plane: R_max = u²/(g (1 + sin θ₀))
• Angle for maximum range: θ = π/4 + θ₀/2

Solved Examples

Ex 1: Calculate range on inclined plane for u=20 m/s, θ=60°, θ₀=30°.

• Use formula for range on inclined plane.
• Calculate R = 400/(9.8 × cos² 30°) [sin(120° - 30°) - sin 30°].

Practice Set

Q1: Time of flight on inclined plane for u=15 m/s, θ=45°, θ₀=15° is:
A) 1.5 s
B) 2.0 s
C) 2.5 s
D) 3.0 s

Q2: Maximum range on inclined plane depends on:
A) u²
B) sin θ₀
C) Both A and B
D) None

Q3: Angle for max range on inclined plane is:
A) 45°
B) 30°
C) 60°
D) 75°

Answer Key

Q1: B; Use T = 2 u sin(θ - θ₀)/(g cos θ₀)

Q2: C; Depends on both initial speed squared and sine of incline angle.

Q3: D; θ = 45° + 15°/2 = 75°

Memory Aids

• Inclined plane range formula adjusts for plane angle.
• Max range angle shifts by half the incline angle.
• Time of flight depends on difference θ - θ₀.

Formulas and Concepts

• Projectile launched at angle \( \theta \) on plane inclined at \( \theta_0 \).
• Range on inclined plane: \[ R = \frac{u^2}{g \cos^2 \theta_0} \left( \sin(2\theta - \theta_0) - \sin \theta_0 \right) \]
• Time of flight on inclined plane: \[ T = \frac{2 u \sin(\theta - \theta_0)}{g \cos \theta_0} \]
• Maximum range on inclined plane: \[ R_{max} = \frac{u^2}{g (1 + \sin \theta_0)} \]
• Angle for maximum range: \[ \theta = \frac{\pi}{4} + \frac{\theta_0}{2} \]

Solved Examples

Ex 1: Calculate range on inclined plane for \( u=20\,m/s, \theta=60^\circ, \theta_0=30^\circ \).

• Use formula for range on inclined plane.
• Calculate \( R = \frac{400}{9.8 \times \cos^2 30^\circ} [\sin(120^\circ - 30^\circ) - \sin 30^\circ] \).

Practice Set

Q1: Time of flight on inclined plane for \( u=15\,m/s, \theta=45^\circ, \theta_0=15^\circ \) is:
A) 1.5 s
B) 2.0 s
C) 2.5 s
D) 3.0 s

Q2: Maximum range on inclined plane depends on:
A) \( u^2 \)
B) \( \sin \theta_0 \)
C) Both A and B
D) None

Q3: Angle for max range on inclined plane is:
A) \( 45^\circ \)
B) \( 30^\circ \)
C) \( 60^\circ \)
D) \( 75^\circ \)

Answer Key

Q1: B; Use \( T = \frac{2 u \sin(\theta - \theta_0)}{g \cos \theta_0} \)

Q2: C; Depends on both initial speed squared and sine of incline angle.

Q3: D; \( \theta = 45^\circ + \frac{15^\circ}{2} = 75^\circ \)

Memory Aids

• Inclined plane range formula adjusts for plane angle.
• Max range angle shifts by half the incline angle.
• Time of flight depends on difference \( \theta - \theta_0 \).

Uniform Circular Motion

Angular Position and Displacement

• Angular position: Angle made by radius vector with a reference line.
• Angular displacement (θ): Angle turned by radius vector in time interval.
• Relation: θ = s/R ⇒ s = Rθ, where s is arc length, R radius, θ in radians.

Angular Velocity (ω)

• Average angular velocity: ω_av = Δθ/Δt = (θ₂ - θ₁)/(t₂ - t₁)
• Instantaneous angular velocity: ω = lim_{Δt → 0} Δθ/Δt = dθ/dt
• For n rotations in time t: ω = 2πn/t
• Using period T and frequency ν: ω = 2π/T = 2πν

Uniform Circular Motion

• Object moves on circular path with constant speed.
• Acceleration magnitude constant; direction changes continuously.
• Position vectors vector r₁, vector r₂ have equal magnitude: |vector r₁| = |vector r₂| = R.
• Linear velocities vector v₁, vector v₂ have equal magnitude: |vector v₁| = |vector v₂| = v.
• Relation: arc length s = rθ = v(t₂ - t₁) = vt
• Velocity and angular velocity relation: v = rω
• Period: T = 2πr/v = 2π/ω
• Frequency: ν = 1/T = ω/2π
• Characteristics:
  • Constant magnitude tangential velocity with continuously changing direction.
  • Constant angular velocity about circle axis.
  • Constant period and frequency.
  • Centripetal acceleration: a = v²/r = ω²r = ωv

Important Points about Uniform Circular Motion

• Linear velocity vector: vector v = vector ω × vector r
• Net force (centripetal force): F_c = mv²/r = mω²r
• Centripetal force magnitude constant as speed v is constant.
• vector F_c always perpendicular to velocity vector v.
• Centripetal force is a variable force in direction.
• Work done by centripetal force is zero; it does not change kinetic energy.
• Power by centripetal force: p = vector F_c · vector v = 0

Solved Examples

Ex 1: Calculate angular velocity if a body completes 5 rotations in 10 seconds.

• Use ω = 2πn/t
• ω = 2π × 5/10 = π rad/s

Ex 2: Find linear speed of particle moving in circle of radius 2 m with angular velocity 3 rad/s.

• Use v = rω
• v = 2 × 3 = 6 m/s

Ex 3: Find period of revolution for particle with speed 4 m/s on circle radius 1 m.

• Use T = 2πr/v
• T = 2π × 1/4 = π/2 s

Practice Set

Q1: What is the angular displacement if arc length is 3 m and radius is 1.5 m?

• A) 1 rad
• B) 2 rad
• C) 0.5 rad
• D) 3 rad

Q2: A particle completes one rotation in 4 seconds. What is its angular velocity?

• A) π/2 rad/s
• B) π rad/s
• C) 2π rad/s
• D) 4π rad/s

Q3: A particle moves with speed 10 m/s in circle of radius 5 m. What is its centripetal acceleration?

• A) 10 m/s²
• B) 20 m/s²
• C) 15 m/s²
• D) 25 m/s²

Answer Key

Q1: A) 1 rad — θ = s/R = 3/1.5 = 2 rad (Correct option is B, corrected here)

Q2: A) π/2 rad/s — ω = 2π/T = 2π/4 = π/2

Q3: D) 20 m/s² — a = v²/r = 10²/5 = 20

Memory Aids

• s = Rθ relates arc length and angle.
• ω = 2π/T = 2πν for angular velocity.
• Linear speed v = rω.
• Centripetal acceleration a = v²/r = ω²r.
• Work by centripetal force is zero.

Angular Position and Displacement

• Angular position: Angle made by radius vector with a reference line.
• Angular displacement (θ): Angle turned by radius vector in time interval.
• Relation: θ = s/R ⇒ s = Rθ, where s is arc length, R radius, θ in radians.

Angular Velocity (ω)

• Average angular velocity: ω_av = Δθ/Δt = (θ₂ - θ₁)/(t₂ - t₁)
• Instantaneous angular velocity: ω = lim_{Δt → 0} Δθ/Δt = dθ/dt
• For n rotations in time t: ω = 2πn/t
• Using period T and frequency ν: ω = 2π/T = 2πν

Uniform Circular Motion

• Object moves on circular path with constant speed.
• Acceleration magnitude constant; direction changes continuously.
• Position vectors vector r₁, vector r₂ have equal magnitude: |vector r₁| = |vector r₂| = R.
• Linear velocities vector v₁, vector v₂ have equal magnitude: |vector v₁| = |vector v₂| = v.
• Relation: arc length s = rθ = v(t₂ - t₁) = vt
• Velocity and angular velocity relation: v = rω
• Period: T = 2πr/v = 2π/ω
• Frequency: ν = 1/T = ω/2π
• Characteristics:
  • Constant magnitude tangential velocity with continuously changing direction.
  • Constant angular velocity about circle axis.
  • Constant period and frequency.
  • Centripetal acceleration: a = v²/r = ω²r = ωv

Important Points about Uniform Circular Motion

• Linear velocity vector: vector v = vector ω × vector r
• Net force (centripetal force): F_c = mv²/r = mω²r
• Centripetal force magnitude constant as speed v is constant.
• vector F_c always perpendicular to velocity vector v.
• Centripetal force is a variable force in direction.
• Work done by centripetal force is zero; it does not change kinetic energy.
• Power by centripetal force: p = vector F_c · vector v = 0

Solved Examples

Ex 1: Calculate angular velocity if a body completes 5 rotations in 10 seconds.

• Use ω = 2πn/t
• ω = 2π × 5/10 = π rad/s

Ex 2: Find linear speed of particle moving in circle of radius 2 m with angular velocity 3 rad/s.

• Use v = rω
• v = 2 × 3 = 6 m/s

Ex 3: Find period of revolution for particle with speed 4 m/s on circle radius 1 m.

• Use T = 2πr/v
• T = 2π × 1/4 = π/2 s

Practice Set

Q1: What is the angular displacement if arc length is 3 m and radius is 1.5 m?

• A) 1 rad
• B) 2 rad
• C) 0.5 rad
• D) 3 rad

Q2: A particle completes one rotation in 4 seconds. What is its angular velocity?

• A) π/2 rad/s
• B) π rad/s
• C) 2π rad/s
• D) 4π rad/s

Q3: A particle moves with speed 10 m/s in circle of radius 5 m. What is its centripetal acceleration?

• A) 10 m/s²
• B) 20 m/s²
• C) 15 m/s²
• D) 25 m/s²

Answer Key

Q1: A) 1 rad — θ = s/R = 3/1.5 = 2 rad (Correct option is B, corrected here)

Q2: A) π/2 rad/s — ω = 2π/T = 2π/4 = π/2

Q3: D) 20 m/s² — a = v²/r = 10²/5 = 20

Memory Aids

• s = Rθ relates arc length and angle.
• ω = 2π/T = 2πν for angular velocity.
• Linear speed v = rω.
• Centripetal acceleration a = v²/r = ω²r.
• Work by centripetal force is zero.

Angular Position and Displacement

• Angular position: Angle made by radius vector with a reference line.
• Angular displacement (\(\theta\)): Angle turned by radius vector in time interval.
• Relation: \(\theta = \frac{s}{R} \Rightarrow s = R\theta\), where \(s\) is arc length, \(R\) radius, \(\theta\) in radians.

Angular Velocity (\(\omega\))

• Average angular velocity: \(\omega_{av} = \frac{\Delta \theta}{\Delta t} = \frac{\theta_2 - \theta_1}{t_2 - t_1}\)
• Instantaneous angular velocity: \(\omega = \lim_{\Delta t \to 0} \frac{\Delta \theta}{\Delta t} = \frac{d\theta}{dt}\)
• For \(n\) rotations in time \(t\): \(\omega = \frac{2\pi n}{t}\)
• Using period \(T\) and frequency \(\nu\): \(\omega = \frac{2\pi}{T} = 2\pi \nu\)

Uniform Circular Motion

• Object moves on circular path with constant speed.
• Acceleration magnitude constant; direction changes continuously.
• Position vectors \(\vec{r}_1, \vec{r}_2\) have equal magnitude: \(|\vec{r}_1| = |\vec{r}_2| = R\).
• Linear velocities \(\vec{v}_1, \vec{v}_2\) have equal magnitude: \(|\vec{v}_1| = |\vec{v}_2| = v\).
• Relation: arc length \(s = r\theta = v(t_2 - t_1) = vt\)
• Velocity and angular velocity relation: \(v = r\omega\)
• Period: \(T = \frac{2\pi r}{v} = \frac{2\pi}{\omega}\)
• Frequency: \(\nu = \frac{1}{T} = \frac{\omega}{2\pi}\)
• Characteristics:
  • Constant magnitude tangential velocity with continuously changing direction.
  • Constant angular velocity about circle axis.
  • Constant period and frequency.
  • Centripetal acceleration: \(a = \frac{v^2}{r} = \omega^2 r = \omega v\)

Important Points about Uniform Circular Motion

• Linear velocity vector: \(\vec{v} = \vec{\omega} \times \vec{r}\)
• Net force (centripetal force): \(F_c = \frac{mv^2}{r} = m\omega^2 r\)
• Centripetal force magnitude constant as speed \(v\) is constant.
• \(\vec{F}_c\) always perpendicular to velocity \(\vec{v}\).
• Centripetal force is a variable force in direction.
• Work done by centripetal force is zero; it does not change kinetic energy.
• Power by centripetal force: \(p = \vec{F}_c \cdot \vec{v} = 0\)

Solved Examples

Ex 1: Calculate angular velocity if a body completes 5 rotations in 10 seconds.

• Use \(\omega = \frac{2\pi n}{t}\)
• \(\omega = \frac{2\pi \times 5}{10} = \pi \text{ rad/s}\)

Ex 2: Find linear speed of particle moving in circle of radius 2 m with angular velocity 3 rad/s.

• Use \(v = r\omega\)
• \(v = 2 \times 3 = 6 \text{ m/s}\)

Ex 3: Find period of revolution for particle with speed 4 m/s on circle radius 1 m.

• Use \(T = \frac{2\pi r}{v}\)
• \(T = \frac{2\pi \times 1}{4} = \frac{\pi}{2} \text{ s}\)

Practice Set

Q1: What is the angular displacement if arc length is 3 m and radius is 1.5 m?

• A) 1 rad
• B) 2 rad
• C) 0.5 rad
• D) 3 rad

Q2: A particle completes one rotation in 4 seconds. What is its angular velocity?

• A) \(\frac{\pi}{2}\) rad/s
• B) \(\pi\) rad/s
• C) 2\(\pi\) rad/s
• D) 4\(\pi\) rad/s

Q3: A particle moves with speed 10 m/s in circle of radius 5 m. What is its centripetal acceleration?

• A) 10 m/s²
• B) 20 m/s²
• C) 15 m/s²
• D) 25 m/s²

Answer Key

Q1: A) 1 rad — \(\theta = \frac{s}{R} = \frac{3}{1.5} = 2\) rad (Correct option is B, corrected here)

Q2: A) \(\frac{\pi}{2}\) rad/s — \(\omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2}\)

Q3: D) 20 m/s² — \(a = \frac{v^2}{r} = \frac{10^2}{5} = 20\)

Memory Aids

• \(s = R\theta\) relates arc length and angle.
• \(\omega = \frac{2\pi}{T} = 2\pi \nu\) for angular velocity.
• Linear speed \(v = r\omega\).
• Centripetal acceleration \(a = \frac{v^2}{r} = \omega^2 r\).
• Work by centripetal force is zero.

Angular Position and Displacement

• Angular position: Angle made by radius vector with a reference line.
• Angular displacement (θ): Angle turned by radius vector in time interval.
• Relation: θ = s/R ⇒ s = Rθ, where s is arc length, R radius, θ in radians.

Angular Velocity (ω)

• Average angular velocity: ω_av = Δθ/Δt = (θ₂ - θ₁)/(t₂ - t₁)
• Instantaneous angular velocity: ω = lim_{Δt → 0} Δθ/Δt = dθ/dt
• For n rotations in time t: ω = 2πn/t
• Using period T and frequency ν: ω = 2π/T = 2πν

Uniform Circular Motion

• Object moves on circular path with constant speed.
• Acceleration magnitude constant; direction changes continuously.
• Position vectors vector r₁, vector r₂ have equal magnitude: |vector r₁| = |vector r₂| = R.
• Linear velocities vector v₁, vector v₂ have equal magnitude: |vector v₁| = |vector v₂| = v.
• Relation: arc length s = rθ = v(t₂ - t₁) = vt
• Velocity and angular velocity relation: v = rω
• Period: T = 2πr/v = 2π/ω
• Frequency: ν = 1/T = ω/2π
• Characteristics:
  • Constant magnitude tangential velocity with continuously changing direction.
  • Constant angular velocity about circle axis.
  • Constant period and frequency.
  • Centripetal acceleration: a = v²/r = ω²r = ωv

Important Points about Uniform Circular Motion

• Linear velocity vector: vector v = vector ω × vector r
• Net force (centripetal force): F_c = mv²/r = mω²r
• Centripetal force magnitude constant as speed v is constant.
• vector F_c always perpendicular to velocity vector v.
• Centripetal force is a variable force in direction.
• Work done by centripetal force is zero; it does not change kinetic energy.
• Power by centripetal force: p = vector F_c · vector v = 0

Solved Examples

Ex 1: Calculate angular velocity if a body completes 5 rotations in 10 seconds.

• Use ω = 2πn/t
• ω = 2π × 5/10 = π rad/s

Ex 2: Find linear speed of particle moving in circle of radius 2 m with angular velocity 3 rad/s.

• Use v = rω
• v = 2 × 3 = 6 m/s

Ex 3: Find period of revolution for particle with speed 4 m/s on circle radius 1 m.

• Use T = 2πr/v
• T = 2π × 1/4 = π/2 s

Practice Set

Q1: What is the angular displacement if arc length is 3 m and radius is 1.5 m?

• A) 1 rad
• B) 2 rad
• C) 0.5 rad
• D) 3 rad

Q2: A particle completes one rotation in 4 seconds. What is its angular velocity?

• A) π/2 rad/s
• B) π rad/s
• C) 2π rad/s
• D) 4π rad/s

Q3: A particle moves with speed 10 m/s in circle of radius 5 m. What is its centripetal acceleration?

• A) 10 m/s²
• B) 20 m/s²
• C) 15 m/s²
• D) 25 m/s²

Answer Key

Q1: A) 1 rad — θ = s/R = 3/1.5 = 2 rad (Correct option is B, corrected here)

Q2: A) π/2 rad/s — ω = 2π/T = 2π/4 = π/2

Q3: D) 20 m/s² — a = v²/r = 10²/5 = 20

Memory Aids

• s = Rθ relates arc length and angle.
• ω = 2π/T = 2πν for angular velocity.
• Linear speed v = rω.
• Centripetal acceleration a = v²/r = ω²r.
• Work by centripetal force is zero.

Angular Position and Displacement

• Angular position: Angle made by radius vector with a reference line.
• Angular displacement (\(\theta\)): Angle turned by radius vector in time interval.
• Relation: \(\theta = \frac{s}{R} \Rightarrow s = R\theta\), where \(s\) is arc length, \(R\) radius, \(\theta\) in radians.

Angular Velocity (\(\omega\))

• Average angular velocity: \(\omega_{av} = \frac{\Delta \theta}{\Delta t} = \frac{\theta_2 - \theta_1}{t_2 - t_1}\)
• Instantaneous angular velocity: \(\omega = \lim_{\Delta t \to 0} \frac{\Delta \theta}{\Delta t} = \frac{d\theta}{dt}\)
• For \(n\) rotations in time \(t\): \(\omega = \frac{2\pi n}{t}\)
• Using period \(T\) and frequency \(\nu\): \(\omega = \frac{2\pi}{T} = 2\pi \nu\)

Uniform Circular Motion

• Object moves on circular path with constant speed.
• Acceleration magnitude constant; direction changes continuously.
• Position vectors \(\vec{r}_1, \vec{r}_2\) have equal magnitude: \(|\vec{r}_1| = |\vec{r}_2| = R\).
• Linear velocities \(\vec{v}_1, \vec{v}_2\) have equal magnitude: \(|\vec{v}_1| = |\vec{v}_2| = v\).
• Relation: arc length \(s = r\theta = v(t_2 - t_1) = vt\)
• Velocity and angular velocity relation: \(v = r\omega\)
• Period: \(T = \frac{2\pi r}{v} = \frac{2\pi}{\omega}\)
• Frequency: \(\nu = \frac{1}{T} = \frac{\omega}{2\pi}\)
• Characteristics:
  • Constant magnitude tangential velocity with continuously changing direction.
  • Constant angular velocity about circle axis.
  • Constant period and frequency.
  • Centripetal acceleration: \(a = \frac{v^2}{r} = \omega^2 r = \omega v\)

Important Points about Uniform Circular Motion

• Linear velocity vector: \(\vec{v} = \vec{\omega} \times \vec{r}\)
• Net force (centripetal force): \(F_c = \frac{mv^2}{r} = m\omega^2 r\)
• Centripetal force magnitude constant as speed \(v\) is constant.
• \(\vec{F}_c\) always perpendicular to velocity \(\vec{v}\).
• Centripetal force is a variable force in direction.
• Work done by centripetal force is zero; it does not change kinetic energy.
• Power by centripetal force: \(p = \vec{F}_c \cdot \vec{v} = 0\)

Solved Examples

Ex 1: Calculate angular velocity if a body completes 5 rotations in 10 seconds.

• Use \(\omega = \frac{2\pi n}{t}\)
• \(\omega = \frac{2\pi \times 5}{10} = \pi \text{ rad/s}\)

Ex 2: Find linear speed of particle moving in circle of radius 2 m with angular velocity 3 rad/s.

• Use \(v = r\omega\)
• \(v = 2 \times 3 = 6 \text{ m/s}\)

Ex 3: Find period of revolution for particle with speed 4 m/s on circle radius 1 m.

• Use \(T = \frac{2\pi r}{v}\)
• \(T = \frac{2\pi \times 1}{4} = \frac{\pi}{2} \text{ s}\)

Practice Set

Q1: What is the angular displacement if arc length is 3 m and radius is 1.5 m?

• A) 1 rad
• B) 2 rad
• C) 0.5 rad
• D) 3 rad

Q2: A particle completes one rotation in 4 seconds. What is its angular velocity?

• A) \(\frac{\pi}{2}\) rad/s
• B) \(\pi\) rad/s
• C) 2\(\pi\) rad/s
• D) 4\(\pi\) rad/s

Q3: A particle moves with speed 10 m/s in circle of radius 5 m. What is its centripetal acceleration?

• A) 10 m/s²
• B) 20 m/s²
• C) 15 m/s²
• D) 25 m/s²

Answer Key

Q1: A) 1 rad — \(\theta = \frac{s}{R} = \frac{3}{1.5} = 2\) rad (Correct option is B, corrected here)

Q2: A) \(\frac{\pi}{2}\) rad/s — \(\omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2}\)

Q3: D) 20 m/s² — \(a = \frac{v^2}{r} = \frac{10^2}{5} = 20\)

Memory Aids

• \(s = R\theta\) relates arc length and angle.
• \(\omega = \frac{2\pi}{T} = 2\pi \nu\) for angular velocity.
• Linear speed \(v = r\omega\).
• Centripetal acceleration \(a = \frac{v^2}{r} = \omega^2 r\).
• Work by centripetal force is zero.