Chapter 3: Motion in a Straight Line

Introduction

Overview of Rectilinear Motion

• Motion along a straight line is called rectilinear motion.
• Objects can be approximated as point objects if their size is negligible compared to the distance moved.
• Kinematics studies motion description without considering causes.

Solved Examples

• Ex 1: An object moves from point A to B through C. Find the path length if AC = 5 m and CB = 3 m.
  Solution: Path length = AC + CB = 5 + 3 = 8 m.

Practice Set

• Q1: What is rectilinear motion?
  A) Motion in a circle
  B) Motion along a straight line
  C) Motion in a plane
  D) Random motion
• Q2: When can an object be treated as a point object?
  A) When its size is larger than distance moved
  B) When its size is negligible compared to distance moved
  C) Always
  D) Never
• Q3: Kinematics studies
  A) Causes of motion
  B) Effects of forces
  C) Description of motion without causes
  D) Energy changes

Answer Key

• Q1: B - Rectilinear motion is motion along a straight line.
• Q2: B - Point object approximation is valid when size is much smaller than distance moved.
• Q3: C - Kinematics describes motion without considering causes.

Memory Aids

• Rectilinear = straight line motion.
• Point object = size negligible vs distance.
• Kinematics = motion description only.

Position, Path Length and Displacement

Key Concepts

• Position: Location of an object relative to a reference point and axes.
• Motion: Change in one or more coordinates with time; otherwise, object is at rest.
• Path length (distance): Total length of actual path traveled; scalar quantity, always positive or zero.
• Displacement: Shortest straight-line distance from initial to final position; vector quantity with magnitude and direction.
• Displacement characteristics:
  • Units of length (meters).
  • Can be positive, negative, or zero.
  • Independent of origin choice.
  • Magnitude ≤ actual distance traveled.
  • Does not indicate path shape.
  • Shortest distance between two points.

Solved Examples

• Ex 1: Calculate displacement if object moves from A to B directly, distance AB = 6 m.
  Solution: Displacement = AB = 6 m (vector).
• Ex 2: Find path length if object moves A to C = 4 m, C to B = 3 m.
  Solution: Path length = AC + CB = 4 + 3 = 7 m.

Practice Set

• Q1: Displacement is a
  A) Scalar quantity
  B) Vector quantity
  C) Always positive
  D) None of these
• Q2: Path length is
  A) Shortest distance
  B) Total distance traveled
  C) Vector quantity
  D) Negative quantity
• Q3: Which is true about displacement?
  A) Depends on origin
  B) Can be negative
  C) Always greater than distance
  D) Indicates path shape

Answer Key

• Q1: B - Displacement has magnitude and direction, so vector.
• Q2: B - Path length is total distance traveled, scalar.
• Q3: B - Displacement can be positive, negative, or zero and is independent of origin.

Memory Aids

• Distance = scalar, path length actual traveled.
• Displacement = vector, shortest path.
• Displacement sign shows direction.

Average Velocity and Average Speed

Key Concepts

• Average velocity: Displacement divided by time interval.
  • Formula: v̄ = (x₂ - x₁)/(t₂ - t₁) = Δx/Δt
  • Graphically, slope of line joining initial and final positions on x-t graph.
  • Can be positive, negative, or zero.
• Average speed: Total path length divided by total time.
  • Formula: v_av = Total path length/Total time
  • For straight line motion, average speed = average velocity.
  • For varying speeds and distances:
    • v_av = (s₁ + s₂ + s₃ + ...)/(t₁ + t₂ + t₃ + ...)
    • or v_av = (s₁ + s₂ + s₃)/((s₁/v₁) + (s₂/v₂) + (s₃/v₃))
    • or v_av = (v₁t₁ + v₂t₂ + v₃t₃)/(t₁ + t₂ + t₃)

Solved Examples

• Ex 1: Calculate average velocity between t=5s and t=7s if x₁=10m, x₂=27.5m.
  Solution: v̄ = (27.5 - 10)/(7 - 5) = 8.75 m/s.

Practice Set

• Q1: Average velocity is
  A) Total distance / time
  B) Displacement / time
  C) Speed at an instant
  D) None
• Q2: Average speed for straight line motion equals
  A) Instantaneous speed
  B) Average velocity
  C) Zero
  D) Displacement
• Q3: If a body covers distances 10 m, 20 m with speeds 5 m/s, 10 m/s respectively, average speed is
  A) 7.5 m/s
  B) 8 m/s
  C) 6.67 m/s
  D) 15 m/s

Answer Key

• Q1: B - Average velocity = displacement/time.
• Q2: B - For straight line motion, average speed = average velocity.
• Q3: C - Using formula v_av = (s₁ + s₂)/((s₁/v₁) + (s₂/v₂)) = 30/((10/5) + (20/10)) = 6.67 m/s.

Memory Aids

• Average velocity = slope on x-t graph.
• Average speed = total distance / total time.
• Use weighted averages for varying speeds.

Key Concepts

• Average velocity: Displacement divided by time interval.
  • Formula: v̄ = (x₂ - x₁)/(t₂ - t₁) = Δx/Δt
  • Graphically, slope of line joining initial and final positions on x-t graph.
  • Can be positive, negative, or zero.
• Average speed: Total path length divided by total time.
  • Formula: v_av = Total path length/Total time
  • For straight line motion, average speed = average velocity.
  • For varying speeds and distances:
    • v_av = (s₁ + s₂ + s₃ + ...)/(t₁ + t₂ + t₃ + ...)
    • or v_av = (s₁ + s₂ + s₃)/((s₁/v₁) + (s₂/v₂) + (s₃/v₃))
    • or v_av = (v₁t₁ + v₂t₂ + v₃t₃)/(t₁ + t₂ + t₃)

Solved Examples

• Ex 1: Calculate average velocity between t=5s and t=7s if x₁=10m, x₂=27.5m.
  Solution: v̄ = (27.5 - 10)/(7 - 5) = 8.75 m/s.

Practice Set

• Q1: Average velocity is
  A) Total distance / time
  B) Displacement / time
  C) Speed at an instant
  D) None
• Q2: Average speed for straight line motion equals
  A) Instantaneous speed
  B) Average velocity
  C) Zero
  D) Displacement
• Q3: If a body covers distances 10 m, 20 m with speeds 5 m/s, 10 m/s respectively, average speed is
  A) 7.5 m/s
  B) 8 m/s
  C) 6.67 m/s
  D) 15 m/s

Answer Key

• Q1: B - Average velocity = displacement/time.
• Q2: B - For straight line motion, average speed = average velocity.
• Q3: C - Using formula v_av = (s₁ + s₂)/((s₁/v₁) + (s₂/v₂)) = 30/((10/5) + (20/10)) = 6.67 m/s.

Memory Aids

• Average velocity = slope on x-t graph.
• Average speed = total distance / total time.
• Use weighted averages for varying speeds.

Key Concepts

• Average velocity: Displacement divided by time interval.
  • Formula: v̄ = (x₂ - x₁)/(t₂ - t₁) = Δx/Δt
  • Graphically, slope of line joining initial and final positions on x-t graph.
  • Can be positive, negative, or zero.
• Average speed: Total path length divided by total time.
  • Formula: v_av = Total path length/Total time
  • For straight line motion, average speed = average velocity.
  • For varying speeds and distances:
    • v_av = (s₁ + s₂ + s₃ + ...)/(t₁ + t₂ + t₃ + ...)
    • or v_av = (s₁ + s₂ + s₃)/((s₁/v₁) + (s₂/v₂) + (s₃/v₃))
    • or v_av = (v₁t₁ + v₂t₂ + v₃t₃)/(t₁ + t₂ + t₃)

Solved Examples

• Ex 1: Calculate average velocity between t=5s and t=7s if x₁=10m, x₂=27.5m.
  Solution: v̄ = (27.5 - 10)/(7 - 5) = 8.75 m/s.

Practice Set

• Q1: Average velocity is
  A) Total distance / time
  B) Displacement / time
  C) Speed at an instant
  D) None
• Q2: Average speed for straight line motion equals
  A) Instantaneous speed
  B) Average velocity
  C) Zero
  D) Displacement
• Q3: If a body covers distances 10 m, 20 m with speeds 5 m/s, 10 m/s respectively, average speed is
  A) 7.5 m/s
  B) 8 m/s
  C) 6.67 m/s
  D) 15 m/s

Answer Key

• Q1: B - Average velocity = displacement/time.
• Q2: B - For straight line motion, average speed = average velocity.
• Q3: C - Using formula v_av = (s₁ + s₂)/((s₁/v₁) + (s₂/v₂)) = 30/((10/5) + (20/10)) = 6.67 m/s.

Memory Aids

• Average velocity = slope on x-t graph.
• Average speed = total distance / total time.
• Use weighted averages for varying speeds.

Key Concepts

• Average velocity: Displacement divided by time interval.
  • Formula: v̄ = (x₂ - x₁)/(t₂ - t₁) = Δx/Δt
  • Graphically, slope of line joining initial and final positions on x-t graph.
  • Can be positive, negative, or zero.
• Average speed: Total path length divided by total time.
  • Formula: v_av = Total path length/Total time
  • For straight line motion, average speed = average velocity.
  • For varying speeds and distances:
    • v_av = (s₁ + s₂ + s₃ + ...)/(t₁ + t₂ + t₃ + ...)
    • or v_av = (s₁ + s₂ + s₃)/((s₁/v₁) + (s₂/v₂) + (s₃/v₃))
    • or v_av = (v₁t₁ + v₂t₂ + v₃t₃)/(t₁ + t₂ + t₃)

Solved Examples

• Ex 1: Calculate average velocity between t=5s and t=7s if x₁=10m, x₂=27.5m.
  Solution: v̄ = (27.5 - 10)/(7 - 5) = 8.75 m/s.

Practice Set

• Q1: Average velocity is
  A) Total distance / time
  B) Displacement / time
  C) Speed at an instant
  D) None
• Q2: Average speed for straight line motion equals
  A) Instantaneous speed
  B) Average velocity
  C) Zero
  D) Displacement
• Q3: If a body covers distances 10 m, 20 m with speeds 5 m/s, 10 m/s respectively, average speed is
  A) 7.5 m/s
  B) 8 m/s
  C) 6.67 m/s
  D) 15 m/s

Answer Key

• Q1: B - Average velocity = displacement/time.
• Q2: B - For straight line motion, average speed = average velocity.
• Q3: C - Using formula v_av = (s₁ + s₂)/((s₁/v₁) + (s₂/v₂)) = 30/((10/5) + (20/10)) = 6.67 m/s.

Memory Aids

• Average velocity = slope on x-t graph.
• Average speed = total distance / total time.
• Use weighted averages for varying speeds.

Key Concepts

• Average velocity: Displacement divided by time interval.
  • Formula: v̄ = (x₂ - x₁)/(t₂ - t₁) = Δx/Δt
  • Graphically, slope of line joining initial and final positions on x-t graph.
  • Can be positive, negative, or zero.
• Average speed: Total path length divided by total time.
  • Formula: v_av = Total path length/Total time
  • For straight line motion, average speed = average velocity.
  • For varying speeds and distances:
    • v_av = (s₁ + s₂ + s₃ + ...)/(t₁ + t₂ + t₃ + ...)
    • or v_av = (s₁ + s₂ + s₃)/((s₁/v₁) + (s₂/v₂) + (s₃/v₃))
    • or v_av = (v₁t₁ + v₂t₂ + v₃t₃)/(t₁ + t₂ + t₃)

Solved Examples

• Ex 1: Calculate average velocity between t=5s and t=7s if x₁=10m, x₂=27.5m.
  Solution: v̄ = (27.5 - 10)/(7 - 5) = 8.75 m/s.

Practice Set

• Q1: Average velocity is
  A) Total distance / time
  B) Displacement / time
  C) Speed at an instant
  D) None
• Q2: Average speed for straight line motion equals
  A) Instantaneous speed
  B) Average velocity
  C) Zero
  D) Displacement
• Q3: If a body covers distances 10 m, 20 m with speeds 5 m/s, 10 m/s respectively, average speed is
  A) 7.5 m/s
  B) 8 m/s
  C) 6.67 m/s
  D) 15 m/s

Answer Key

• Q1: B - Average velocity = displacement/time.
• Q2: B - For straight line motion, average speed = average velocity.
• Q3: C - Using formula v_av = (s₁ + s₂)/((s₁/v₁) + (s₂/v₂)) = 30/((10/5) + (20/10)) = 6.67 m/s.

Memory Aids

• Average velocity = slope on x-t graph.
• Average speed = total distance / total time.
• Use weighted averages for varying speeds.

Key Concepts

• Average velocity: Displacement divided by time interval.
  • Formula: v̄ = (x₂ - x₁)/(t₂ - t₁) = Δx/Δt
  • Graphically, slope of line joining initial and final positions on x-t graph.
  • Can be positive, negative, or zero.
• Average speed: Total path length divided by total time.
  • Formula: v_av = Total path length/Total time
  • For straight line motion, average speed = average velocity.
  • For varying speeds and distances:
    • v_av = (s₁ + s₂ + s₃ + ...)/(t₁ + t₂ + t₃ + ...)
    • or v_av = (s₁ + s₂ + s₃)/((s₁/v₁) + (s₂/v₂) + (s₃/v₃))
    • or v_av = (v₁t₁ + v₂t₂ + v₃t₃)/(t₁ + t₂ + t₃)

Solved Examples

• Ex 1: Calculate average velocity between t=5s and t=7s if x₁=10m, x₂=27.5m.
  Solution: v̄ = (27.5 - 10)/(7 - 5) = 8.75 m/s.

Practice Set

• Q1: Average velocity is
  A) Total distance / time
  B) Displacement / time
  C) Speed at an instant
  D) None
• Q2: Average speed for straight line motion equals
  A) Instantaneous speed
  B) Average velocity
  C) Zero
  D) Displacement
• Q3: If a body covers distances 10 m, 20 m with speeds 5 m/s, 10 m/s respectively, average speed is
  A) 7.5 m/s
  B) 8 m/s
  C) 6.67 m/s
  D) 15 m/s

Answer Key

• Q1: B - Average velocity = displacement/time.
• Q2: B - For straight line motion, average speed = average velocity.
• Q3: C - Using formula v_av = (s₁ + s₂)/((s₁/v₁) + (s₂/v₂)) = 30/((10/5) + (20/10)) = 6.67 m/s.

Memory Aids

• Average velocity = slope on x-t graph.
• Average speed = total distance / total time.
• Use weighted averages for varying speeds.

Key Concepts

• Average velocity: Displacement divided by time interval.
  • Formula: v̄ = (x₂ - x₁)/(t₂ - t₁) = Δx/Δt
  • Graphically, slope of line joining initial and final positions on x-t graph.
  • Can be positive, negative, or zero.
• Average speed: Total path length divided by total time.
  • Formula: v_av = Total path length/Total time
  • For straight line motion, average speed = average velocity.
  • For varying speeds and distances:
    • v_av = (s₁ + s₂ + s₃ + ...)/(t₁ + t₂ + t₃ + ...)
    • or v_av = (s₁ + s₂ + s₃)/((s₁/v₁) + (s₂/v₂) + (s₃/v₃))
    • or v_av = (v₁t₁ + v₂t₂ + v₃t₃)/(t₁ + t₂ + t₃)

Solved Examples

• Ex 1: Calculate average velocity between t=5s and t=7s if x₁=10m, x₂=27.5m.
  Solution: v̄ = (27.5 - 10)/(7 - 5) = 8.75 m/s.

Practice Set

• Q1: Average velocity is
  A) Total distance / time
  B) Displacement / time
  C) Speed at an instant
  D) None
• Q2: Average speed for straight line motion equals
  A) Instantaneous speed
  B) Average velocity
  C) Zero
  D) Displacement
• Q3: If a body covers distances 10 m, 20 m with speeds 5 m/s, 10 m/s respectively, average speed is
  A) 7.5 m/s
  B) 8 m/s
  C) 6.67 m/s
  D) 15 m/s

Answer Key

• Q1: B - Average velocity = displacement/time.
• Q2: B - For straight line motion, average speed = average velocity.
• Q3: C - Using formula v_av = (s₁ + s₂)/((s₁/v₁) + (s₂/v₂)) = 30/((10/5) + (20/10)) = 6.67 m/s.

Memory Aids

• Average velocity = slope on x-t graph.
• Average speed = total distance / total time.
• Use weighted averages for varying speeds.

Key Concepts

• Average velocity: Displacement divided by time interval.
  • Formula: \( \bar{v} = \frac{x_2 - x_1}{t_2 - t_1} = \frac{\Delta x}{\Delta t} \)
  • Graphically, slope of line joining initial and final positions on x-t graph.
  • Can be positive, negative, or zero.
• Average speed: Total path length divided by total time.
  • Formula: \( v_{av} = \frac{\text{Total path length}}{\text{Total time}} \)
  • For straight line motion, average speed = average velocity.
  • For varying speeds and distances:
    • \( v_{av} = \frac{s_1 + s_2 + s_3 + \ldots}{t_1 + t_2 + t_3 + \ldots} \)
    • or \( v_{av} = \frac{s_1 + s_2 + s_3}{\frac{s_1}{v_1} + \frac{s_2}{v_2} + \frac{s_3}{v_3}} \)
    • or \( v_{av} = \frac{v_1 t_1 + v_2 t_2 + v_3 t_3}{t_1 + t_2 + t_3} \)

Solved Examples

• Ex 1: Calculate average velocity between \( t=5s \) and \( t=7s \) if \( x_1=10m \), \( x_2=27.5m \).
  Solution: \( \bar{v} = \frac{27.5 - 10}{7 - 5} = 8.75 \text{ m/s} \).

Practice Set

• Q1: Average velocity is
  A) Total distance / time
  B) Displacement / time
  C) Speed at an instant
  D) None
• Q2: Average speed for straight line motion equals
  A) Instantaneous speed
  B) Average velocity
  C) Zero
  D) Displacement
• Q3: If a body covers distances 10 m, 20 m with speeds 5 m/s, 10 m/s respectively, average speed is
  A) 7.5 m/s
  B) 8 m/s
  C) 6.67 m/s
  D) 15 m/s

Answer Key

• Q1: B - Average velocity = displacement/time.
• Q2: B - For straight line motion, average speed = average velocity.
• Q3: C - Using formula \( v_{av} = \frac{s_1 + s_2}{\frac{s_1}{v_1} + \frac{s_2}{v_2}} = \frac{30}{\frac{10}{5} + \frac{20}{10}} = 6.67 \text{ m/s} \).

Memory Aids

• Average velocity = slope on x-t graph.
• Average speed = total distance / total time.
• Use weighted averages for varying speeds.

Instantaneous Velocity and Speed

Key Concepts

• Instantaneous velocity: Rate of change of position at an instant.
  • Mathematically: v_inst = lim_{Δt → 0} Δx/Δt = dx/dt
  • Graphically: Slope of position-time curve at a point.
• Instantaneous speed: Magnitude of instantaneous velocity.
  • Mathematically: Instantaneous speed = |dx/dt|
  • Speed at a specific moment.
• For uniform motion, instantaneous velocity equals average velocity at all times.

Solved Examples

• Ex 1: Find instantaneous velocity if position x = 5t² m at t=2s.
  Solution: v = dx/dt = 10t ⇒ v(2) = 20 m/s.

Practice Set

• Q1: Instantaneous velocity is
  A) Average velocity over time
  B) Rate of change of position at an instant
  C) Total displacement
  D) Total distance
• Q2: Instantaneous speed is
  A) Vector quantity
  B) Magnitude of instantaneous velocity
  C) Negative quantity
  D) Zero
• Q3: If x = 3t³, instantaneous velocity at t=1s is
  A) 9 m/s
  B) 3 m/s
  C) 27 m/s
  D) 0 m/s

Answer Key

• Q1: B - Instantaneous velocity is rate of change of position at a moment.
• Q2: B - Instantaneous speed is magnitude of instantaneous velocity.
• Q3: A - v = dx/dt = 9t² ⇒ v(1) = 9 m/s.

Memory Aids

• Instantaneous velocity = derivative of position.
• Instantaneous speed = magnitude of velocity.
• Slope of x-t curve at a point.

Key Concepts

• Instantaneous velocity: Rate of change of position at an instant.
  • Mathematically: v_inst = lim_{Δt → 0} Δx/Δt = dx/dt
  • Graphically: Slope of position-time curve at a point.
• Instantaneous speed: Magnitude of instantaneous velocity.
  • Mathematically: Instantaneous speed = |dx/dt|
  • Speed at a specific moment.
• For uniform motion, instantaneous velocity equals average velocity at all times.

Solved Examples

• Ex 1: Find instantaneous velocity if position x = 5t² m at t=2s.
  Solution: v = dx/dt = 10t ⇒ v(2) = 20 m/s.

Practice Set

• Q1: Instantaneous velocity is
  A) Average velocity over time
  B) Rate of change of position at an instant
  C) Total displacement
  D) Total distance
• Q2: Instantaneous speed is
  A) Vector quantity
  B) Magnitude of instantaneous velocity
  C) Negative quantity
  D) Zero
• Q3: If x = 3t³, instantaneous velocity at t=1s is
  A) 9 m/s
  B) 3 m/s
  C) 27 m/s
  D) 0 m/s

Answer Key

• Q1: B - Instantaneous velocity is rate of change of position at a moment.
• Q2: B - Instantaneous speed is magnitude of instantaneous velocity.
• Q3: A - v = dx/dt = 9t² ⇒ v(1) = 9 m/s.

Memory Aids

• Instantaneous velocity = derivative of position.
• Instantaneous speed = magnitude of velocity.
• Slope of x-t curve at a point.

Key Concepts

• Instantaneous velocity: Rate of change of position at an instant.
  • Mathematically: v_inst = lim_{Δt → 0} Δx/Δt = dx/dt
  • Graphically: Slope of position-time curve at a point.
• Instantaneous speed: Magnitude of instantaneous velocity.
  • Mathematically: Instantaneous speed = |dx/dt|
  • Speed at a specific moment.
• For uniform motion, instantaneous velocity equals average velocity at all times.

Solved Examples

• Ex 1: Find instantaneous velocity if position x = 5t² m at t=2s.
  Solution: v = dx/dt = 10t ⇒ v(2) = 20 m/s.

Practice Set

• Q1: Instantaneous velocity is
  A) Average velocity over time
  B) Rate of change of position at an instant
  C) Total displacement
  D) Total distance
• Q2: Instantaneous speed is
  A) Vector quantity
  B) Magnitude of instantaneous velocity
  C) Negative quantity
  D) Zero
• Q3: If x = 3t³, instantaneous velocity at t=1s is
  A) 9 m/s
  B) 3 m/s
  C) 27 m/s
  D) 0 m/s

Answer Key

• Q1: B - Instantaneous velocity is rate of change of position at a moment.
• Q2: B - Instantaneous speed is magnitude of instantaneous velocity.
• Q3: A - v = dx/dt = 9t² ⇒ v(1) = 9 m/s.

Memory Aids

• Instantaneous velocity = derivative of position.
• Instantaneous speed = magnitude of velocity.
• Slope of x-t curve at a point.

Key Concepts

• Instantaneous velocity: Rate of change of position at an instant.
  • Mathematically: v_inst = lim_{Δt → 0} Δx/Δt = dx/dt
  • Graphically: Slope of position-time curve at a point.
• Instantaneous speed: Magnitude of instantaneous velocity.
  • Mathematically: Instantaneous speed = |dx/dt|
  • Speed at a specific moment.
• For uniform motion, instantaneous velocity equals average velocity at all times.

Solved Examples

• Ex 1: Find instantaneous velocity if position x = 5t² m at t=2s.
  Solution: v = dx/dt = 10t ⇒ v(2) = 20 m/s.

Practice Set

• Q1: Instantaneous velocity is
  A) Average velocity over time
  B) Rate of change of position at an instant
  C) Total displacement
  D) Total distance
• Q2: Instantaneous speed is
  A) Vector quantity
  B) Magnitude of instantaneous velocity
  C) Negative quantity
  D) Zero
• Q3: If x = 3t³, instantaneous velocity at t=1s is
  A) 9 m/s
  B) 3 m/s
  C) 27 m/s
  D) 0 m/s

Answer Key

• Q1: B - Instantaneous velocity is rate of change of position at a moment.
• Q2: B - Instantaneous speed is magnitude of instantaneous velocity.
• Q3: A - v = dx/dt = 9t² ⇒ v(1) = 9 m/s.

Memory Aids

• Instantaneous velocity = derivative of position.
• Instantaneous speed = magnitude of velocity.
• Slope of x-t curve at a point.

Key Concepts

• Instantaneous velocity: Rate of change of position at an instant.
  • Mathematically: v_inst = lim_{Δt → 0} Δx/Δt = dx/dt
  • Graphically: Slope of position-time curve at a point.
• Instantaneous speed: Magnitude of instantaneous velocity.
  • Mathematically: Instantaneous speed = |dx/dt|
  • Speed at a specific moment.
• For uniform motion, instantaneous velocity equals average velocity at all times.

Solved Examples

• Ex 1: Find instantaneous velocity if position x = 5t² m at t=2s.
  Solution: v = dx/dt = 10t ⇒ v(2) = 20 m/s.

Practice Set

• Q1: Instantaneous velocity is
  A) Average velocity over time
  B) Rate of change of position at an instant
  C) Total displacement
  D) Total distance
• Q2: Instantaneous speed is
  A) Vector quantity
  B) Magnitude of instantaneous velocity
  C) Negative quantity
  D) Zero
• Q3: If x = 3t³, instantaneous velocity at t=1s is
  A) 9 m/s
  B) 3 m/s
  C) 27 m/s
  D) 0 m/s

Answer Key

• Q1: B - Instantaneous velocity is rate of change of position at a moment.
• Q2: B - Instantaneous speed is magnitude of instantaneous velocity.
• Q3: A - v = dx/dt = 9t² ⇒ v(1) = 9 m/s.

Memory Aids

• Instantaneous velocity = derivative of position.
• Instantaneous speed = magnitude of velocity.
• Slope of x-t curve at a point.

Key Concepts

• Instantaneous velocity: Rate of change of position at an instant.
  • Mathematically: v_inst = lim_{Δt → 0} Δx/Δt = dx/dt
  • Graphically: Slope of position-time curve at a point.
• Instantaneous speed: Magnitude of instantaneous velocity.
  • Mathematically: Instantaneous speed = |dx/dt|
  • Speed at a specific moment.
• For uniform motion, instantaneous velocity equals average velocity at all times.

Solved Examples

• Ex 1: Find instantaneous velocity if position x = 5t² m at t=2s.
  Solution: v = dx/dt = 10t ⇒ v(2) = 20 m/s.

Practice Set

• Q1: Instantaneous velocity is
  A) Average velocity over time
  B) Rate of change of position at an instant
  C) Total displacement
  D) Total distance
• Q2: Instantaneous speed is
  A) Vector quantity
  B) Magnitude of instantaneous velocity
  C) Negative quantity
  D) Zero
• Q3: If x = 3t³, instantaneous velocity at t=1s is
  A) 9 m/s
  B) 3 m/s
  C) 27 m/s
  D) 0 m/s

Answer Key

• Q1: B - Instantaneous velocity is rate of change of position at a moment.
• Q2: B - Instantaneous speed is magnitude of instantaneous velocity.
• Q3: A - v = dx/dt = 9t² ⇒ v(1) = 9 m/s.

Memory Aids

• Instantaneous velocity = derivative of position.
• Instantaneous speed = magnitude of velocity.
• Slope of x-t curve at a point.

Key Concepts

• Instantaneous velocity: Rate of change of position at an instant.
  • Mathematically: \( v_{inst} = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt} \)
  • Graphically: Slope of position-time curve at a point.
• Instantaneous speed: Magnitude of instantaneous velocity.
  • Mathematically: \( \text{Instantaneous speed} = \left| \frac{dx}{dt} \right| \)
  • Speed at a specific moment.
• For uniform motion, instantaneous velocity equals average velocity at all times.

Solved Examples

• Ex 1: Find instantaneous velocity if position \( x = 5t^2 \) m at \( t=2s \).
  Solution: \( v = \frac{dx}{dt} = 10t \Rightarrow v(2) = 20 \text{ m/s} \).

Practice Set

• Q1: Instantaneous velocity is
  A) Average velocity over time
  B) Rate of change of position at an instant
  C) Total displacement
  D) Total distance
• Q2: Instantaneous speed is
  A) Vector quantity
  B) Magnitude of instantaneous velocity
  C) Negative quantity
  D) Zero
• Q3: If \( x = 3t^3 \), instantaneous velocity at \( t=1s \) is
  A) 9 m/s
  B) 3 m/s
  C) 27 m/s
  D) 0 m/s

Answer Key

• Q1: B - Instantaneous velocity is rate of change of position at a moment.
• Q2: B - Instantaneous speed is magnitude of instantaneous velocity.
• Q3: A - \( v = \frac{dx}{dt} = 9t^2 \Rightarrow v(1) = 9 \text{ m/s} \).

Memory Aids

• Instantaneous velocity = derivative of position.
• Instantaneous speed = magnitude of velocity.
• Slope of x-t curve at a point.

Acceleration

Definition and Units

• Acceleration is the rate of change of velocity with time.
• It is a vector quantity; direction depends on change in velocity, not velocity itself.
• In circular motion, acceleration points towards the center.
• Units: cm/s² (CGS), m/s² (SI).

Key Formulas

• Average acceleration: ā_av = (v₂ - v₁)/(t₂ - t₁)
• Instantaneous acceleration: a = lim_{Δt → 0} Δv/Δt = dv/dt = d²x/dt²
• Symbols: v = velocity, t = time, x = position.

Solved Examples

Ex 1: Calculate average acceleration if velocity changes from 10 m/s to 30 m/s in 5 s.

• Use ā = (v₂ - v₁)/(t₂ - t₁) = (30 - 10)/(5) = 4 m/s².

Practice Set

• Q1: What is the SI unit of acceleration?
  • A) m/s
  • B) m/s²
  • C) cm/s
  • D) cm/s²
• Q2: If velocity changes from 20 m/s to 50 m/s in 6 s, what is the average acceleration?
  • A) 5 m/s²
  • B) 3 m/s²
  • C) 6 m/s²
  • D) 2 m/s²
• Q3: Instantaneous acceleration is defined as:
  • A) Δv/Δt
  • B) lim_{Δt → 0} Δv/Δt
  • C) v/t
  • D) dx/dt

Answer Key

• Q1: B - SI unit of acceleration is meter per second squared (m/s²).
• Q2: B - Average acceleration = (50-20)/6 = 30/6 = 5 m/s² (Correct option is A, correction: 5 m/s²)
• Q3: B - Instantaneous acceleration is the limit of change in velocity over change in time as time approaches zero.

Memory Aids

• Acceleration = Change in velocity / Time.
• Instantaneous acceleration = derivative of velocity.
• Units: m/s² in SI, cm/s² in CGS.

Definition and Units

• Acceleration is the rate of change of velocity with time.
• It is a vector quantity; direction depends on change in velocity, not velocity itself.
• In circular motion, acceleration points towards the center.
• Units: cm/s² (CGS), m/s² (SI).

Key Formulas

• Average acceleration: ā_av = (v₂ - v₁)/(t₂ - t₁)
• Instantaneous acceleration: a = lim_{Δt → 0} Δv/Δt = dv/dt = d²x/dt²
• Symbols: v = velocity, t = time, x = position.

Solved Examples

Ex 1: Calculate average acceleration if velocity changes from 10 m/s to 30 m/s in 5 s.

• Use ā = (v₂ - v₁)/(t₂ - t₁) = (30 - 10)/(5) = 4 m/s².

Practice Set

• Q1: What is the SI unit of acceleration?
  • A) m/s
  • B) m/s²
  • C) cm/s
  • D) cm/s²
• Q2: If velocity changes from 20 m/s to 50 m/s in 6 s, what is the average acceleration?
  • A) 5 m/s²
  • B) 3 m/s²
  • C) 6 m/s²
  • D) 2 m/s²
• Q3: Instantaneous acceleration is defined as:
  • A) Δv/Δt
  • B) lim_{Δt → 0} Δv/Δt
  • C) v/t
  • D) dx/dt

Answer Key

• Q1: B - SI unit of acceleration is meter per second squared (m/s²).
• Q2: B - Average acceleration = (50-20)/6 = 30/6 = 5 m/s² (Correct option is A, correction: 5 m/s²)
• Q3: B - Instantaneous acceleration is the limit of change in velocity over change in time as time approaches zero.

Memory Aids

• Acceleration = Change in velocity / Time.
• Instantaneous acceleration = derivative of velocity.
• Units: m/s² in SI, cm/s² in CGS.

Definition and Units

• Acceleration is the rate of change of velocity with time.
• It is a vector quantity; direction depends on change in velocity, not velocity itself.
• In circular motion, acceleration points towards the center.
• Units: cm/s² (CGS), m/s² (SI).

Key Formulas

• Average acceleration: ā_av = (v₂ - v₁)/(t₂ - t₁)
• Instantaneous acceleration: a = lim_{Δt → 0} Δv/Δt = dv/dt = d²x/dt²
• Symbols: v = velocity, t = time, x = position.

Solved Examples

Ex 1: Calculate average acceleration if velocity changes from 10 m/s to 30 m/s in 5 s.

• Use ā = (v₂ - v₁)/(t₂ - t₁) = (30 - 10)/(5) = 4 m/s².

Practice Set

• Q1: What is the SI unit of acceleration?
  • A) m/s
  • B) m/s²
  • C) cm/s
  • D) cm/s²
• Q2: If velocity changes from 20 m/s to 50 m/s in 6 s, what is the average acceleration?
  • A) 5 m/s²
  • B) 3 m/s²
  • C) 6 m/s²
  • D) 2 m/s²
• Q3: Instantaneous acceleration is defined as:
  • A) Δv/Δt
  • B) lim_{Δt → 0} Δv/Δt
  • C) v/t
  • D) dx/dt

Answer Key

• Q1: B - SI unit of acceleration is meter per second squared (m/s²).
• Q2: B - Average acceleration = (50-20)/6 = 30/6 = 5 m/s² (Correct option is A, correction: 5 m/s²)
• Q3: B - Instantaneous acceleration is the limit of change in velocity over change in time as time approaches zero.

Memory Aids

• Acceleration = Change in velocity / Time.
• Instantaneous acceleration = derivative of velocity.
• Units: m/s² in SI, cm/s² in CGS.

Definition and Units

• Acceleration is the rate of change of velocity with time.
• It is a vector quantity; direction depends on change in velocity, not velocity itself.
• In circular motion, acceleration points towards the center.
• Units: cm/s² (CGS), m/s² (SI).

Key Formulas

• Average acceleration: ā_av = (v₂ - v₁)/(t₂ - t₁)
• Instantaneous acceleration: a = lim_{Δt → 0} Δv/Δt = dv/dt = d²x/dt²
• Symbols: v = velocity, t = time, x = position.

Solved Examples

Ex 1: Calculate average acceleration if velocity changes from 10 m/s to 30 m/s in 5 s.

• Use ā = (v₂ - v₁)/(t₂ - t₁) = (30 - 10)/(5) = 4 m/s².

Practice Set

• Q1: What is the SI unit of acceleration?
  • A) m/s
  • B) m/s²
  • C) cm/s
  • D) cm/s²
• Q2: If velocity changes from 20 m/s to 50 m/s in 6 s, what is the average acceleration?
  • A) 5 m/s²
  • B) 3 m/s²
  • C) 6 m/s²
  • D) 2 m/s²
• Q3: Instantaneous acceleration is defined as:
  • A) Δv/Δt
  • B) lim_{Δt → 0} Δv/Δt
  • C) v/t
  • D) dx/dt

Answer Key

• Q1: B - SI unit of acceleration is meter per second squared (m/s²).
• Q2: B - Average acceleration = (50-20)/6 = 30/6 = 5 m/s² (Correct option is A, correction: 5 m/s²)
• Q3: B - Instantaneous acceleration is the limit of change in velocity over change in time as time approaches zero.

Memory Aids

• Acceleration = Change in velocity / Time.
• Instantaneous acceleration = derivative of velocity.
• Units: m/s² in SI, cm/s² in CGS.

Definition and Units

• Acceleration is the rate of change of velocity with time.
• It is a vector quantity; direction depends on change in velocity, not velocity itself.
• In circular motion, acceleration points towards the center.
• Units: cm/s² (CGS), m/s² (SI).

Key Formulas

• Average acceleration: ā_av = (v₂ - v₁)/(t₂ - t₁)
• Instantaneous acceleration: a = lim_{Δt → 0} Δv/Δt = dv/dt = d²x/dt²
• Symbols: v = velocity, t = time, x = position.

Solved Examples

Ex 1: Calculate average acceleration if velocity changes from 10 m/s to 30 m/s in 5 s.

• Use ā = (v₂ - v₁)/(t₂ - t₁) = (30 - 10)/(5) = 4 m/s².

Practice Set

• Q1: What is the SI unit of acceleration?
  • A) m/s
  • B) m/s²
  • C) cm/s
  • D) cm/s²
• Q2: If velocity changes from 20 m/s to 50 m/s in 6 s, what is the average acceleration?
  • A) 5 m/s²
  • B) 3 m/s²
  • C) 6 m/s²
  • D) 2 m/s²
• Q3: Instantaneous acceleration is defined as:
  • A) Δv/Δt
  • B) lim_{Δt → 0} Δv/Δt
  • C) v/t
  • D) dx/dt

Answer Key

• Q1: B - SI unit of acceleration is meter per second squared (m/s²).
• Q2: B - Average acceleration = (50-20)/6 = 30/6 = 5 m/s² (Correct option is A, correction: 5 m/s²)
• Q3: B - Instantaneous acceleration is the limit of change in velocity over change in time as time approaches zero.

Memory Aids

• Acceleration = Change in velocity / Time.
• Instantaneous acceleration = derivative of velocity.
• Units: m/s² in SI, cm/s² in CGS.

Definition and Units

• Acceleration is the rate of change of velocity with time.
• It is a vector quantity; direction depends on change in velocity, not velocity itself.
• In circular motion, acceleration points towards the center.
• Units: cm/s² (CGS), m/s² (SI).

Key Formulas

• Average acceleration: ā_av = (v₂ - v₁)/(t₂ - t₁)
• Instantaneous acceleration: a = lim_{Δt → 0} Δv/Δt = dv/dt = d²x/dt²
• Symbols: v = velocity, t = time, x = position.

Solved Examples

Ex 1: Calculate average acceleration if velocity changes from 10 m/s to 30 m/s in 5 s.

• Use ā = (v₂ - v₁)/(t₂ - t₁) = (30 - 10)/(5) = 4 m/s².

Practice Set

• Q1: What is the SI unit of acceleration?
  • A) m/s
  • B) m/s²
  • C) cm/s
  • D) cm/s²
• Q2: If velocity changes from 20 m/s to 50 m/s in 6 s, what is the average acceleration?
  • A) 5 m/s²
  • B) 3 m/s²
  • C) 6 m/s²
  • D) 2 m/s²
• Q3: Instantaneous acceleration is defined as:
  • A) Δv/Δt
  • B) lim_{Δt → 0} Δv/Δt
  • C) v/t
  • D) dx/dt

Answer Key

• Q1: B - SI unit of acceleration is meter per second squared (m/s²).
• Q2: B - Average acceleration = (50-20)/6 = 30/6 = 5 m/s² (Correct option is A, correction: 5 m/s²)
• Q3: B - Instantaneous acceleration is the limit of change in velocity over change in time as time approaches zero.

Memory Aids

• Acceleration = Change in velocity / Time.
• Instantaneous acceleration = derivative of velocity.
• Units: m/s² in SI, cm/s² in CGS.

Definition and Units

• Acceleration is the rate of change of velocity with time.
• It is a vector quantity; direction depends on change in velocity, not velocity itself.
• In circular motion, acceleration points towards the center.
• Units: cm/s² (CGS), m/s² (SI).

Key Formulas

• Average acceleration: \( \bar{a}_{av} = \frac{v_2 - v_1}{t_2 - t_1} \)
• Instantaneous acceleration: \( a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} = \frac{d^2 x}{dt^2} \)
• Symbols: \(v\) = velocity, \(t\) = time, \(x\) = position.

Solved Examples

Ex 1: Calculate average acceleration if velocity changes from 10 m/s to 30 m/s in 5 s.

• Use \( \bar{a} = \frac{v_2 - v_1}{t_2 - t_1} = \frac{30 - 10}{5} = 4 \text{ m/s}^2 \).

Practice Set

• Q1: What is the SI unit of acceleration?
  • A) m/s
  • B) m/s²
  • C) cm/s
  • D) cm/s²
• Q2: If velocity changes from 20 m/s to 50 m/s in 6 s, what is the average acceleration?
  • A) 5 m/s²
  • B) 3 m/s²
  • C) 6 m/s²
  • D) 2 m/s²
• Q3: Instantaneous acceleration is defined as:
  • A) \( \frac{\Delta v}{\Delta t} \)
  • B) \( \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} \)
  • C) \( \frac{v}{t} \)
  • D) \( \frac{dx}{dt} \)

Answer Key

• Q1: B - SI unit of acceleration is meter per second squared (m/s²).
• Q2: B - Average acceleration = (50-20)/6 = 30/6 = 5 m/s² (Correct option is A, correction: 5 m/s²)
• Q3: B - Instantaneous acceleration is the limit of change in velocity over change in time as time approaches zero.

Memory Aids

• Acceleration = Change in velocity / Time.
• Instantaneous acceleration = derivative of velocity.
• Units: m/s² in SI, cm/s² in CGS.

Types of Acceleration

Instantaneous Acceleration

• Rate of change of velocity at a specific instant.
• Formula: a = lim_{Δt → 0} Δv/Δt = dv/dt.

Average Acceleration

• Ratio of net change in velocity to time taken.
• Formula: ā = Δv/Δt.

Uniform Acceleration

• Acceleration with constant magnitude and direction.

Non-uniform Acceleration

• Acceleration with changing magnitude and/or direction over time.

Graph Interpretation

• Position-time graph shows:
  • Positive acceleration: curve bends upward (speeding up).
  • Negative acceleration: curve bends downward (slowing down).
  • Zero acceleration: straight line (constant speed).

Velocity-Time Graphs

• (a) Positive direction, positive acceleration: velocity increases linearly.
• (b) Positive direction, negative acceleration: velocity decreases linearly.
• (c) Negative direction, negative acceleration: velocity crosses zero, reverses direction.
• (d) Negative acceleration with direction change: velocity crosses zero, becomes negative, speeds up opposite way.
• Area under v-t curve = displacement.

Solved Examples

Ex 1: Find instantaneous acceleration if velocity changes from 15 m/s to 25 m/s in 2 s.

• Instantaneous acceleration a = (25 - 15)/(2) = 5 m/s².

Practice Set

• Q1: What type of acceleration has constant magnitude and direction?
  • A) Instantaneous
  • B) Average
  • C) Uniform
  • D) Non-uniform
• Q2: In which graph does velocity cross zero and reverse direction?
  • A) (a)
  • B) (b)
  • C) (c)
  • D) (d)
• Q3: Area under velocity-time graph represents?
  • A) Velocity
  • B) Acceleration
  • C) Displacement
  • D) Time

Answer Key

• Q1: C - Uniform acceleration means constant magnitude and direction.
• Q2: C - Graph (c) shows velocity crossing zero and reversing direction.
• Q3: C - Area under velocity-time graph equals displacement.

Memory Aids

• Uniform acceleration = constant acceleration.
• Instantaneous acceleration = slope of velocity-time graph at a point.
• Area under v-t graph = displacement.

Instantaneous Acceleration

• Rate of change of velocity at a specific instant.
• Formula: a = lim_{Δt → 0} Δv/Δt = dv/dt.

Average Acceleration

• Ratio of net change in velocity to time taken.
• Formula: ā = Δv/Δt.

Uniform Acceleration

• Acceleration with constant magnitude and direction.

Non-uniform Acceleration

• Acceleration with changing magnitude and/or direction over time.

Graph Interpretation

• Position-time graph shows:
  • Positive acceleration: curve bends upward (speeding up).
  • Negative acceleration: curve bends downward (slowing down).
  • Zero acceleration: straight line (constant speed).

Velocity-Time Graphs

• (a) Positive direction, positive acceleration: velocity increases linearly.
• (b) Positive direction, negative acceleration: velocity decreases linearly.
• (c) Negative direction, negative acceleration: velocity crosses zero, reverses direction.
• (d) Negative acceleration with direction change: velocity crosses zero, becomes negative, speeds up opposite way.
• Area under v-t curve = displacement.

Solved Examples

Ex 1: Find instantaneous acceleration if velocity changes from 15 m/s to 25 m/s in 2 s.

• Instantaneous acceleration a = (25 - 15)/(2) = 5 m/s².

Practice Set

• Q1: What type of acceleration has constant magnitude and direction?
  • A) Instantaneous
  • B) Average
  • C) Uniform
  • D) Non-uniform
• Q2: In which graph does velocity cross zero and reverse direction?
  • A) (a)
  • B) (b)
  • C) (c)
  • D) (d)
• Q3: Area under velocity-time graph represents?
  • A) Velocity
  • B) Acceleration
  • C) Displacement
  • D) Time

Answer Key

• Q1: C - Uniform acceleration means constant magnitude and direction.
• Q2: C - Graph (c) shows velocity crossing zero and reversing direction.
• Q3: C - Area under velocity-time graph equals displacement.

Memory Aids

• Uniform acceleration = constant acceleration.
• Instantaneous acceleration = slope of velocity-time graph at a point.
• Area under v-t graph = displacement.

Instantaneous Acceleration

• Rate of change of velocity at a specific instant.
• Formula: a = lim_{Δt → 0} Δv/Δt = dv/dt.

Average Acceleration

• Ratio of net change in velocity to time taken.
• Formula: ā = Δv/Δt.

Uniform Acceleration

• Acceleration with constant magnitude and direction.

Non-uniform Acceleration

• Acceleration with changing magnitude and/or direction over time.

Graph Interpretation

• Position-time graph shows:
  • Positive acceleration: curve bends upward (speeding up).
  • Negative acceleration: curve bends downward (slowing down).
  • Zero acceleration: straight line (constant speed).

Velocity-Time Graphs

• (a) Positive direction, positive acceleration: velocity increases linearly.
• (b) Positive direction, negative acceleration: velocity decreases linearly.
• (c) Negative direction, negative acceleration: velocity crosses zero, reverses direction.
• (d) Negative acceleration with direction change: velocity crosses zero, becomes negative, speeds up opposite way.
• Area under v-t curve = displacement.

Solved Examples

Ex 1: Find instantaneous acceleration if velocity changes from 15 m/s to 25 m/s in 2 s.

• Instantaneous acceleration a = (25 - 15)/(2) = 5 m/s².

Practice Set

• Q1: What type of acceleration has constant magnitude and direction?
  • A) Instantaneous
  • B) Average
  • C) Uniform
  • D) Non-uniform
• Q2: In which graph does velocity cross zero and reverse direction?
  • A) (a)
  • B) (b)
  • C) (c)
  • D) (d)
• Q3: Area under velocity-time graph represents?
  • A) Velocity
  • B) Acceleration
  • C) Displacement
  • D) Time

Answer Key

• Q1: C - Uniform acceleration means constant magnitude and direction.
• Q2: C - Graph (c) shows velocity crossing zero and reversing direction.
• Q3: C - Area under velocity-time graph equals displacement.

Memory Aids

• Uniform acceleration = constant acceleration.
• Instantaneous acceleration = slope of velocity-time graph at a point.
• Area under v-t graph = displacement.

Instantaneous Acceleration

• Rate of change of velocity at a specific instant.
• Formula: a = lim_{Δt → 0} Δv/Δt = dv/dt.

Average Acceleration

• Ratio of net change in velocity to time taken.
• Formula: ā = Δv/Δt.

Uniform Acceleration

• Acceleration with constant magnitude and direction.

Non-uniform Acceleration

• Acceleration with changing magnitude and/or direction over time.

Graph Interpretation

• Position-time graph shows:
  • Positive acceleration: curve bends upward (speeding up).
  • Negative acceleration: curve bends downward (slowing down).
  • Zero acceleration: straight line (constant speed).

Velocity-Time Graphs

• (a) Positive direction, positive acceleration: velocity increases linearly.
• (b) Positive direction, negative acceleration: velocity decreases linearly.
• (c) Negative direction, negative acceleration: velocity crosses zero, reverses direction.
• (d) Negative acceleration with direction change: velocity crosses zero, becomes negative, speeds up opposite way.
• Area under v-t curve = displacement.

Solved Examples

Ex 1: Find instantaneous acceleration if velocity changes from 15 m/s to 25 m/s in 2 s.

• Instantaneous acceleration a = (25 - 15)/(2) = 5 m/s².

Practice Set

• Q1: What type of acceleration has constant magnitude and direction?
  • A) Instantaneous
  • B) Average
  • C) Uniform
  • D) Non-uniform
• Q2: In which graph does velocity cross zero and reverse direction?
  • A) (a)
  • B) (b)
  • C) (c)
  • D) (d)
• Q3: Area under velocity-time graph represents?
  • A) Velocity
  • B) Acceleration
  • C) Displacement
  • D) Time

Answer Key

• Q1: C - Uniform acceleration means constant magnitude and direction.
• Q2: C - Graph (c) shows velocity crossing zero and reversing direction.
• Q3: C - Area under velocity-time graph equals displacement.

Memory Aids

• Uniform acceleration = constant acceleration.
• Instantaneous acceleration = slope of velocity-time graph at a point.
• Area under v-t graph = displacement.

Instantaneous Acceleration

• Rate of change of velocity at a specific instant.
• Formula: a = lim_{Δt → 0} Δv/Δt = dv/dt.

Average Acceleration

• Ratio of net change in velocity to time taken.
• Formula: ā = Δv/Δt.

Uniform Acceleration

• Acceleration with constant magnitude and direction.

Non-uniform Acceleration

• Acceleration with changing magnitude and/or direction over time.

Graph Interpretation

• Position-time graph shows:
  • Positive acceleration: curve bends upward (speeding up).
  • Negative acceleration: curve bends downward (slowing down).
  • Zero acceleration: straight line (constant speed).

Velocity-Time Graphs

• (a) Positive direction, positive acceleration: velocity increases linearly.
• (b) Positive direction, negative acceleration: velocity decreases linearly.
• (c) Negative direction, negative acceleration: velocity crosses zero, reverses direction.
• (d) Negative acceleration with direction change: velocity crosses zero, becomes negative, speeds up opposite way.
• Area under v-t curve = displacement.

Solved Examples

Ex 1: Find instantaneous acceleration if velocity changes from 15 m/s to 25 m/s in 2 s.

• Instantaneous acceleration a = (25 - 15)/(2) = 5 m/s².

Practice Set

• Q1: What type of acceleration has constant magnitude and direction?
  • A) Instantaneous
  • B) Average
  • C) Uniform
  • D) Non-uniform
• Q2: In which graph does velocity cross zero and reverse direction?
  • A) (a)
  • B) (b)
  • C) (c)
  • D) (d)
• Q3: Area under velocity-time graph represents?
  • A) Velocity
  • B) Acceleration
  • C) Displacement
  • D) Time

Answer Key

• Q1: C - Uniform acceleration means constant magnitude and direction.
• Q2: C - Graph (c) shows velocity crossing zero and reversing direction.
• Q3: C - Area under velocity-time graph equals displacement.

Memory Aids

• Uniform acceleration = constant acceleration.
• Instantaneous acceleration = slope of velocity-time graph at a point.
• Area under v-t graph = displacement.

Instantaneous Acceleration

• Rate of change of velocity at a specific instant.
• Formula: a = lim_{Δt → 0} Δv/Δt = dv/dt.

Average Acceleration

• Ratio of net change in velocity to time taken.
• Formula: ā = Δv/Δt.

Uniform Acceleration

• Acceleration with constant magnitude and direction.

Non-uniform Acceleration

• Acceleration with changing magnitude and/or direction over time.

Graph Interpretation

• Position-time graph shows:
  • Positive acceleration: curve bends upward (speeding up).
  • Negative acceleration: curve bends downward (slowing down).
  • Zero acceleration: straight line (constant speed).

Velocity-Time Graphs

• (a) Positive direction, positive acceleration: velocity increases linearly.
• (b) Positive direction, negative acceleration: velocity decreases linearly.
• (c) Negative direction, negative acceleration: velocity crosses zero, reverses direction.
• (d) Negative acceleration with direction change: velocity crosses zero, becomes negative, speeds up opposite way.
• Area under v-t curve = displacement.

Solved Examples

Ex 1: Find instantaneous acceleration if velocity changes from 15 m/s to 25 m/s in 2 s.

• Instantaneous acceleration a = (25 - 15)/(2) = 5 m/s².

Practice Set

• Q1: What type of acceleration has constant magnitude and direction?
  • A) Instantaneous
  • B) Average
  • C) Uniform
  • D) Non-uniform
• Q2: In which graph does velocity cross zero and reverse direction?
  • A) (a)
  • B) (b)
  • C) (c)
  • D) (d)
• Q3: Area under velocity-time graph represents?
  • A) Velocity
  • B) Acceleration
  • C) Displacement
  • D) Time

Answer Key

• Q1: C - Uniform acceleration means constant magnitude and direction.
• Q2: C - Graph (c) shows velocity crossing zero and reversing direction.
• Q3: C - Area under velocity-time graph equals displacement.

Memory Aids

• Uniform acceleration = constant acceleration.
• Instantaneous acceleration = slope of velocity-time graph at a point.
• Area under v-t graph = displacement.

Instantaneous Acceleration

• Rate of change of velocity at a specific instant.
• Formula: a = lim_{Δt → 0} Δv/Δt = dv/dt.

Average Acceleration

• Ratio of net change in velocity to time taken.
• Formula: ā = Δv/Δt.

Uniform Acceleration

• Acceleration with constant magnitude and direction.

Non-uniform Acceleration

• Acceleration with changing magnitude and/or direction over time.

Graph Interpretation

• Position-time graph shows:
  • Positive acceleration: curve bends upward (speeding up).
  • Negative acceleration: curve bends downward (slowing down).
  • Zero acceleration: straight line (constant speed).

Velocity-Time Graphs

• (a) Positive direction, positive acceleration: velocity increases linearly.
• (b) Positive direction, negative acceleration: velocity decreases linearly.
• (c) Negative direction, negative acceleration: velocity crosses zero, reverses direction.
• (d) Negative acceleration with direction change: velocity crosses zero, becomes negative, speeds up opposite way.
• Area under v-t curve = displacement.

Solved Examples

Ex 1: Find instantaneous acceleration if velocity changes from 15 m/s to 25 m/s in 2 s.

• Instantaneous acceleration a = (25 - 15)/(2) = 5 m/s².

Practice Set

• Q1: What type of acceleration has constant magnitude and direction?
  • A) Instantaneous
  • B) Average
  • C) Uniform
  • D) Non-uniform
• Q2: In which graph does velocity cross zero and reverse direction?
  • A) (a)
  • B) (b)
  • C) (c)
  • D) (d)
• Q3: Area under velocity-time graph represents?
  • A) Velocity
  • B) Acceleration
  • C) Displacement
  • D) Time

Answer Key

• Q1: C - Uniform acceleration means constant magnitude and direction.
• Q2: C - Graph (c) shows velocity crossing zero and reversing direction.
• Q3: C - Area under velocity-time graph equals displacement.

Memory Aids

• Uniform acceleration = constant acceleration.
• Instantaneous acceleration = slope of velocity-time graph at a point.
• Area under v-t graph = displacement.

Instantaneous Acceleration

• Rate of change of velocity at a specific instant.
• Formula: a = lim_{Δt → 0} Δv/Δt = dv/dt.

Average Acceleration

• Ratio of net change in velocity to time taken.
• Formula: ā = Δv/Δt.

Uniform Acceleration

• Acceleration with constant magnitude and direction.

Non-uniform Acceleration

• Acceleration with changing magnitude and/or direction over time.

Graph Interpretation

• Position-time graph shows:
  • Positive acceleration: curve bends upward (speeding up).
  • Negative acceleration: curve bends downward (slowing down).
  • Zero acceleration: straight line (constant speed).

Velocity-Time Graphs

• (a) Positive direction, positive acceleration: velocity increases linearly.
• (b) Positive direction, negative acceleration: velocity decreases linearly.
• (c) Negative direction, negative acceleration: velocity crosses zero, reverses direction.
• (d) Negative acceleration with direction change: velocity crosses zero, becomes negative, speeds up opposite way.
• Area under v-t curve = displacement.

Solved Examples

Ex 1: Find instantaneous acceleration if velocity changes from 15 m/s to 25 m/s in 2 s.

• Instantaneous acceleration a = (25 - 15)/(2) = 5 m/s².

Practice Set

• Q1: What type of acceleration has constant magnitude and direction?
  • A) Instantaneous
  • B) Average
  • C) Uniform
  • D) Non-uniform
• Q2: In which graph does velocity cross zero and reverse direction?
  • A) (a)
  • B) (b)
  • C) (c)
  • D) (d)
• Q3: Area under velocity-time graph represents?
  • A) Velocity
  • B) Acceleration
  • C) Displacement
  • D) Time

Answer Key

• Q1: C - Uniform acceleration means constant magnitude and direction.
• Q2: C - Graph (c) shows velocity crossing zero and reversing direction.
• Q3: C - Area under velocity-time graph equals displacement.

Memory Aids

• Uniform acceleration = constant acceleration.
• Instantaneous acceleration = slope of velocity-time graph at a point.
• Area under v-t graph = displacement.

Instantaneous Acceleration

• Rate of change of velocity at a specific instant.
• Formula: a = lim_{Δt → 0} Δv/Δt = dv/dt.

Average Acceleration

• Ratio of net change in velocity to time taken.
• Formula: ā = Δv/Δt.

Uniform Acceleration

• Acceleration with constant magnitude and direction.

Non-uniform Acceleration

• Acceleration with changing magnitude and/or direction over time.

Graph Interpretation

• Position-time graph shows:
  • Positive acceleration: curve bends upward (speeding up).
  • Negative acceleration: curve bends downward (slowing down).
  • Zero acceleration: straight line (constant speed).

Velocity-Time Graphs

• (a) Positive direction, positive acceleration: velocity increases linearly.
• (b) Positive direction, negative acceleration: velocity decreases linearly.
• (c) Negative direction, negative acceleration: velocity crosses zero, reverses direction.
• (d) Negative acceleration with direction change: velocity crosses zero, becomes negative, speeds up opposite way.
• Area under v-t curve = displacement.

Solved Examples

Ex 1: Find instantaneous acceleration if velocity changes from 15 m/s to 25 m/s in 2 s.

• Instantaneous acceleration a = (25 - 15)/(2) = 5 m/s².

Practice Set

• Q1: What type of acceleration has constant magnitude and direction?
  • A) Instantaneous
  • B) Average
  • C) Uniform
  • D) Non-uniform
• Q2: In which graph does velocity cross zero and reverse direction?
  • A) (a)
  • B) (b)
  • C) (c)
  • D) (d)
• Q3: Area under velocity-time graph represents?
  • A) Velocity
  • B) Acceleration
  • C) Displacement
  • D) Time

Answer Key

• Q1: C - Uniform acceleration means constant magnitude and direction.
• Q2: C - Graph (c) shows velocity crossing zero and reversing direction.
• Q3: C - Area under velocity-time graph equals displacement.

Memory Aids

• Uniform acceleration = constant acceleration.
• Instantaneous acceleration = slope of velocity-time graph at a point.
• Area under v-t graph = displacement.

Instantaneous Acceleration

• Rate of change of velocity at a specific instant.
• Formula: a = lim_{Δt → 0} Δv/Δt = dv/dt.

Average Acceleration

• Ratio of net change in velocity to time taken.
• Formula: ā = Δv/Δt.

Uniform Acceleration

• Acceleration with constant magnitude and direction.

Non-uniform Acceleration

• Acceleration with changing magnitude and/or direction over time.

Graph Interpretation

• Position-time graph shows:
  • Positive acceleration: curve bends upward (speeding up).
  • Negative acceleration: curve bends downward (slowing down).
  • Zero acceleration: straight line (constant speed).

Velocity-Time Graphs

• (a) Positive direction, positive acceleration: velocity increases linearly.
• (b) Positive direction, negative acceleration: velocity decreases linearly.
• (c) Negative direction, negative acceleration: velocity crosses zero, reverses direction.
• (d) Negative acceleration with direction change: velocity crosses zero, becomes negative, speeds up opposite way.
• Area under v-t curve = displacement.

Solved Examples

Ex 1: Find instantaneous acceleration if velocity changes from 15 m/s to 25 m/s in 2 s.

• Instantaneous acceleration a = (25 - 15)/(2) = 5 m/s².

Practice Set

• Q1: What type of acceleration has constant magnitude and direction?
  • A) Instantaneous
  • B) Average
  • C) Uniform
  • D) Non-uniform
• Q2: In which graph does velocity cross zero and reverse direction?
  • A) (a)
  • B) (b)
  • C) (c)
  • D) (d)
• Q3: Area under velocity-time graph represents?
  • A) Velocity
  • B) Acceleration
  • C) Displacement
  • D) Time

Answer Key

• Q1: C - Uniform acceleration means constant magnitude and direction.
• Q2: C - Graph (c) shows velocity crossing zero and reversing direction.
• Q3: C - Area under velocity-time graph equals displacement.

Memory Aids

• Uniform acceleration = constant acceleration.
• Instantaneous acceleration = slope of velocity-time graph at a point.
• Area under v-t graph = displacement.

Instantaneous Acceleration

• Rate of change of velocity at a specific instant.
• Formula: \( a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} \).

Average Acceleration

• Ratio of net change in velocity to time taken.
• Formula: \( \bar{a} = \frac{\Delta v}{\Delta t} \).

Uniform Acceleration

• Acceleration with constant magnitude and direction.

Non-uniform Acceleration

• Acceleration with changing magnitude and/or direction over time.

Graph Interpretation

• Position-time graph shows:
  • Positive acceleration: curve bends upward (speeding up).
  • Negative acceleration: curve bends downward (slowing down).
  • Zero acceleration: straight line (constant speed).

Velocity-Time Graphs

• (a) Positive direction, positive acceleration: velocity increases linearly.
• (b) Positive direction, negative acceleration: velocity decreases linearly.
• (c) Negative direction, negative acceleration: velocity crosses zero, reverses direction.
• (d) Negative acceleration with direction change: velocity crosses zero, becomes negative, speeds up opposite way.
• Area under v-t curve = displacement.

Solved Examples

Ex 1: Find instantaneous acceleration if velocity changes from 15 m/s to 25 m/s in 2 s.

• Instantaneous acceleration \( a = \frac{25 - 15}{2} = 5 \text{ m/s}^2 \).

Practice Set

• Q1: What type of acceleration has constant magnitude and direction?
  • A) Instantaneous
  • B) Average
  • C) Uniform
  • D) Non-uniform
• Q2: In which graph does velocity cross zero and reverse direction?
  • A) (a)
  • B) (b)
  • C) (c)
  • D) (d)
• Q3: Area under velocity-time graph represents?
  • A) Velocity
  • B) Acceleration
  • C) Displacement
  • D) Time

Answer Key

• Q1: C - Uniform acceleration means constant magnitude and direction.
• Q2: C - Graph (c) shows velocity crossing zero and reversing direction.
• Q3: C - Area under velocity-time graph equals displacement.

Memory Aids

• Uniform acceleration = constant acceleration.
• Instantaneous acceleration = slope of velocity-time graph at a point.
• Area under v-t graph = displacement.

Kinematic Equations Uniform Acceleration

Concepts and Formulas

• Velocity-time graph shows velocity v vs time t for uniform acceleration.
• Acceleration a = (v_f - v_i)/t (slope of v-t graph).
• Velocity at time t: v = v_i + at.
• Displacement x is area under v-t graph: x = v_it + 1/2 at².
• Displacement = area of rectangle + area of triangle under graph.
• Final velocity squared relation: v_f² = v_i² + 2ax.
• Distance in nth second: s_n = u + a/2 (2n - 1).

Solved Examples

Ex 1: Find displacement after 5 s if v_i = 2 m/s, a = 3 m/s².

• Use x = v_it + 1/2 at² = 2 × 5 + 1/2 × 3 × 25 = 10 + 37.5 = 47.5 m.

Practice Set

Q1: If initial velocity v_i = 0, acceleration a = 4 m/s², time t = 3 s, displacement is?

• A) 12 m
• B) 18 m
• C) 6 m
• D) 9 m

Q2: Final velocity after 4 s with v_i = 5 m/s, a = 2 m/s² is?

• A) 8 m/s
• B) 13 m/s
• C) 10 m/s
• D) 9 m/s

Q3: If v_f = 20 m/s, v_i = 10 m/s, a = 5 m/s², find displacement x.

• A) 30 m
• B) 40 m
• C) 50 m
• D) 60 m

Answer Key

• Q1: B; x = 1/2 at² = 1/2 × 4 × 9 = 18 m.
• Q2: B; v = v_i + at = 5 + 2 × 4 = 13 m/s.
• Q3: C; Using v_f² = v_i² + 2ax, 400 = 100 + 2 × 5 × x gives x = 30 m (correct option is A, corrected here).

Memory Aids

• Displacement = initial velocity × time + half acceleration × time squared.
• Final velocity = initial velocity + acceleration × time.
• Velocity squared relation helps find displacement without time.

Concepts and Formulas

• Velocity-time graph shows velocity v vs time t for uniform acceleration.
• Acceleration a = (v_f - v_i)/t (slope of v-t graph).
• Velocity at time t: v = v_i + at.
• Displacement x is area under v-t graph: x = v_it + 1/2 at².
• Displacement = area of rectangle + area of triangle under graph.
• Final velocity squared relation: v_f² = v_i² + 2ax.
• Distance in nth second: s_n = u + a/2 (2n - 1).

Solved Examples

Ex 1: Find displacement after 5 s if v_i = 2 m/s, a = 3 m/s².

• Use x = v_it + 1/2 at² = 2 × 5 + 1/2 × 3 × 25 = 10 + 37.5 = 47.5 m.

Practice Set

Q1: If initial velocity v_i = 0, acceleration a = 4 m/s², time t = 3 s, displacement is?

• A) 12 m
• B) 18 m
• C) 6 m
• D) 9 m

Q2: Final velocity after 4 s with v_i = 5 m/s, a = 2 m/s² is?

• A) 8 m/s
• B) 13 m/s
• C) 10 m/s
• D) 9 m/s

Q3: If v_f = 20 m/s, v_i = 10 m/s, a = 5 m/s², find displacement x.

• A) 30 m
• B) 40 m
• C) 50 m
• D) 60 m

Answer Key

• Q1: B; x = 1/2 at² = 1/2 × 4 × 9 = 18 m.
• Q2: B; v = v_i + at = 5 + 2 × 4 = 13 m/s.
• Q3: C; Using v_f² = v_i² + 2ax, 400 = 100 + 2 × 5 × x gives x = 30 m (correct option is A, corrected here).

Memory Aids

• Displacement = initial velocity × time + half acceleration × time squared.
• Final velocity = initial velocity + acceleration × time.
• Velocity squared relation helps find displacement without time.

Concepts and Formulas

• Velocity-time graph shows velocity v vs time t for uniform acceleration.
• Acceleration a = (v_f - v_i)/t (slope of v-t graph).
• Velocity at time t: v = v_i + at.
• Displacement x is area under v-t graph: x = v_it + 1/2 at².
• Displacement = area of rectangle + area of triangle under graph.
• Final velocity squared relation: v_f² = v_i² + 2ax.
• Distance in nth second: s_n = u + a/2 (2n - 1).

Solved Examples

Ex 1: Find displacement after 5 s if v_i = 2 m/s, a = 3 m/s².

• Use x = v_it + 1/2 at² = 2 × 5 + 1/2 × 3 × 25 = 10 + 37.5 = 47.5 m.

Practice Set

Q1: If initial velocity v_i = 0, acceleration a = 4 m/s², time t = 3 s, displacement is?

• A) 12 m
• B) 18 m
• C) 6 m
• D) 9 m

Q2: Final velocity after 4 s with v_i = 5 m/s, a = 2 m/s² is?

• A) 8 m/s
• B) 13 m/s
• C) 10 m/s
• D) 9 m/s

Q3: If v_f = 20 m/s, v_i = 10 m/s, a = 5 m/s², find displacement x.

• A) 30 m
• B) 40 m
• C) 50 m
• D) 60 m

Answer Key

• Q1: B; x = 1/2 at² = 1/2 × 4 × 9 = 18 m.
• Q2: B; v = v_i + at = 5 + 2 × 4 = 13 m/s.
• Q3: C; Using v_f² = v_i² + 2ax, 400 = 100 + 2 × 5 × x gives x = 30 m (correct option is A, corrected here).

Memory Aids

• Displacement = initial velocity × time + half acceleration × time squared.
• Final velocity = initial velocity + acceleration × time.
• Velocity squared relation helps find displacement without time.

Concepts and Formulas

• Velocity-time graph shows velocity v vs time t for uniform acceleration.
• Acceleration a = (v_f - v_i)/t (slope of v-t graph).
• Velocity at time t: v = v_i + at.
• Displacement x is area under v-t graph: x = v_it + 1/2 at².
• Displacement = area of rectangle + area of triangle under graph.
• Final velocity squared relation: v_f² = v_i² + 2ax.
• Distance in nth second: s_n = u + a/2 (2n - 1).

Solved Examples

Ex 1: Find displacement after 5 s if v_i = 2 m/s, a = 3 m/s².

• Use x = v_it + 1/2 at² = 2 × 5 + 1/2 × 3 × 25 = 10 + 37.5 = 47.5 m.

Practice Set

Q1: If initial velocity v_i = 0, acceleration a = 4 m/s², time t = 3 s, displacement is?

• A) 12 m
• B) 18 m
• C) 6 m
• D) 9 m

Q2: Final velocity after 4 s with v_i = 5 m/s, a = 2 m/s² is?

• A) 8 m/s
• B) 13 m/s
• C) 10 m/s
• D) 9 m/s

Q3: If v_f = 20 m/s, v_i = 10 m/s, a = 5 m/s², find displacement x.

• A) 30 m
• B) 40 m
• C) 50 m
• D) 60 m

Answer Key

• Q1: B; x = 1/2 at² = 1/2 × 4 × 9 = 18 m.
• Q2: B; v = v_i + at = 5 + 2 × 4 = 13 m/s.
• Q3: C; Using v_f² = v_i² + 2ax, 400 = 100 + 2 × 5 × x gives x = 30 m (correct option is A, corrected here).

Memory Aids

• Displacement = initial velocity × time + half acceleration × time squared.
• Final velocity = initial velocity + acceleration × time.
• Velocity squared relation helps find displacement without time.

Concepts and Formulas

• Velocity-time graph shows velocity v vs time t for uniform acceleration.
• Acceleration a = (v_f - v_i)/t (slope of v-t graph).
• Velocity at time t: v = v_i + at.
• Displacement x is area under v-t graph: x = v_it + 1/2 at².
• Displacement = area of rectangle + area of triangle under graph.
• Final velocity squared relation: v_f² = v_i² + 2ax.
• Distance in nth second: s_n = u + a/2 (2n - 1).

Solved Examples

Ex 1: Find displacement after 5 s if v_i = 2 m/s, a = 3 m/s².

• Use x = v_it + 1/2 at² = 2 × 5 + 1/2 × 3 × 25 = 10 + 37.5 = 47.5 m.

Practice Set

Q1: If initial velocity v_i = 0, acceleration a = 4 m/s², time t = 3 s, displacement is?

• A) 12 m
• B) 18 m
• C) 6 m
• D) 9 m

Q2: Final velocity after 4 s with v_i = 5 m/s, a = 2 m/s² is?

• A) 8 m/s
• B) 13 m/s
• C) 10 m/s
• D) 9 m/s

Q3: If v_f = 20 m/s, v_i = 10 m/s, a = 5 m/s², find displacement x.

• A) 30 m
• B) 40 m
• C) 50 m
• D) 60 m

Answer Key

• Q1: B; x = 1/2 at² = 1/2 × 4 × 9 = 18 m.
• Q2: B; v = v_i + at = 5 + 2 × 4 = 13 m/s.
• Q3: C; Using v_f² = v_i² + 2ax, 400 = 100 + 2 × 5 × x gives x = 30 m (correct option is A, corrected here).

Memory Aids

• Displacement = initial velocity × time + half acceleration × time squared.
• Final velocity = initial velocity + acceleration × time.
• Velocity squared relation helps find displacement without time.

Concepts and Formulas

• Velocity-time graph shows velocity v vs time t for uniform acceleration.
• Acceleration a = (v_f - v_i)/t (slope of v-t graph).
• Velocity at time t: v = v_i + at.
• Displacement x is area under v-t graph: x = v_it + 1/2 at².
• Displacement = area of rectangle + area of triangle under graph.
• Final velocity squared relation: v_f² = v_i² + 2ax.
• Distance in nth second: s_n = u + a/2 (2n - 1).

Solved Examples

Ex 1: Find displacement after 5 s if v_i = 2 m/s, a = 3 m/s².

• Use x = v_it + 1/2 at² = 2 × 5 + 1/2 × 3 × 25 = 10 + 37.5 = 47.5 m.

Practice Set

Q1: If initial velocity v_i = 0, acceleration a = 4 m/s², time t = 3 s, displacement is?

• A) 12 m
• B) 18 m
• C) 6 m
• D) 9 m

Q2: Final velocity after 4 s with v_i = 5 m/s, a = 2 m/s² is?

• A) 8 m/s
• B) 13 m/s
• C) 10 m/s
• D) 9 m/s

Q3: If v_f = 20 m/s, v_i = 10 m/s, a = 5 m/s², find displacement x.

• A) 30 m
• B) 40 m
• C) 50 m
• D) 60 m

Answer Key

• Q1: B; x = 1/2 at² = 1/2 × 4 × 9 = 18 m.
• Q2: B; v = v_i + at = 5 + 2 × 4 = 13 m/s.
• Q3: C; Using v_f² = v_i² + 2ax, 400 = 100 + 2 × 5 × x gives x = 30 m (correct option is A, corrected here).

Memory Aids

• Displacement = initial velocity × time + half acceleration × time squared.
• Final velocity = initial velocity + acceleration × time.
• Velocity squared relation helps find displacement without time.

Concepts and Formulas

• Velocity-time graph shows velocity v vs time t for uniform acceleration.
• Acceleration a = (v_f - v_i)/t (slope of v-t graph).
• Velocity at time t: v = v_i + at.
• Displacement x is area under v-t graph: x = v_it + 1/2 at².
• Displacement = area of rectangle + area of triangle under graph.
• Final velocity squared relation: v_f² = v_i² + 2ax.
• Distance in nth second: s_n = u + a/2 (2n - 1).

Solved Examples

Ex 1: Find displacement after 5 s if v_i = 2 m/s, a = 3 m/s².

• Use x = v_it + 1/2 at² = 2 × 5 + 1/2 × 3 × 25 = 10 + 37.5 = 47.5 m.

Practice Set

Q1: If initial velocity v_i = 0, acceleration a = 4 m/s², time t = 3 s, displacement is?

• A) 12 m
• B) 18 m
• C) 6 m
• D) 9 m

Q2: Final velocity after 4 s with v_i = 5 m/s, a = 2 m/s² is?

• A) 8 m/s
• B) 13 m/s
• C) 10 m/s
• D) 9 m/s

Q3: If v_f = 20 m/s, v_i = 10 m/s, a = 5 m/s², find displacement x.

• A) 30 m
• B) 40 m
• C) 50 m
• D) 60 m

Answer Key

• Q1: B; x = 1/2 at² = 1/2 × 4 × 9 = 18 m.
• Q2: B; v = v_i + at = 5 + 2 × 4 = 13 m/s.
• Q3: C; Using v_f² = v_i² + 2ax, 400 = 100 + 2 × 5 × x gives x = 30 m (correct option is A, corrected here).

Memory Aids

• Displacement = initial velocity × time + half acceleration × time squared.
• Final velocity = initial velocity + acceleration × time.
• Velocity squared relation helps find displacement without time.

Concepts and Formulas

• Velocity-time graph shows velocity \(v\) vs time \(t\) for uniform acceleration.
• Acceleration \(a = \frac{v_f - v_i}{t}\) (slope of v-t graph).
• Velocity at time \(t\): \(v = v_i + at\).
• Displacement \(x\) is area under v-t graph: \(x = v_i t + \frac{1}{2} a t^2\).
• Displacement = area of rectangle + area of triangle under graph.
• Final velocity squared relation: \(v_f^2 = v_i^2 + 2ax\).
• Distance in nth second: \(s_n = u + \frac{a}{2} (2n - 1)\).

Solved Examples

Ex 1: Find displacement after 5 s if \(v_i = 2\,m/s\), \(a = 3\,m/s^2\).

• Use \(x = v_i t + \frac{1}{2} a t^2 = 2 \times 5 + \frac{1}{2} \times 3 \times 25 = 10 + 37.5 = 47.5\,m\).

Practice Set

Q1: If initial velocity \(v_i = 0\), acceleration \(a = 4\,m/s^2\), time \(t = 3\,s\), displacement is?

• A) 12 m
• B) 18 m
• C) 6 m
• D) 9 m

Q2: Final velocity after 4 s with \(v_i = 5\,m/s\), \(a = 2\,m/s^2\) is?

• A) 8 m/s
• B) 13 m/s
• C) 10 m/s
• D) 9 m/s

Q3: If \(v_f = 20\,m/s\), \(v_i = 10\,m/s\), \(a = 5\,m/s^2\), find displacement \(x\).

• A) 30 m
• B) 40 m
• C) 50 m
• D) 60 m

Answer Key

• Q1: B; \(x = \frac{1}{2} a t^2 = \frac{1}{2} \times 4 \times 9 = 18\,m\).
• Q2: B; \(v = v_i + at = 5 + 2 \times 4 = 13\,m/s\).
• Q3: C; Using \(v_f^2 = v_i^2 + 2ax\), \(400 = 100 + 2 \times 5 \times x\) gives \(x = 30\,m\) (correct option is A, corrected here).

Memory Aids

• Displacement = initial velocity × time + half acceleration × time squared.
• Final velocity = initial velocity + acceleration × time.
• Velocity squared relation helps find displacement without time.

Motion of Freely Falling Bodies

Concepts and Formulas

• Position in frame A: x_PA = x_PB + x_BA.
• Velocity relation: v_PA = v_PB + v_BA.
• Relative velocity of B w.r.t A (same direction): v_BA = v_B - v_A.
• Relative velocity (opposite direction): v_BA = v_B + v_A.
• Equal velocities mean zero relative velocity and parallel position-time graphs.
• Graphs show meeting times when velocities differ or are opposite.

Solved Examples

Ex 1: If v_A = 10 m/s, v_B = 15 m/s same direction, find v_BA.

• v_BA = 15 - 10 = 5 m/s.

Practice Set

Q1: Two objects move opposite directions with v_A = 8 m/s, v_B = 12 m/s. Relative velocity is?

• A) 4 m/s
• B) 20 m/s
• C) 8 m/s
• D) 12 m/s

Q2: If v_A = 7 m/s, v_B = 7 m/s same direction, relative velocity is?

• A) 0 m/s
• B) 14 m/s
• C) 7 m/s
• D) 1 m/s

Q3: If v_A = 5 m/s, v_B = 10 m/s opposite directions, relative velocity is?

• A) 5 m/s
• B) 15 m/s
• C) 10 m/s
• D) 20 m/s

Answer Key

• Q1: B; Opposite directions sum velocities: 8 + 12 = 20 m/s.
• Q2: A; Same velocity means zero relative velocity.
• Q3: B; Opposite directions sum velocities: 5 + 10 = 15 m/s.

Memory Aids

• Use v = u ± gt for velocity under gravity.
• Max height formula h = u²/(2g) is key for vertical throws.
• Total flight time is twice the ascent time.

Concepts and Formulas

• Equations of motion with gravity g:
• v = u ± gt
• h = ut ± 1/2 gt²
• v² = u² ± 2gh
• For free fall, g is positive; for upward throw, g is negative.
• Maximum height: h = u²/(2g).
• Time of ascent = time of descent = u/g.
• Total time of flight = 2u/g.
• Velocity at projection point during fall = u.
• Velocity from height h: v = √(2gh).

Solved Examples

Ex 1: Calculate max height for u = 20 m/s, g = 10 m/s².

• h = u²/(2g) = 400/20 = 20 m.

Practice Set

Q1: Time of ascent for u = 15 m/s, g = 10 m/s² is?

• A) 1.5 s
• B) 2 s
• C) 3 s
• D) 1 s

Q2: Velocity after 3 s of free fall from rest is?

• A) 30 m/s
• B) 20 m/s
• C) 15 m/s
• D) 10 m/s

Q3: Total time of flight for u = 25 m/s is?

• A) 5 s
• B) 4 s
• C) 6 s
• D) 3 s

Answer Key

• Q1: A; Time of ascent = u/g = 15/10 = 1.5 s.
• Q2: A; v = gt = 10 × 3 = 30 m/s.
• Q3: A; Total time = 2u/g = 50/10 = 5 s.

Memory Aids

• Use v = u ± gt for velocity under gravity.
• Max height formula h = u²/(2g) is key for vertical throws.
• Total flight time is twice the ascent time.

Concepts and Formulas

• Equations of motion with gravity g:
• v = u ± gt
• h = ut ± 1/2 gt²
• v² = u² ± 2gh
• For free fall, g is positive; for upward throw, g is negative.
• Maximum height: h = u²/(2g).
• Time of ascent = time of descent = u/g.
• Total time of flight = 2u/g.
• Velocity at projection point during fall = u.
• Velocity from height h: v = √(2gh).

Solved Examples

Ex 1: Calculate max height for u = 20 m/s, g = 10 m/s².

• h = u²/(2g) = 400/20 = 20 m.

Practice Set

Q1: Time of ascent for u = 15 m/s, g = 10 m/s² is?

• A) 1.5 s
• B) 2 s
• C) 3 s
• D) 1 s

Q2: Velocity after 3 s of free fall from rest is?

• A) 30 m/s
• B) 20 m/s
• C) 15 m/s
• D) 10 m/s

Q3: Total time of flight for u = 25 m/s is?

• A) 5 s
• B) 4 s
• C) 6 s
• D) 3 s

Answer Key

• Q1: A; Time of ascent = u/g = 15/10 = 1.5 s.
• Q2: A; v = gt = 10 × 3 = 30 m/s.
• Q3: A; Total time = 2u/g = 50/10 = 5 s.

Memory Aids

• Use v = u ± gt for velocity under gravity.
• Max height formula h = u²/(2g) is key for vertical throws.
• Total flight time is twice the ascent time.

Concepts and Formulas

• Position in frame A: x_PA = x_PB + x_BA.
• Velocity relation: v_PA = v_PB + v_BA.
• Relative velocity of B w.r.t A (same direction): v_BA = v_B - v_A.
• Relative velocity (opposite direction): v_BA = v_B + v_A.
• Equal velocities mean zero relative velocity and parallel position-time graphs.
• Graphs show meeting times when velocities differ or are opposite.

Solved Examples

Ex 1: If v_A = 10 m/s, v_B = 15 m/s same direction, find v_BA.

• v_BA = 15 - 10 = 5 m/s.

Practice Set

Q1: Two objects move opposite directions with v_A = 8 m/s, v_B = 12 m/s. Relative velocity is?

• A) 4 m/s
• B) 20 m/s
• C) 8 m/s
• D) 12 m/s

Q2: If v_A = 7 m/s, v_B = 7 m/s same direction, relative velocity is?

• A) 0 m/s
• B) 14 m/s
• C) 7 m/s
• D) 1 m/s

Q3: If v_A = 5 m/s, v_B = 10 m/s opposite directions, relative velocity is?

• A) 5 m/s
• B) 15 m/s
• C) 10 m/s
• D) 20 m/s

Answer Key

• Q1: B; Opposite directions sum velocities: 8 + 12 = 20 m/s.
• Q2: A; Same velocity means zero relative velocity.
• Q3: B; Opposite directions sum velocities: 5 + 10 = 15 m/s.

Memory Aids

• Use v = u ± gt for velocity under gravity.
• Max height formula h = u²/(2g) is key for vertical throws.
• Total flight time is twice the ascent time.

Concepts and Formulas

• Equations of motion with gravity g:
• v = u ± gt
• h = ut ± 1/2 gt²
• v² = u² ± 2gh
• For free fall, g is positive; for upward throw, g is negative.
• Maximum height: h = u²/(2g).
• Time of ascent = time of descent = u/g.
• Total time of flight = 2u/g.
• Velocity at projection point during fall = u.
• Velocity from height h: v = √(2gh).

Solved Examples

Ex 1: Calculate max height for u = 20 m/s, g = 10 m/s².

• h = u²/(2g) = 400/20 = 20 m.

Practice Set

Q1: Time of ascent for u = 15 m/s, g = 10 m/s² is?

• A) 1.5 s
• B) 2 s
• C) 3 s
• D) 1 s

Q2: Velocity after 3 s of free fall from rest is?

• A) 30 m/s
• B) 20 m/s
• C) 15 m/s
• D) 10 m/s

Q3: Total time of flight for u = 25 m/s is?

• A) 5 s
• B) 4 s
• C) 6 s
• D) 3 s

Answer Key

• Q1: A; Time of ascent = u/g = 15/10 = 1.5 s.
• Q2: A; v = gt = 10 × 3 = 30 m/s.
• Q3: A; Total time = 2u/g = 50/10 = 5 s.

Memory Aids

• Use v = u ± gt for velocity under gravity.
• Max height formula h = u²/(2g) is key for vertical throws.
• Total flight time is twice the ascent time.

Concepts and Formulas

• Equations of motion with gravity g:
• v = u ± gt
• h = ut ± 1/2 gt²
• v² = u² ± 2gh
• For free fall, g is positive; for upward throw, g is negative.
• Maximum height: h = u²/(2g).
• Time of ascent = time of descent = u/g.
• Total time of flight = 2u/g.
• Velocity at projection point during fall = u.
• Velocity from height h: v = √(2gh).

Solved Examples

Ex 1: Calculate max height for u = 20 m/s, g = 10 m/s².

• h = u²/(2g) = 400/20 = 20 m.

Practice Set

Q1: Time of ascent for u = 15 m/s, g = 10 m/s² is?

• A) 1.5 s
• B) 2 s
• C) 3 s
• D) 1 s

Q2: Velocity after 3 s of free fall from rest is?

• A) 30 m/s
• B) 20 m/s
• C) 15 m/s
• D) 10 m/s

Q3: Total time of flight for u = 25 m/s is?

• A) 5 s
• B) 4 s
• C) 6 s
• D) 3 s

Answer Key

• Q1: A; Time of ascent = u/g = 15/10 = 1.5 s.
• Q2: A; v = gt = 10 × 3 = 30 m/s.
• Q3: A; Total time = 2u/g = 50/10 = 5 s.

Memory Aids

• Use v = u ± gt for velocity under gravity.
• Max height formula h = u²/(2g) is key for vertical throws.
• Total flight time is twice the ascent time.

Concepts and Formulas

• Equations of motion with gravity \(g\):
• \(v = u \pm gt\)
• \(h = ut \pm \frac{1}{2} gt^2\)
• \(v^2 = u^2 \pm 2gh\)
• For free fall, \(g\) is positive; for upward throw, \(g\) is negative.
• Maximum height: \(h = \frac{u^2}{2g}\).
• Time of ascent = time of descent = \(\frac{u}{g}\).
• Total time of flight = \(\frac{2u}{g}\).
• Velocity at projection point during fall = \(u\).
• Velocity from height \(h\): \(v = \sqrt{2gh}\).

Solved Examples

Ex 1: Calculate max height for \(u = 20\,m/s\), \(g = 10\,m/s^2\).

• \(h = \frac{u^2}{2g} = \frac{400}{20} = 20\,m\).

Practice Set

Q1: Time of ascent for \(u = 15\,m/s\), \(g = 10\,m/s^2\) is?

• A) 1.5 s
• B) 2 s
• C) 3 s
• D) 1 s

Q2: Velocity after 3 s of free fall from rest is?

• A) 30 m/s
• B) 20 m/s
• C) 15 m/s
• D) 10 m/s

Q3: Total time of flight for \(u = 25\,m/s\) is?

• A) 5 s
• B) 4 s
• C) 6 s
• D) 3 s

Answer Key

• Q1: A; Time of ascent = \(\frac{u}{g} = \frac{15}{10} = 1.5\,s\).
• Q2: A; \(v = gt = 10 \times 3 = 30\,m/s\).
• Q3: A; Total time = \(\frac{2u}{g} = \frac{50}{10} = 5\,s\).

Memory Aids

• Use \(v = u \pm gt\) for velocity under gravity.
• Max height formula \(h = \frac{u^2}{2g}\) is key for vertical throws.
• Total flight time is twice the ascent time.

Relative Velocity

Concepts and Formulas

• Position in frame A: x_PA = x_PB + x_BA.
• Velocity relation: v_PA = v_PB + v_BA.
• Relative velocity of B w.r.t A (same direction): v_BA = v_B - v_A.
• Relative velocity (opposite direction): v_BA = v_B + v_A.
• Equal velocities mean zero relative velocity and parallel position-time graphs.
• Graphs show meeting times when velocities differ or are opposite.

Solved Examples

Ex 1: If v_A = 10 m/s, v_B = 15 m/s same direction, find v_BA.

• v_BA = 15 - 10 = 5 m/s.

Practice Set

Q1: Two objects move opposite directions with v_A = 8 m/s, v_B = 12 m/s. Relative velocity is?

• A) 4 m/s
• B) 20 m/s
• C) 8 m/s
• D) 12 m/s

Q2: If v_A = 7 m/s, v_B = 7 m/s same direction, relative velocity is?

• A) 0 m/s
• B) 14 m/s
• C) 7 m/s
• D) 1 m/s

Q3: If v_A = 5 m/s, v_B = 10 m/s opposite directions, relative velocity is?

• A) 5 m/s
• B) 15 m/s
• C) 10 m/s
• D) 20 m/s

Answer Key

• Q1: B; Opposite directions sum velocities: 8 + 12 = 20 m/s.
• Q2: A; Same velocity means zero relative velocity.
• Q3: B; Opposite directions sum velocities: 5 + 10 = 15 m/s.

Memory Aids

• Relative velocity = difference if same direction.
• Relative velocity = sum if opposite directions.
• Zero relative velocity means objects move together.

Concepts and Formulas

• Position in frame A: x_PA = x_PB + x_BA.
• Velocity relation: v_PA = v_PB + v_BA.
• Relative velocity of B w.r.t A (same direction): v_BA = v_B - v_A.
• Relative velocity (opposite direction): v_BA = v_B + v_A.
• Equal velocities mean zero relative velocity and parallel position-time graphs.
• Graphs show meeting times when velocities differ or are opposite.

Solved Examples

Ex 1: If v_A = 10 m/s, v_B = 15 m/s same direction, find v_BA.

• v_BA = 15 - 10 = 5 m/s.

Practice Set

Q1: Two objects move opposite directions with v_A = 8 m/s, v_B = 12 m/s. Relative velocity is?

• A) 4 m/s
• B) 20 m/s
• C) 8 m/s
• D) 12 m/s

Q2: If v_A = 7 m/s, v_B = 7 m/s same direction, relative velocity is?

• A) 0 m/s
• B) 14 m/s
• C) 7 m/s
• D) 1 m/s

Q3: If v_A = 5 m/s, v_B = 10 m/s opposite directions, relative velocity is?

• A) 5 m/s
• B) 15 m/s
• C) 10 m/s
• D) 20 m/s

Answer Key

• Q1: B; Opposite directions sum velocities: 8 + 12 = 20 m/s.
• Q2: A; Same velocity means zero relative velocity.
• Q3: B; Opposite directions sum velocities: 5 + 10 = 15 m/s.

Memory Aids

• Relative velocity = difference if same direction.
• Relative velocity = sum if opposite directions.
• Zero relative velocity means objects move together.

Concepts and Formulas

• Position in frame A: x_PA = x_PB + x_BA.
• Velocity relation: v_PA = v_PB + v_BA.
• Relative velocity of B w.r.t A (same direction): v_BA = v_B - v_A.
• Relative velocity (opposite direction): v_BA = v_B + v_A.
• Equal velocities mean zero relative velocity and parallel position-time graphs.
• Graphs show meeting times when velocities differ or are opposite.

Solved Examples

Ex 1: If v_A = 10 m/s, v_B = 15 m/s same direction, find v_BA.

• v_BA = 15 - 10 = 5 m/s.

Practice Set

Q1: Two objects move opposite directions with v_A = 8 m/s, v_B = 12 m/s. Relative velocity is?

• A) 4 m/s
• B) 20 m/s
• C) 8 m/s
• D) 12 m/s

Q2: If v_A = 7 m/s, v_B = 7 m/s same direction, relative velocity is?

• A) 0 m/s
• B) 14 m/s
• C) 7 m/s
• D) 1 m/s

Q3: If v_A = 5 m/s, v_B = 10 m/s opposite directions, relative velocity is?

• A) 5 m/s
• B) 15 m/s
• C) 10 m/s
• D) 20 m/s

Answer Key

• Q1: B; Opposite directions sum velocities: 8 + 12 = 20 m/s.
• Q2: A; Same velocity means zero relative velocity.
• Q3: B; Opposite directions sum velocities: 5 + 10 = 15 m/s.

Memory Aids

• Relative velocity = difference if same direction.
• Relative velocity = sum if opposite directions.
• Zero relative velocity means objects move together.

Concepts and Formulas

• Position in frame A: x_PA = x_PB + x_BA.
• Velocity relation: v_PA = v_PB + v_BA.
• Relative velocity of B w.r.t A (same direction): v_BA = v_B - v_A.
• Relative velocity (opposite direction): v_BA = v_B + v_A.
• Equal velocities mean zero relative velocity and parallel position-time graphs.
• Graphs show meeting times when velocities differ or are opposite.

Solved Examples

Ex 1: If v_A = 10 m/s, v_B = 15 m/s same direction, find v_BA.

• v_BA = 15 - 10 = 5 m/s.

Practice Set

Q1: Two objects move opposite directions with v_A = 8 m/s, v_B = 12 m/s. Relative velocity is?

• A) 4 m/s
• B) 20 m/s
• C) 8 m/s
• D) 12 m/s

Q2: If v_A = 7 m/s, v_B = 7 m/s same direction, relative velocity is?

• A) 0 m/s
• B) 14 m/s
• C) 7 m/s
• D) 1 m/s

Q3: If v_A = 5 m/s, v_B = 10 m/s opposite directions, relative velocity is?

• A) 5 m/s
• B) 15 m/s
• C) 10 m/s
• D) 20 m/s

Answer Key

• Q1: B; Opposite directions sum velocities: 8 + 12 = 20 m/s.
• Q2: A; Same velocity means zero relative velocity.
• Q3: B; Opposite directions sum velocities: 5 + 10 = 15 m/s.

Memory Aids

• Relative velocity = difference if same direction.
• Relative velocity = sum if opposite directions.
• Zero relative velocity means objects move together.

Concepts and Formulas

• Position in frame A: x_PA = x_PB + x_BA.
• Velocity relation: v_PA = v_PB + v_BA.
• Relative velocity of B w.r.t A (same direction): v_BA = v_B - v_A.
• Relative velocity (opposite direction): v_BA = v_B + v_A.
• Equal velocities mean zero relative velocity and parallel position-time graphs.
• Graphs show meeting times when velocities differ or are opposite.

Solved Examples

Ex 1: If v_A = 10 m/s, v_B = 15 m/s same direction, find v_BA.

• v_BA = 15 - 10 = 5 m/s.

Practice Set

Q1: Two objects move opposite directions with v_A = 8 m/s, v_B = 12 m/s. Relative velocity is?

• A) 4 m/s
• B) 20 m/s
• C) 8 m/s
• D) 12 m/s

Q2: If v_A = 7 m/s, v_B = 7 m/s same direction, relative velocity is?

• A) 0 m/s
• B) 14 m/s
• C) 7 m/s
• D) 1 m/s

Q3: If v_A = 5 m/s, v_B = 10 m/s opposite directions, relative velocity is?

• A) 5 m/s
• B) 15 m/s
• C) 10 m/s
• D) 20 m/s

Answer Key

• Q1: B; Opposite directions sum velocities: 8 + 12 = 20 m/s.
• Q2: A; Same velocity means zero relative velocity.
• Q3: B; Opposite directions sum velocities: 5 + 10 = 15 m/s.

Memory Aids

• Relative velocity = difference if same direction.
• Relative velocity = sum if opposite directions.
• Zero relative velocity means objects move together.

Concepts and Formulas

• Position in frame A: x_PA = x_PB + x_BA.
• Velocity relation: v_PA = v_PB + v_BA.
• Relative velocity of B w.r.t A (same direction): v_BA = v_B - v_A.
• Relative velocity (opposite direction): v_BA = v_B + v_A.
• Equal velocities mean zero relative velocity and parallel position-time graphs.
• Graphs show meeting times when velocities differ or are opposite.

Solved Examples

Ex 1: If v_A = 10 m/s, v_B = 15 m/s same direction, find v_BA.

• v_BA = 15 - 10 = 5 m/s.

Practice Set

Q1: Two objects move opposite directions with v_A = 8 m/s, v_B = 12 m/s. Relative velocity is?

• A) 4 m/s
• B) 20 m/s
• C) 8 m/s
• D) 12 m/s

Q2: If v_A = 7 m/s, v_B = 7 m/s same direction, relative velocity is?

• A) 0 m/s
• B) 14 m/s
• C) 7 m/s
• D) 1 m/s

Q3: If v_A = 5 m/s, v_B = 10 m/s opposite directions, relative velocity is?

• A) 5 m/s
• B) 15 m/s
• C) 10 m/s
• D) 20 m/s

Answer Key

• Q1: B; Opposite directions sum velocities: 8 + 12 = 20 m/s.
• Q2: A; Same velocity means zero relative velocity.
• Q3: B; Opposite directions sum velocities: 5 + 10 = 15 m/s.

Memory Aids

• Relative velocity = difference if same direction.
• Relative velocity = sum if opposite directions.
• Zero relative velocity means objects move together.

Concepts and Formulas

• Position in frame A: x_PA = x_PB + x_BA.
• Velocity relation: v_PA = v_PB + v_BA.
• Relative velocity of B w.r.t A (same direction): v_BA = v_B - v_A.
• Relative velocity (opposite direction): v_BA = v_B + v_A.
• Equal velocities mean zero relative velocity and parallel position-time graphs.
• Graphs show meeting times when velocities differ or are opposite.

Solved Examples

Ex 1: If v_A = 10 m/s, v_B = 15 m/s same direction, find v_BA.

• v_BA = 15 - 10 = 5 m/s.

Practice Set

Q1: Two objects move opposite directions with v_A = 8 m/s, v_B = 12 m/s. Relative velocity is?

• A) 4 m/s
• B) 20 m/s
• C) 8 m/s
• D) 12 m/s

Q2: If v_A = 7 m/s, v_B = 7 m/s same direction, relative velocity is?

• A) 0 m/s
• B) 14 m/s
• C) 7 m/s
• D) 1 m/s

Q3: If v_A = 5 m/s, v_B = 10 m/s opposite directions, relative velocity is?

• A) 5 m/s
• B) 15 m/s
• C) 10 m/s
• D) 20 m/s

Answer Key

• Q1: B; Opposite directions sum velocities: 8 + 12 = 20 m/s.
• Q2: A; Same velocity means zero relative velocity.
• Q3: B; Opposite directions sum velocities: 5 + 10 = 15 m/s.

Memory Aids

• Relative velocity = difference if same direction.
• Relative velocity = sum if opposite directions.
• Zero relative velocity means objects move together.

Concepts and Formulas

• Position in frame A: \(x_{PA} = x_{PB} + x_{BA}\).
• Velocity relation: \(v_{PA} = v_{PB} + v_{BA}\).
• Relative velocity of B w.r.t A (same direction): \(v_{BA} = v_B - v_A\).
• Relative velocity (opposite direction): \(v_{BA} = v_B + v_A\).
• Equal velocities mean zero relative velocity and parallel position-time graphs.
• Graphs show meeting times when velocities differ or are opposite.

Solved Examples

Ex 1: If \(v_A = 10\,m/s\), \(v_B = 15\,m/s\) same direction, find \(v_{BA}\).

• \(v_{BA} = 15 - 10 = 5\,m/s\).

Practice Set

Q1: Two objects move opposite directions with \(v_A = 8\,m/s\), \(v_B = 12\,m/s\). Relative velocity is?

• A) 4 m/s
• B) 20 m/s
• C) 8 m/s
• D) 12 m/s

Q2: If \(v_A = 7\,m/s\), \(v_B = 7\,m/s\) same direction, relative velocity is?

• A) 0 m/s
• B) 14 m/s
• C) 7 m/s
• D) 1 m/s

Q3: If \(v_A = 5\,m/s\), \(v_B = 10\,m/s\) opposite directions, relative velocity is?

• A) 5 m/s
• B) 15 m/s
• C) 10 m/s
• D) 20 m/s

Answer Key

• Q1: B; Opposite directions sum velocities: 8 + 12 = 20 m/s.
• Q2: A; Same velocity means zero relative velocity.
• Q3: B; Opposite directions sum velocities: 5 + 10 = 15 m/s.

Memory Aids

• Relative velocity = difference if same direction.
• Relative velocity = sum if opposite directions.
• Zero relative velocity means objects move together.