Chapter 03: Motion in a Straight Line
Physics XI : Complete NCERT Exercise Solutions
Practice these examination-oriented questions to master the concepts of rectilinear motion, velocity, acceleration, and kinematic equations. Each question includes a detailed solution to enhance your understanding.
Question 3.1
In which of the following examples of motion, can the body be considered approximately a point object:
Answer & Explanation:
Correct Options: (a) and (b)
Explanation:
An object can be treated as a point object when its size is much smaller than the distance it moves in a reasonable time.
• (a) Railway carriage: The distance between stations is much larger than the size of the carriage, so it can be considered a point object.
• (b) Monkey on a cyclist: For motion along a circular track, if we're considering the motion of the cyclist+monkey system along the track, the size is negligible compared to the circumference of the track.
• (c) Spinning cricket ball: The spin and sharp turn involve rotational motion where the size and shape matter, so it cannot be treated as a point object.
• (d) Tumbling beaker: The tumbling motion involves rotation and the shape/size affects the motion, so it cannot be treated as a point object.
Explanation:
An object can be treated as a point object when its size is much smaller than the distance it moves in a reasonable time.
• (a) Railway carriage: The distance between stations is much larger than the size of the carriage, so it can be considered a point object.
• (b) Monkey on a cyclist: For motion along a circular track, if we're considering the motion of the cyclist+monkey system along the track, the size is negligible compared to the circumference of the track.
• (c) Spinning cricket ball: The spin and sharp turn involve rotational motion where the size and shape matter, so it cannot be treated as a point object.
• (d) Tumbling beaker: The tumbling motion involves rotation and the shape/size affects the motion, so it cannot be treated as a point object.
Question 3.2
The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 3.19. Choose the correct entries in the brackets below:
Answer & Explanation:
Answers:
(a) A lives closer to the school than B
(b) A starts from the school earlier than B
(c) B walks faster than A
(d) A and B reach home at the same time
(e) B overtakes A on the road once
Explanation:
From the x-t graph:
• A's home is at a smaller distance from school than B's home.
• A starts at time t=0 while B starts later.
• B's graph has a steeper slope, indicating higher speed.
• Both reach their homes at the same time.
• B's graph intersects A's graph once, indicating one overtaking.
(a) A lives closer to the school than B
(b) A starts from the school earlier than B
(c) B walks faster than A
(d) A and B reach home at the same time
(e) B overtakes A on the road once
Explanation:
From the x-t graph:
• A's home is at a smaller distance from school than B's home.
• A starts at time t=0 while B starts later.
• B's graph has a steeper slope, indicating higher speed.
• Both reach their homes at the same time.
• B's graph intersects A's graph once, indicating one overtaking.
Question 3.3
A woman starts from her home at 9.00 am, walks with a speed of 5 km h⁻¹ on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h⁻¹. Choose suitable scales and plot the x-t graph of her motion.
Answer & Explanation:
Graph Description:
• Time scale: 1 hour = 1-2 cm on graph paper
• Distance scale: 1 km = 1-2 cm on graph paper
Calculations:
1. Going to office: Time = Distance/Speed = 2.5/5 = 0.5 hour = 30 min.
She reaches office at 9:30 am.
2. At office: From 9:30 am to 5:00 pm = 7.5 hours stationary at x = 2.5 km.
3. Returning home: Time = 2.5/25 = 0.1 hour = 6 min.
She reaches home at 5:06 pm.
Graph Features:
• From (9:00, 0) to (9:30, 2.5): Straight line with positive slope (5 km/h)
• From (9:30, 2.5) to (17:00, 2.5): Horizontal line (stationary)
• From (17:00, 2.5) to (17:06, 0): Straight line with negative slope (25 km/h)
• Time scale: 1 hour = 1-2 cm on graph paper
• Distance scale: 1 km = 1-2 cm on graph paper
Calculations:
1. Going to office: Time = Distance/Speed = 2.5/5 = 0.5 hour = 30 min.
She reaches office at 9:30 am.
2. At office: From 9:30 am to 5:00 pm = 7.5 hours stationary at x = 2.5 km.
3. Returning home: Time = 2.5/25 = 0.1 hour = 6 min.
She reaches home at 5:06 pm.
Graph Features:
• From (9:00, 0) to (9:30, 2.5): Straight line with positive slope (5 km/h)
• From (9:30, 2.5) to (17:00, 2.5): Horizontal line (stationary)
• From (17:00, 2.5) to (17:06, 0): Straight line with negative slope (25 km/h)
Question 3.4
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Answer & Explanation:
Analysis:
• Each cycle: 5 steps forward + 3 steps backward = net 2 steps forward = 2 m forward
• Time per cycle: 5 + 3 = 8 seconds
Position after each cycle:
After 1st cycle: 2 m at t = 8 s
After 2nd cycle: 4 m at t = 16 s
After 3rd cycle: 6 m at t = 24 s
After 4th cycle: 8 m at t = 32 s
After 5th cycle: 10 m at t = 40 s
To reach 13 m:
After 5 cycles: He's at 10 m at t = 40 s
Now he takes 5 steps forward (5 m) = reaches 15 m
But pit is at 13 m, so he falls during these 5 steps.
Time calculation:
From 10 m to 13 m = 3 steps forward = 3 seconds
Total time = 40 + 3 = 43 seconds
Graph: Zigzag pattern with net forward drift, reaching 13 m at t = 43 s.
• Each cycle: 5 steps forward + 3 steps backward = net 2 steps forward = 2 m forward
• Time per cycle: 5 + 3 = 8 seconds
Position after each cycle:
After 1st cycle: 2 m at t = 8 s
After 2nd cycle: 4 m at t = 16 s
After 3rd cycle: 6 m at t = 24 s
After 4th cycle: 8 m at t = 32 s
After 5th cycle: 10 m at t = 40 s
To reach 13 m:
After 5 cycles: He's at 10 m at t = 40 s
Now he takes 5 steps forward (5 m) = reaches 15 m
But pit is at 13 m, so he falls during these 5 steps.
Time calculation:
From 10 m to 13 m = 3 steps forward = 3 seconds
Total time = 40 + 3 = 43 seconds
Graph: Zigzag pattern with net forward drift, reaching 13 m at t = 43 s.
Question 3.5
A jet airplane travelling at the speed of 500 km h⁻¹ ejects its products of combustion at the speed of 1500 km h⁻¹ relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
Answer & Explanation:
Given:
• Speed of jet plane, vJ = 500 km/h (relative to ground)
• Speed of combustion products relative to jet plane = 1500 km/h
Relative velocity concept:
vPG = vPJ + vJG
Where:
vPG = velocity of products relative to ground
vPJ = velocity of products relative to jet = 1500 km/h
vJG = velocity of jet relative to ground = 500 km/h
Calculation:
Since the products are ejected backwards (opposite to plane's motion):
vPJ = -1500 km/h (relative to jet)
vPG = (-1500) + 500 = -1000 km/h
Answer: The combustion products move at 1000 km/h in the opposite direction to the jet's motion relative to ground.
• Speed of jet plane, vJ = 500 km/h (relative to ground)
• Speed of combustion products relative to jet plane = 1500 km/h
Relative velocity concept:
vPG = vPJ + vJG
Where:
vPG = velocity of products relative to ground
vPJ = velocity of products relative to jet = 1500 km/h
vJG = velocity of jet relative to ground = 500 km/h
Calculation:
Since the products are ejected backwards (opposite to plane's motion):
vPJ = -1500 km/h (relative to jet)
vPG = (-1500) + 500 = -1000 km/h
Answer: The combustion products move at 1000 km/h in the opposite direction to the jet's motion relative to ground.
Question 3.6
A car moving along a straight highway with speed of 126 km h⁻¹ is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
Answer & Explanation:
Given:
• Initial speed, u = 126 km/h = 126 × (1000/3600) = 35 m/s
• Final speed, v = 0 m/s
• Distance, s = 200 m
• Retardation (negative acceleration), a = ?
• Time, t = ?
Using kinematic equation: v² = u² + 2as
0 = (35)² + 2 × a × 200
0 = 1225 + 400a
a = -1225/400 = -3.0625 m/s²
Retardation magnitude: 3.0625 m/s²
Time to stop: v = u + at
0 = 35 + (-3.0625) × t
t = 35/3.0625 ≈ 11.43 s
Answer: Retardation = 3.0625 m/s², Time to stop = 11.43 s
• Initial speed, u = 126 km/h = 126 × (1000/3600) = 35 m/s
• Final speed, v = 0 m/s
• Distance, s = 200 m
• Retardation (negative acceleration), a = ?
• Time, t = ?
Using kinematic equation: v² = u² + 2as
0 = (35)² + 2 × a × 200
0 = 1225 + 400a
a = -1225/400 = -3.0625 m/s²
Retardation magnitude: 3.0625 m/s²
Time to stop: v = u + at
0 = 35 + (-3.0625) × t
t = 35/3.0625 ≈ 11.43 s
Answer: Retardation = 3.0625 m/s², Time to stop = 11.43 s
Question 3.7
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h⁻¹ in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s⁻². If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?
Answer & Explanation:
Given:
• Length of each train = 400 m
• Initial speed of both trains = 72 km/h = 20 m/s
• Acceleration of train B = 1 m/s²
• Time for overtaking = 50 s
Concept: For B to just brush past A, B must cover:
Distance = Original separation + Length of train A + Length of train B
= d + 400 + 400 = d + 800 m
Relative motion approach:
Initial relative velocity = 0 (both have same speed)
Relative acceleration = 1 m/s²
Relative distance to be covered = d + 800 m
Using: s = ut + ½at² (in relative terms)
d + 800 = 0 × 50 + ½ × 1 × (50)²
d + 800 = ½ × 2500 = 1250
d = 1250 - 800 = 450 m
Answer: Original distance between trains = 450 m
• Length of each train = 400 m
• Initial speed of both trains = 72 km/h = 20 m/s
• Acceleration of train B = 1 m/s²
• Time for overtaking = 50 s
Concept: For B to just brush past A, B must cover:
Distance = Original separation + Length of train A + Length of train B
= d + 400 + 400 = d + 800 m
Relative motion approach:
Initial relative velocity = 0 (both have same speed)
Relative acceleration = 1 m/s²
Relative distance to be covered = d + 800 m
Using: s = ut + ½at² (in relative terms)
d + 800 = 0 × 50 + ½ × 1 × (50)²
d + 800 = ½ × 2500 = 1250
d = 1250 - 800 = 450 m
Answer: Original distance between trains = 450 m
Question 3.8
On a two-lane road, car A is travelling with a speed of 36 km h⁻¹. Two cars B and C approach car A in opposite directions with a speed of 54 km h⁻¹ each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
Answer & Explanation:
Given:
• Speed of A, vA = 36 km/h = 10 m/s
• Speed of B and C, vB = vC = 54 km/h = 15 m/s
• Initial distances: AB = AC = 1 km = 1000 m
• B and C are on opposite sides of A
Analysis:
Time for C to reach A: tC = Distance/Relative speed
Relative speed of C wrt A = 15 + 10 = 25 m/s (opposite directions)
tC = 1000/25 = 40 s
B must overtake A in less than 40 seconds to avoid collision with C.
For B to overtake A:
Relative initial speed of B wrt A = 15 - 10 = 5 m/s
Relative distance to be covered for overtaking = length of cars (say L each)
But for safe overtaking, B should get sufficiently ahead of A.
Let's assume B needs to be at least 100 m ahead of A after overtaking.
So total relative distance = 100 m (safe gap)
Using: s = ut + ½at²
100 = 5 × 40 + ½ × a × (40)²
100 = 200 + ½ × a × 1600
100 = 200 + 800a
800a = -100
a = -0.125 m/s² (This is negative, meaning our assumption needs correction)
Correct approach: B must cover 1000 m relative to A in less than 40 s:
1000 = 5 × t + ½ × a × t², with t < 40
Also, after overtaking, B should be ahead when C arrives.
This requires solving two conditions simultaneously.
Simplified answer: The minimum acceleration needed is approximately 1 m/s² (detailed calculation involves quadratic equations and inequality conditions).
• Speed of A, vA = 36 km/h = 10 m/s
• Speed of B and C, vB = vC = 54 km/h = 15 m/s
• Initial distances: AB = AC = 1 km = 1000 m
• B and C are on opposite sides of A
Analysis:
Time for C to reach A: tC = Distance/Relative speed
Relative speed of C wrt A = 15 + 10 = 25 m/s (opposite directions)
tC = 1000/25 = 40 s
B must overtake A in less than 40 seconds to avoid collision with C.
For B to overtake A:
Relative initial speed of B wrt A = 15 - 10 = 5 m/s
Relative distance to be covered for overtaking = length of cars (say L each)
But for safe overtaking, B should get sufficiently ahead of A.
Let's assume B needs to be at least 100 m ahead of A after overtaking.
So total relative distance = 100 m (safe gap)
Using: s = ut + ½at²
100 = 5 × 40 + ½ × a × (40)²
100 = 200 + ½ × a × 1600
100 = 200 + 800a
800a = -100
a = -0.125 m/s² (This is negative, meaning our assumption needs correction)
Correct approach: B must cover 1000 m relative to A in less than 40 s:
1000 = 5 × t + ½ × a × t², with t < 40
Also, after overtaking, B should be ahead when C arrives.
This requires solving two conditions simultaneously.
Simplified answer: The minimum acceleration needed is approximately 1 m/s² (detailed calculation involves quadratic equations and inequality conditions).
Question 3.9
Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h⁻¹ in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?
Answer & Explanation:
Given:
• Cyclist speed, vc = 20 km/h
• Time between buses in same direction = 18 min = 18/60 = 0.3 h
• Time between buses in opposite direction = 6 min = 6/60 = 0.1 h
• Bus period = T minutes = T/60 hours
• Bus speed = vb (to find)
Relative speed concepts:
1. Buses in same direction: Relative speed = vb - vc
Distance between consecutive buses = vb × (T/60)
This distance is covered at relative speed (vb - vc) in 0.3 h:
vb × (T/60) = (vb - vc) × 0.3 ...(1)
2. Buses in opposite direction: Relative speed = vb + vc
Distance between consecutive buses = vb × (T/60)
This distance is covered at relative speed (vb + vc) in 0.1 h:
vb × (T/60) = (vb + vc) × 0.1 ...(2)
From (1) and (2):
(vb - 20) × 0.3 = (vb + 20) × 0.1
0.3vb - 6 = 0.1vb + 2
0.2vb = 8
vb = 40 km/h
From equation (2):
40 × (T/60) = (40 + 20) × 0.1
(40T/60) = 60 × 0.1 = 6
2T/3 = 6
T = 9 minutes
Answer: Bus speed = 40 km/h, Period T = 9 minutes
• Cyclist speed, vc = 20 km/h
• Time between buses in same direction = 18 min = 18/60 = 0.3 h
• Time between buses in opposite direction = 6 min = 6/60 = 0.1 h
• Bus period = T minutes = T/60 hours
• Bus speed = vb (to find)
Relative speed concepts:
1. Buses in same direction: Relative speed = vb - vc
Distance between consecutive buses = vb × (T/60)
This distance is covered at relative speed (vb - vc) in 0.3 h:
vb × (T/60) = (vb - vc) × 0.3 ...(1)
2. Buses in opposite direction: Relative speed = vb + vc
Distance between consecutive buses = vb × (T/60)
This distance is covered at relative speed (vb + vc) in 0.1 h:
vb × (T/60) = (vb + vc) × 0.1 ...(2)
From (1) and (2):
(vb - 20) × 0.3 = (vb + 20) × 0.1
0.3vb - 6 = 0.1vb + 2
0.2vb = 8
vb = 40 km/h
From equation (2):
40 × (T/60) = (40 + 20) × 0.1
(40T/60) = 60 × 0.1 = 6
2T/3 = 6
T = 9 minutes
Answer: Bus speed = 40 km/h, Period T = 9 minutes
Question 3.10
A player throws a ball upwards with an initial speed of 29.4 m s⁻¹.
Answer & Explanation:
(a) Direction of acceleration during upward motion:
Downward (acceleration due to gravity always acts downward)
(b) At highest point:
• Velocity = 0 m/s (momentarily at rest)
• Acceleration = 9.8 m/s² downward
(c) Signs with downward as positive:
During upward motion:
• Position: Negative (ball is above reference point)
• Velocity: Negative (moving upward, opposite to positive direction)
• Acceleration: Positive (gravity acts downward, which is positive direction)
During downward motion:
• Position: Negative initially, becoming less negative
• Velocity: Positive (moving downward, same as positive direction)
• Acceleration: Positive (gravity acts downward)
(d) Calculations:
• Maximum height: h = u²/(2g) = (29.4)²/(2 × 9.8) = 864.36/19.6 = 44.1 m
• Time to return: Total time = 2u/g = 2 × 29.4/9.8 = 58.8/9.8 = 6 s
Answer: Height = 44.1 m, Time to return = 6 s
Downward (acceleration due to gravity always acts downward)
(b) At highest point:
• Velocity = 0 m/s (momentarily at rest)
• Acceleration = 9.8 m/s² downward
(c) Signs with downward as positive:
During upward motion:
• Position: Negative (ball is above reference point)
• Velocity: Negative (moving upward, opposite to positive direction)
• Acceleration: Positive (gravity acts downward, which is positive direction)
During downward motion:
• Position: Negative initially, becoming less negative
• Velocity: Positive (moving downward, same as positive direction)
• Acceleration: Positive (gravity acts downward)
(d) Calculations:
• Maximum height: h = u²/(2g) = (29.4)²/(2 × 9.8) = 864.36/19.6 = 44.1 m
• Time to return: Total time = 2u/g = 2 × 29.4/9.8 = 58.8/9.8 = 6 s
Answer: Height = 44.1 m, Time to return = 6 s
Question 3.11
Read each statement below carefully and state with reasons and examples, if it is true or false:
A particle in one-dimensional motion:
A particle in one-dimensional motion:
Answer & Explanation:
(a) True
Example: A ball thrown vertically upward has zero speed at the highest point, but acceleration due to gravity (9.8 m/s² downward) is still acting.
(b) False
Reason: Speed is the magnitude of velocity. Zero speed means magnitude is zero, so velocity must be zero.
(c) False
Example: Uniform circular motion has constant speed but non-zero acceleration (centripetal acceleration). Even in 1D, if direction changes (like a ball bouncing back), speed can be constant but acceleration exists at turning points.
(d) False
Reason: In 1D motion, whether an object is speeding up depends on the sign of both velocity and acceleration. If velocity is negative and acceleration is positive, the object is actually slowing down.
• Speeding up: v and a have same sign
• Slowing down: v and a have opposite signs
Example: A ball thrown vertically upward has zero speed at the highest point, but acceleration due to gravity (9.8 m/s² downward) is still acting.
(b) False
Reason: Speed is the magnitude of velocity. Zero speed means magnitude is zero, so velocity must be zero.
(c) False
Example: Uniform circular motion has constant speed but non-zero acceleration (centripetal acceleration). Even in 1D, if direction changes (like a ball bouncing back), speed can be constant but acceleration exists at turning points.
(d) False
Reason: In 1D motion, whether an object is speeding up depends on the sign of both velocity and acceleration. If velocity is negative and acceleration is positive, the object is actually slowing down.
• Speeding up: v and a have same sign
• Slowing down: v and a have opposite signs
Question 3.12
A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Answer & Explanation:
Step 1: First fall
Initial height = 90 m, u = 0, a = g = 9.8 m/s²
v² = u² + 2as
v² = 0 + 2 × 9.8 × 90 = 1764
v = √1764 = 42 m/s (speed just before first impact)
Time for first fall: v = u + at ⇒ 42 = 0 + 9.8t ⇒ t = 42/9.8 ≈ 4.29 s
Step 2: First bounce
Speed after collision = 42 × 0.9 = 37.8 m/s (upward)
Time to reach maximum height: v = u + at ⇒ 0 = 37.8 - 9.8t ⇒ t = 3.86 s
Maximum height: h = ut + ½at² = 37.8×3.86 - ½×9.8×(3.86)² ≈ 72.9 m
Step 3: Second fall and subsequent bounces
Continue similarly with decreasing speeds.
After first bounce: new speed = 37.8 m/s downward
Time for second fall from 72.9 m: t = √(2h/g) = √(2×72.9/9.8) ≈ 3.86 s
Speed before second impact: v = √(2×9.8×72.9) = 37.8 m/s
After second collision: speed = 37.8 × 0.9 = 34.02 m/s upward
Graph Features:
• Straight lines with slope ±9.8 m/s²
• Positive slopes during downward motion
• Negative slopes during upward motion
• Decreasing peaks with each bounce
• Total time for multiple bounces within 12 s
Initial height = 90 m, u = 0, a = g = 9.8 m/s²
v² = u² + 2as
v² = 0 + 2 × 9.8 × 90 = 1764
v = √1764 = 42 m/s (speed just before first impact)
Time for first fall: v = u + at ⇒ 42 = 0 + 9.8t ⇒ t = 42/9.8 ≈ 4.29 s
Step 2: First bounce
Speed after collision = 42 × 0.9 = 37.8 m/s (upward)
Time to reach maximum height: v = u + at ⇒ 0 = 37.8 - 9.8t ⇒ t = 3.86 s
Maximum height: h = ut + ½at² = 37.8×3.86 - ½×9.8×(3.86)² ≈ 72.9 m
Step 3: Second fall and subsequent bounces
Continue similarly with decreasing speeds.
After first bounce: new speed = 37.8 m/s downward
Time for second fall from 72.9 m: t = √(2h/g) = √(2×72.9/9.8) ≈ 3.86 s
Speed before second impact: v = √(2×9.8×72.9) = 37.8 m/s
After second collision: speed = 37.8 × 0.9 = 34.02 m/s upward
Graph Features:
• Straight lines with slope ±9.8 m/s²
• Positive slopes during downward motion
• Negative slopes during upward motion
• Decreasing peaks with each bounce
• Total time for multiple bounces within 12 s
Question 3.13
Explain clearly, with examples, the distinction between:
Answer & Explanation:
(a) Magnitude of displacement vs. Path length:
• Magnitude of displacement: Shortest distance between initial and final positions. It's a straight-line distance regardless of actual path.
• Path length: Actual total distance traveled along the path.
Example: A car goes from point A to point B (10 km apart) via a circular route of 15 km and returns to A.
• Magnitude of displacement = 0 km (returns to starting point)
• Path length = 30 km (15 km each way)
(b) Magnitude of average velocity vs. Average speed:
• Magnitude of average velocity: |Displacement|/Time interval
• Average speed: Path length/Time interval
Example: Same car example above taking 1 hour total.
• |Average velocity| = 0/1 = 0 km/h
• Average speed = 30/1 = 30 km/h
Equality condition: Both quantities are equal only when motion is in a straight line without change in direction.
• Magnitude of displacement: Shortest distance between initial and final positions. It's a straight-line distance regardless of actual path.
• Path length: Actual total distance traveled along the path.
Example: A car goes from point A to point B (10 km apart) via a circular route of 15 km and returns to A.
• Magnitude of displacement = 0 km (returns to starting point)
• Path length = 30 km (15 km each way)
(b) Magnitude of average velocity vs. Average speed:
• Magnitude of average velocity: |Displacement|/Time interval
• Average speed: Path length/Time interval
Example: Same car example above taking 1 hour total.
• |Average velocity| = 0/1 = 0 km/h
• Average speed = 30/1 = 30 km/h
Equality condition: Both quantities are equal only when motion is in a straight line without change in direction.
Question 3.14
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h⁻¹. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h⁻¹. What is the:
Answer & Explanation:
Given:
• Distance to market = 2.5 km
• Speed to market = 5 km/h
• Speed returning = 7.5 km/h
Time calculations:
Time to market = 2.5/5 = 0.5 h = 30 min
Time to return = 2.5/7.5 = 1/3 h = 20 min
Total time for round trip = 30 + 20 = 50 min
(i) 0 to 30 min (reaches market):
• Displacement = 2.5 km from home
• Path length = 2.5 km
• Time = 30 min = 0.5 h
(a) |Average velocity| = 2.5/0.5 = 5 km/h
(b) Average speed = 2.5/0.5 = 5 km/h
(ii) 0 to 50 min (returns home):
• Displacement = 0 km (back at home)
• Path length = 5 km (2.5 + 2.5)
• Time = 50 min = 5/6 h
(a) |Average velocity| = 0/(5/6) = 0 km/h
(b) Average speed = 5/(5/6) = 6 km/h
(iii) 0 to 40 min (on return journey):
At 40 min: 30 min to market + 10 min returning
Distance returned in 10 min = 7.5 × (10/60) = 1.25 km
Position from home = 2.5 - 1.25 = 1.25 km
Path length = 2.5 + 1.25 = 3.75 km
Time = 40 min = 2/3 h
(a) |Average velocity| = 1.25/(2/3) = 1.875 km/h
(b) Average speed = 3.75/(2/3) = 5.625 km/h
• Distance to market = 2.5 km
• Speed to market = 5 km/h
• Speed returning = 7.5 km/h
Time calculations:
Time to market = 2.5/5 = 0.5 h = 30 min
Time to return = 2.5/7.5 = 1/3 h = 20 min
Total time for round trip = 30 + 20 = 50 min
(i) 0 to 30 min (reaches market):
• Displacement = 2.5 km from home
• Path length = 2.5 km
• Time = 30 min = 0.5 h
(a) |Average velocity| = 2.5/0.5 = 5 km/h
(b) Average speed = 2.5/0.5 = 5 km/h
(ii) 0 to 50 min (returns home):
• Displacement = 0 km (back at home)
• Path length = 5 km (2.5 + 2.5)
• Time = 50 min = 5/6 h
(a) |Average velocity| = 0/(5/6) = 0 km/h
(b) Average speed = 5/(5/6) = 6 km/h
(iii) 0 to 40 min (on return journey):
At 40 min: 30 min to market + 10 min returning
Distance returned in 10 min = 7.5 × (10/60) = 1.25 km
Position from home = 2.5 - 1.25 = 1.25 km
Path length = 2.5 + 1.25 = 3.75 km
Time = 40 min = 2/3 h
(a) |Average velocity| = 1.25/(2/3) = 1.875 km/h
(b) Average speed = 3.75/(2/3) = 5.625 km/h
Question 3.15
In Exercises 3.13 and 3.14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
Answer & Explanation:
Reason:
Instantaneous speed is defined as the magnitude of instantaneous velocity.
Mathematically:
• Instantaneous velocity, v = dx/dt (in 1D, with sign indicating direction)
• Instantaneous speed = |v| = |dx/dt|
Conceptual explanation:
At an instant, we consider an infinitesimally small time interval. Over such a small interval:
1. The direction of motion doesn't change
2. The path is essentially straight
3. The distance traveled equals the magnitude of displacement
Example: In circular motion, even though the average speed and |average velocity| differ over a full circle, at any instant, the instantaneous speed equals the magnitude of instantaneous velocity because we're considering motion along the tangent direction at that point.
Instantaneous speed is defined as the magnitude of instantaneous velocity.
Mathematically:
• Instantaneous velocity, v = dx/dt (in 1D, with sign indicating direction)
• Instantaneous speed = |v| = |dx/dt|
Conceptual explanation:
At an instant, we consider an infinitesimally small time interval. Over such a small interval:
1. The direction of motion doesn't change
2. The path is essentially straight
3. The distance traveled equals the magnitude of displacement
Example: In circular motion, even though the average speed and |average velocity| differ over a full circle, at any instant, the instantaneous speed equals the magnitude of instantaneous velocity because we're considering motion along the tangent direction at that point.
Question 3.16
Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
Answer & Explanation:
(b), (c), and (d) cannot represent 1D motion
Analysis:
(a) Speed-time graph with negative speed: Possible. Speed is magnitude, so cannot be negative. But if the graph shows "speed" as negative, it's actually showing velocity with sign. This is possible for 1D motion.
(b) Total path length decreasing with time: Impossible. Path length is cumulative distance traveled, so it can never decrease. It either increases or stays constant (when particle is at rest).
(c) Velocity-time graph with two different velocities at same time: Impossible. At any instant, a particle can have only one velocity in 1D motion.
(d) Particle at two different positions at same time: Impossible. A particle cannot be at two different positions at the same instant in time.
Analysis:
(a) Speed-time graph with negative speed: Possible. Speed is magnitude, so cannot be negative. But if the graph shows "speed" as negative, it's actually showing velocity with sign. This is possible for 1D motion.
(b) Total path length decreasing with time: Impossible. Path length is cumulative distance traveled, so it can never decrease. It either increases or stays constant (when particle is at rest).
(c) Velocity-time graph with two different velocities at same time: Impossible. At any instant, a particle can have only one velocity in 1D motion.
(d) Particle at two different positions at same time: Impossible. A particle cannot be at two different positions at the same instant in time.
Question 3.17
Figure 3.21 shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.
Answer & Explanation:
No, it's not correct.
Reason: An x-t graph shows position vs. time, not the actual path of the particle. The particle is moving in one dimension (straight line) throughout.
What the graph shows:
• For t < 0: Straight line in x-t graph ⇒ uniform motion (constant velocity)
• For t > 0: Parabolic curve in x-t graph ⇒ uniformly accelerated motion
Suitable physical context:
A car moving on a straight road:
1. Before t = 0: Moving with constant velocity
2. At t = 0: Starts accelerating uniformly (maybe pressing accelerator)
3. For t > 0: Moving with constant acceleration
Equation form:
For t < 0: x = x₀ + vt (linear)
For t > 0: x = x₀ + v₀t + ½at² (parabolic)
Reason: An x-t graph shows position vs. time, not the actual path of the particle. The particle is moving in one dimension (straight line) throughout.
What the graph shows:
• For t < 0: Straight line in x-t graph ⇒ uniform motion (constant velocity)
• For t > 0: Parabolic curve in x-t graph ⇒ uniformly accelerated motion
Suitable physical context:
A car moving on a straight road:
1. Before t = 0: Moving with constant velocity
2. At t = 0: Starts accelerating uniformly (maybe pressing accelerator)
3. For t > 0: Moving with constant acceleration
Equation form:
For t < 0: x = x₀ + vt (linear)
For t > 0: x = x₀ + v₀t + ½at² (parabolic)
Question 3.18
A police van moving on a highway with a speed of 30 km h⁻¹ fires a bullet at a thief's car speeding away in the same direction with a speed of 192 km h⁻¹. If the muzzle speed of the bullet is 150 m s⁻¹, with what speed does the bullet hit the thief's car? (Note: Obtain that speed which is relevant for damaging the thief's car)
Answer & Explanation:
Given:
• Police van speed, vp = 30 km/h = 30 × (1000/3600) = 25/3 ≈ 8.33 m/s
• Thief's car speed, vt = 192 km/h = 192 × (1000/3600) = 160/3 ≈ 53.33 m/s
• Muzzle speed of bullet relative to police van = 150 m/s
Relative velocity approach:
Speed of bullet relative to ground:
vbg = vp + 150 = 8.33 + 150 = 158.33 m/s
Speed of bullet relative to thief's car:
vbt = vbg - vt = 158.33 - 53.33 = 105 m/s
Answer: The bullet hits the thief's car with a speed of 105 m/s relative to the car.
Note: This relative speed (105 m/s) is what matters for damage, as it determines the impact energy relative to the car.
• Police van speed, vp = 30 km/h = 30 × (1000/3600) = 25/3 ≈ 8.33 m/s
• Thief's car speed, vt = 192 km/h = 192 × (1000/3600) = 160/3 ≈ 53.33 m/s
• Muzzle speed of bullet relative to police van = 150 m/s
Relative velocity approach:
Speed of bullet relative to ground:
vbg = vp + 150 = 8.33 + 150 = 158.33 m/s
Speed of bullet relative to thief's car:
vbt = vbg - vt = 158.33 - 53.33 = 105 m/s
Answer: The bullet hits the thief's car with a speed of 105 m/s relative to the car.
Note: This relative speed (105 m/s) is what matters for damage, as it determines the impact energy relative to the car.
Question 3.19
Suggest a suitable physical situation for each of the following graphs (Fig 3.22):
Answer & Explanation:
(a) x-t graph with positive curvature:
Situation: A car starting from rest and accelerating uniformly in positive direction.
Equation: x = ½at² (parabolic curve opening upward)
(b) v-t graph with positive constant acceleration then zero acceleration:
Situation: A rocket launching:
1. First phase: Accelerating upward with constant thrust (positive constant slope)
2. Second phase: Engine cuts off, moving with constant velocity (zero slope)
(c) a-t graph with positive constant acceleration then negative constant acceleration:
Situation: A car on a straight road:
1. Accelerating uniformly from rest (constant positive a)
2. Then applying brakes for uniform deceleration (constant negative a)
Alternative for (c): A ball thrown upward:
1. During upward motion: Constant acceleration = -g (if upward is positive)
2. During downward motion: Constant acceleration = -g (same)
(This would show constant negative a throughout, not matching the described graph)
Situation: A car starting from rest and accelerating uniformly in positive direction.
Equation: x = ½at² (parabolic curve opening upward)
(b) v-t graph with positive constant acceleration then zero acceleration:
Situation: A rocket launching:
1. First phase: Accelerating upward with constant thrust (positive constant slope)
2. Second phase: Engine cuts off, moving with constant velocity (zero slope)
(c) a-t graph with positive constant acceleration then negative constant acceleration:
Situation: A car on a straight road:
1. Accelerating uniformly from rest (constant positive a)
2. Then applying brakes for uniform deceleration (constant negative a)
Alternative for (c): A ball thrown upward:
1. During upward motion: Constant acceleration = -g (if upward is positive)
2. During downward motion: Constant acceleration = -g (same)
(This would show constant negative a throughout, not matching the described graph)
Question 3.20
Figure 3.23 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, -1.2 s.
Answer & Explanation:
For Simple Harmonic Motion (SHM):
• x = A sin(ωt) or x = A cos(ωt)
• v = dx/dt
• a = dv/dt = -ω²x (acceleration proportional to displacement and opposite in direction)
Assuming x = A sin(ωt) with A > 0:
At t = 0.3 s (assuming it's in first quarter period):
• Position (x): Positive (sin positive in 0 to π/2)
• Velocity (v): Positive (cos positive in 0 to π/2)
• Acceleration (a): Negative (a = -ω²x, opposite to x)
At t = 1.2 s (assuming it's in second quarter period):
• Position (x): Positive (sin positive in π/2 to π)
• Velocity (v): Negative (cos negative in π/2 to π)
• Acceleration (a): Negative (a = -ω²x, opposite to x)
At t = -1.2 s (negative time, going backward):
For x = A sin(ωt), sin(-θ) = -sin(θ)
• Position (x): Negative
• Velocity (v): cos(-θ) = cos(θ) ⇒ Positive (if in fourth quadrant)
• Acceleration (a): Positive (a = -ω²x, opposite to negative x gives positive)
• x = A sin(ωt) or x = A cos(ωt)
• v = dx/dt
• a = dv/dt = -ω²x (acceleration proportional to displacement and opposite in direction)
Assuming x = A sin(ωt) with A > 0:
At t = 0.3 s (assuming it's in first quarter period):
• Position (x): Positive (sin positive in 0 to π/2)
• Velocity (v): Positive (cos positive in 0 to π/2)
• Acceleration (a): Negative (a = -ω²x, opposite to x)
At t = 1.2 s (assuming it's in second quarter period):
• Position (x): Positive (sin positive in π/2 to π)
• Velocity (v): Negative (cos negative in π/2 to π)
• Acceleration (a): Negative (a = -ω²x, opposite to x)
At t = -1.2 s (negative time, going backward):
For x = A sin(ωt), sin(-θ) = -sin(θ)
• Position (x): Negative
• Velocity (v): cos(-θ) = cos(θ) ⇒ Positive (if in fourth quadrant)
• Acceleration (a): Positive (a = -ω²x, opposite to negative x gives positive)
Question 3.21
Figure 3.24 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.
Answer & Explanation:
Analysis from x-t graph:
Average speed = Total path length/Time interval
|Average velocity| = |Displacement|/Time interval
Comparing three equal time intervals:
Interval with greatest average speed: Where the magnitude of slope changes most (steepest parts of curve), indicating highest instantaneous speed.
Interval with least average speed: Where the curve is flattest (smallest slope magnitude).
Sign of average velocity:
• Positive average velocity: When x increases with time (positive slope)
• Negative average velocity: When x decreases with time (negative slope)
• Zero average velocity: When net displacement is zero
Without the actual figure, general rules:
1. Interval with largest |slope| variation → Greatest average speed
2. Interval with smallest |slope| variation → Least average speed
3. Sign of average velocity = sign of net displacement in that interval
Average speed = Total path length/Time interval
|Average velocity| = |Displacement|/Time interval
Comparing three equal time intervals:
Interval with greatest average speed: Where the magnitude of slope changes most (steepest parts of curve), indicating highest instantaneous speed.
Interval with least average speed: Where the curve is flattest (smallest slope magnitude).
Sign of average velocity:
• Positive average velocity: When x increases with time (positive slope)
• Negative average velocity: When x decreases with time (negative slope)
• Zero average velocity: When net displacement is zero
Without the actual figure, general rules:
1. Interval with largest |slope| variation → Greatest average speed
2. Interval with smallest |slope| variation → Least average speed
3. Sign of average velocity = sign of net displacement in that interval
Question 3.22
Figure 3.25 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D?
Answer & Explanation:
From a speed-time graph:
• Speed is always positive (magnitude of velocity)
• Acceleration = slope of v-t graph (with sign)
Average acceleration magnitude:
Greatest in interval with steepest slope (largest change in speed per time)
Average speed:
Greatest in interval with highest average speed value
Signs (positive direction as motion direction):
• v: Always positive (speed is magnitude, motion is in positive direction)
• a: Positive when speed increases (positive slope)
• a: Negative when speed decreases (negative slope)
• a: Zero when speed constant (zero slope)
At specific points:
• Point A: Start of graph - depends on initial slope
• Point B: Where slope changes - acceleration changes
• Point C: Maximum or minimum point - zero acceleration (slope = 0)
• Point D: End point - depends on final slope
Without actual figure: These are the general principles to apply when analyzing the graph.
• Speed is always positive (magnitude of velocity)
• Acceleration = slope of v-t graph (with sign)
Average acceleration magnitude:
Greatest in interval with steepest slope (largest change in speed per time)
Average speed:
Greatest in interval with highest average speed value
Signs (positive direction as motion direction):
• v: Always positive (speed is magnitude, motion is in positive direction)
• a: Positive when speed increases (positive slope)
• a: Negative when speed decreases (negative slope)
• a: Zero when speed constant (zero slope)
At specific points:
• Point A: Start of graph - depends on initial slope
• Point B: Where slope changes - acceleration changes
• Point C: Maximum or minimum point - zero acceleration (slope = 0)
• Point D: End point - depends on final slope
Without actual figure: These are the general principles to apply when analyzing the graph.
Question 3.23
A three-wheeler starts from rest, accelerates uniformly with 1 m s⁻² on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nᵗʰ second (n = 1,2,3,...) versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?
Answer & Explanation:
Given:
• Initial velocity, u = 0
• Acceleration, a = 1 m/s² for first 10 s
• Then uniform velocity
Distance in nᵗʰ second formula:
sn = u + a/2 (2n - 1)
For u = 0, a = 1: sn = ½(2n - 1) = n - 0.5
For n = 1 to 10 (accelerated motion):
• n = 1: s₁ = 1 - 0.5 = 0.5 m
• n = 2: s₂ = 2 - 0.5 = 1.5 m
• n = 3: s₃ = 2.5 m
• n = 10: s₁₀ = 9.5 m
This is linear: sn = n - 0.5
For n > 10 (uniform motion):
Velocity after 10 s: v = u + at = 0 + 1×10 = 10 m/s
Distance in each second = 10 m (constant)
Plot:
• For n = 1 to 10: Straight line with slope 1
• For n > 10: Horizontal line at sn = 10 m
Answer: During accelerated motion, the plot of distance in nᵗʰ second vs n is a straight line, not a parabola.
• Initial velocity, u = 0
• Acceleration, a = 1 m/s² for first 10 s
• Then uniform velocity
Distance in nᵗʰ second formula:
sn = u + a/2 (2n - 1)
For u = 0, a = 1: sn = ½(2n - 1) = n - 0.5
For n = 1 to 10 (accelerated motion):
• n = 1: s₁ = 1 - 0.5 = 0.5 m
• n = 2: s₂ = 2 - 0.5 = 1.5 m
• n = 3: s₃ = 2.5 m
• n = 10: s₁₀ = 9.5 m
This is linear: sn = n - 0.5
For n > 10 (uniform motion):
Velocity after 10 s: v = u + at = 0 + 1×10 = 10 m/s
Distance in each second = 10 m (constant)
Plot:
• For n = 1 to 10: Straight line with slope 1
• For n > 10: Horizontal line at sn = 10 m
Answer: During accelerated motion, the plot of distance in nᵗʰ second vs n is a straight line, not a parabola.
Question 3.24
A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s⁻¹. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m s⁻¹ and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?
Answer & Explanation:
Case 1: Stationary lift
• Initial speed, u = 49 m/s upward
• Acceleration, a = -g = -9.8 m/s²
• Displacement when returns to hand = 0
Using: s = ut + ½at²
0 = 49t - ½ × 9.8 × t²
0 = 49t - 4.9t²
4.9t² = 49t
t = 0 (initial) or t = 49/4.9 = 10 s
Time for stationary lift: 10 seconds
Case 2: Lift moving up with constant speed 5 m/s
• Relative speed of ball wrt lift = 49 m/s (same as before)
• But lift is moving up at 5 m/s
• Speed of ball relative to ground = 49 + 5 = 54 m/s
• However, when ball returns, lift has moved up
• Better approach: Work in lift's frame
In lift's frame:
• Initial velocity of ball relative to lift = 49 m/s upward
• Acceleration relative to lift = -g = -9.8 m/s² (gravity acts same in both frames as lift has constant velocity)
• The boy and lift are at rest relative to each other
• So the problem reduces to same as Case 1 in lift's frame
Using relative motion equation in lift's frame:
0 = 49t - 4.9t²
t = 10 s (same as before)
Key Insight: Since the lift moves with constant velocity, it's an inertial frame. The time of flight doesn't change in a frame moving with constant velocity relative to ground.
Answer: Both cases take 10 seconds for the ball to return to the boy's hands.
• Initial speed, u = 49 m/s upward
• Acceleration, a = -g = -9.8 m/s²
• Displacement when returns to hand = 0
Using: s = ut + ½at²
0 = 49t - ½ × 9.8 × t²
0 = 49t - 4.9t²
4.9t² = 49t
t = 0 (initial) or t = 49/4.9 = 10 s
Time for stationary lift: 10 seconds
Case 2: Lift moving up with constant speed 5 m/s
• Relative speed of ball wrt lift = 49 m/s (same as before)
• But lift is moving up at 5 m/s
• Speed of ball relative to ground = 49 + 5 = 54 m/s
• However, when ball returns, lift has moved up
• Better approach: Work in lift's frame
In lift's frame:
• Initial velocity of ball relative to lift = 49 m/s upward
• Acceleration relative to lift = -g = -9.8 m/s² (gravity acts same in both frames as lift has constant velocity)
• The boy and lift are at rest relative to each other
• So the problem reduces to same as Case 1 in lift's frame
Using relative motion equation in lift's frame:
0 = 49t - 4.9t²
t = 10 s (same as before)
Key Insight: Since the lift moves with constant velocity, it's an inertial frame. The time of flight doesn't change in a frame moving with constant velocity relative to ground.
Answer: Both cases take 10 seconds for the ball to return to the boy's hands.
Question 3.25
On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 km h⁻¹ (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h⁻¹. For an observer on a stationary platform outside, what is the:
Answer & Explanation:
Given:
• Child's speed relative to belt, vcb = 9 km/h
• Belt speed relative to ground, vbg = 4 km/h
• Distance between parents on belt = 50 m = 0.05 km
(a) Child running in direction of belt motion:
Speed relative to ground, vcg = vcb + vbg = 9 + 4 = 13 km/h
(b) Child running opposite to belt motion:
Speed relative to ground, vcg = vcb - vbg = 9 - 4 = 5 km/h
(c) Time taken:
Distance = 0.05 km
For motion in belt's direction:
Time = Distance/Speed = 0.05/13 hours
= (0.05/13) × 3600 seconds
= 180/13 ≈ 13.85 seconds
For motion opposite to belt's direction:
Time = 0.05/5 hours
= (0.05/5) × 3600 seconds
= 0.01 × 3600 = 36 seconds
For parent's frame (on the belt):
• Distance = 50 m = 0.05 km
• Child's speed = 9 km/h (relative to them)
• Time = 0.05/9 hours = (0.05/9)×3600 = 20 seconds (both directions)
Summary:
(a) 13 km/h
(b) 5 km/h
(c) 13.85 s (with belt), 36 s (against belt)
• Child's speed relative to belt, vcb = 9 km/h
• Belt speed relative to ground, vbg = 4 km/h
• Distance between parents on belt = 50 m = 0.05 km
(a) Child running in direction of belt motion:
Speed relative to ground, vcg = vcb + vbg = 9 + 4 = 13 km/h
(b) Child running opposite to belt motion:
Speed relative to ground, vcg = vcb - vbg = 9 - 4 = 5 km/h
(c) Time taken:
Distance = 0.05 km
For motion in belt's direction:
Time = Distance/Speed = 0.05/13 hours
= (0.05/13) × 3600 seconds
= 180/13 ≈ 13.85 seconds
For motion opposite to belt's direction:
Time = 0.05/5 hours
= (0.05/5) × 3600 seconds
= 0.01 × 3600 = 36 seconds
For parent's frame (on the belt):
• Distance = 50 m = 0.05 km
• Child's speed = 9 km/h (relative to them)
• Time = 0.05/9 hours = (0.05/9)×3600 = 20 seconds (both directions)
Summary:
(a) 13 km/h
(b) 5 km/h
(c) 13.85 s (with belt), 36 s (against belt)
Question 3.26
Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m s⁻¹ and 30 m s⁻¹. Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 m s⁻². Give the equations for the linear and curved parts of the plot.
Answer & Explanation:
Given:
• Cliff height = 200 m
• Stone 1: u₁ = 15 m/s upward
• Stone 2: u₂ = 30 m/s upward
• g = 10 m/s²
Equations of motion (taking upward positive):
For Stone 1: y₁ = 200 + 15t - 5t²
For Stone 2: y₂ = 200 + 30t - 5t²
Relative position: y₂ - y₁ = (200 + 30t - 5t²) - (200 + 15t - 5t²)
y₂ - y₁ = 15t
This is valid until one stone hits the ground.
Time when stones hit ground:
For Stone 1: 0 = 200 + 15t - 5t²
5t² - 15t - 200 = 0
t² - 3t - 40 = 0
(t - 8)(t + 5) = 0
t = 8 s (positive root)
For Stone 2: 0 = 200 + 30t - 5t²
5t² - 30t - 200 = 0
t² - 6t - 40 = 0
(t - 10)(t + 4) = 0
t = 10 s (positive root)
Graph analysis:
• For 0 ≤ t ≤ 8 s: Both stones in air
Relative position = 15t (linear)
• For 8 s < t ≤ 10 s: Stone 1 has hit ground (y₁ = 0)
Stone 2 still in air: y₂ = 200 + 30t - 5t²
Relative position: y₂ - y₁ = y₂ - 0 = 200 + 30t - 5t² (parabolic)
• For t > 10 s: Both stones on ground
y₁ = y₂ = 0 ⇒ Relative position = 0
Equations for plot:
1. Linear part (0 to 8 s): y₂ - y₁ = 15t
2. Curved part (8 to 10 s): y₂ - y₁ = 200 + 30t - 5t²
3. Horizontal part (t > 10 s): y₂ - y₁ = 0
Verification: The graph in Fig. 3.27 should show a straight line from (0,0) to (8,120), then a parabola from (8,120) to (10,0), then zero.
• Cliff height = 200 m
• Stone 1: u₁ = 15 m/s upward
• Stone 2: u₂ = 30 m/s upward
• g = 10 m/s²
Equations of motion (taking upward positive):
For Stone 1: y₁ = 200 + 15t - 5t²
For Stone 2: y₂ = 200 + 30t - 5t²
Relative position: y₂ - y₁ = (200 + 30t - 5t²) - (200 + 15t - 5t²)
y₂ - y₁ = 15t
This is valid until one stone hits the ground.
Time when stones hit ground:
For Stone 1: 0 = 200 + 15t - 5t²
5t² - 15t - 200 = 0
t² - 3t - 40 = 0
(t - 8)(t + 5) = 0
t = 8 s (positive root)
For Stone 2: 0 = 200 + 30t - 5t²
5t² - 30t - 200 = 0
t² - 6t - 40 = 0
(t - 10)(t + 4) = 0
t = 10 s (positive root)
Graph analysis:
• For 0 ≤ t ≤ 8 s: Both stones in air
Relative position = 15t (linear)
• For 8 s < t ≤ 10 s: Stone 1 has hit ground (y₁ = 0)
Stone 2 still in air: y₂ = 200 + 30t - 5t²
Relative position: y₂ - y₁ = y₂ - 0 = 200 + 30t - 5t² (parabolic)
• For t > 10 s: Both stones on ground
y₁ = y₂ = 0 ⇒ Relative position = 0
Equations for plot:
1. Linear part (0 to 8 s): y₂ - y₁ = 15t
2. Curved part (8 to 10 s): y₂ - y₁ = 200 + 30t - 5t²
3. Horizontal part (t > 10 s): y₂ - y₁ = 0
Verification: The graph in Fig. 3.27 should show a straight line from (0,0) to (8,120), then a parabola from (8,120) to (10,0), then zero.
Question 3.27
The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between:
Answer & Explanation:
From speed-time graph (assuming typical shape from textbook):
The distance traveled = area under speed-time curve
Typical graph interpretation (without actual figure):
Usually Fig. 3.28 shows a graph where:
• From t = 0 to 5 s: Straight line from 0 to some maximum speed
• From t = 5 to 10 s: Straight line back to 0
• Triangular shape overall
Assuming triangular graph:
Maximum speed = vmax (say 20 m/s at t = 5 s)
Total area = ½ × base × height = ½ × 10 × vmax
(a) t = 0 to 10 s:
Distance = Total area under graph
If vmax = 20 m/s: Distance = ½ × 10 × 20 = 100 m
(b) t = 2 s to 6 s:
Need to calculate area between these times.
Assuming linear increase and decrease:
• At t = 2 s: Speed = (2/5) × vmax = 0.4vmax
• At t = 6 s: Speed = (4/5) × vmax = 0.8vmax (since decreasing linearly from t=5 to t=10)
Area from t=2 to t=5: Trapezium area
= ½ × (0.4vmax + vmax) × 3 = ½ × 1.4vmax × 3 = 2.1vmax
Area from t=5 to t=6: Trapezium area
= ½ × (vmax + 0.8vmax) × 1 = ½ × 1.8vmax = 0.9vmax
Total area from t=2 to t=6 = 2.1vmax + 0.9vmax = 3vmax
If vmax = 20 m/s: Distance = 3 × 20 = 60 m
Average speeds:
(a) 0-10 s: Average speed = Total distance/Total time = 100/10 = 10 m/s
(b) 2-6 s: Average speed = Distance/Time = 60/4 = 15 m/s
The distance traveled = area under speed-time curve
Typical graph interpretation (without actual figure):
Usually Fig. 3.28 shows a graph where:
• From t = 0 to 5 s: Straight line from 0 to some maximum speed
• From t = 5 to 10 s: Straight line back to 0
• Triangular shape overall
Assuming triangular graph:
Maximum speed = vmax (say 20 m/s at t = 5 s)
Total area = ½ × base × height = ½ × 10 × vmax
(a) t = 0 to 10 s:
Distance = Total area under graph
If vmax = 20 m/s: Distance = ½ × 10 × 20 = 100 m
(b) t = 2 s to 6 s:
Need to calculate area between these times.
Assuming linear increase and decrease:
• At t = 2 s: Speed = (2/5) × vmax = 0.4vmax
• At t = 6 s: Speed = (4/5) × vmax = 0.8vmax (since decreasing linearly from t=5 to t=10)
Area from t=2 to t=5: Trapezium area
= ½ × (0.4vmax + vmax) × 3 = ½ × 1.4vmax × 3 = 2.1vmax
Area from t=5 to t=6: Trapezium area
= ½ × (vmax + 0.8vmax) × 1 = ½ × 1.8vmax = 0.9vmax
Total area from t=2 to t=6 = 2.1vmax + 0.9vmax = 3vmax
If vmax = 20 m/s: Distance = 3 × 20 = 60 m
Average speeds:
(a) 0-10 s: Average speed = Total distance/Total time = 100/10 = 10 m/s
(b) 2-6 s: Average speed = Distance/Time = 60/4 = 15 m/s
Question 3.28
The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29. Which of the following formulae are correct for describing the motion of the particle over the time-interval t₁ to t₂?
Answer & Explanation:
Analysis of each formula:
(a) x(t₂) = x(t₁) + v(t₁)(t₂ - t₁) + ½ a(t₂ - t₁)²
Correct only if: Acceleration is constant.
From the graph (curved v-t), acceleration is not constant.
❌ Incorrect (for non-uniform acceleration)
(b) v(t₂) = v(t₁) + a(t₂ - t₁)
Correct only if: Acceleration is constant (a = average acceleration).
From the graph, acceleration varies.
❌ Incorrect (for non-uniform acceleration)
(c) vaverage = [x(t₂) - x(t₁)]/(t₂ - t₁)
Always correct: This is the definition of average velocity.
✅ Correct (for any motion)
(d) aaverage = [v(t₂) - v(t₁)]/(t₂ - t₁)
Always correct: This is the definition of average acceleration.
✅ Correct (for any motion)
(e) x(t₂) = x(t₁) + vaverage(t₂ - t₁) + ½ aaverage(t₂ - t₁)²
Correct only if: Acceleration is constant (so aaverage = instantaneous a).
This is like equation (a) but with average quantities.
❌ Incorrect (for non-uniform acceleration)
(f) x(t₂) - x(t₁) = area under the v-t curve between t₁ and t₂
Always correct: Area under v-t curve gives displacement.
✅ Correct (for any motion)
Summary of correct formulas: (c), (d), and (f)
Key points:
• Formulas (a), (b), (e) require constant acceleration
• Formulas (c), (d), (f) are definitions that work for any motion
• The v-t graph in Fig. 3.29 shows curved line ⇒ non-uniform acceleration
(a) x(t₂) = x(t₁) + v(t₁)(t₂ - t₁) + ½ a(t₂ - t₁)²
Correct only if: Acceleration is constant.
From the graph (curved v-t), acceleration is not constant.
❌ Incorrect (for non-uniform acceleration)
(b) v(t₂) = v(t₁) + a(t₂ - t₁)
Correct only if: Acceleration is constant (a = average acceleration).
From the graph, acceleration varies.
❌ Incorrect (for non-uniform acceleration)
(c) vaverage = [x(t₂) - x(t₁)]/(t₂ - t₁)
Always correct: This is the definition of average velocity.
✅ Correct (for any motion)
(d) aaverage = [v(t₂) - v(t₁)]/(t₂ - t₁)
Always correct: This is the definition of average acceleration.
✅ Correct (for any motion)
(e) x(t₂) = x(t₁) + vaverage(t₂ - t₁) + ½ aaverage(t₂ - t₁)²
Correct only if: Acceleration is constant (so aaverage = instantaneous a).
This is like equation (a) but with average quantities.
❌ Incorrect (for non-uniform acceleration)
(f) x(t₂) - x(t₁) = area under the v-t curve between t₁ and t₂
Always correct: Area under v-t curve gives displacement.
✅ Correct (for any motion)
Summary of correct formulas: (c), (d), and (f)
Key points:
• Formulas (a), (b), (e) require constant acceleration
• Formulas (c), (d), (f) are definitions that work for any motion
• The v-t graph in Fig. 3.29 shows curved line ⇒ non-uniform acceleration
📘 Exam Preparation Tip:
Practice these problems thoroughly to master the concepts of kinematics, which form the foundation for more advanced physics topics. Remember to focus on graph interpretation and proper application of kinematic equations.