NCERT Solutions - Class 11 Physics Chapter 2: Motion in a Straight Line
Exercises
Q2.1 In which of the following examples of motion, can the body be considered approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.
Answer -
- The railway carriage is much smaller than the distance between two stations, so it can be treated as a point object.
- The monkey and the cyclist can together be approximated as a point object as the track's radius is large compared to their size.
- The spinning cricket ball cannot be considered a point object, since its rotational motion and structure are relevant.
- The tumbling beaker cannot be treated as a point object due to its irregular motion and rotation.
Q2.2 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. 2.9. Choose the correct entries in the brackets below:
(a) (A/B) lives closer to the school than (B/A)
(b) (A/B) starts from the school earlier than (B/A)
(c) (A/B) walks faster than (B/A)
(d) A and B reach home at the (same/different) time
(e) (A/B) overtakes (B/A) on the road (once/twice).
(a) (A/B) lives closer to the school than (B/A)
(b) (A/B) starts from the school earlier than (B/A)
(c) (A/B) walks faster than (B/A)
(d) A and B reach home at the (same/different) time
(e) (A/B) overtakes (B/A) on the road (once/twice).
Answer - Sorry, the answer for this question is not available yet but will be added soon.
Q2.3 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 -
The motion consists of three parts: walking to the office, staying at the office, and returning home.
The graph is currently unavailable but will be added soon.
The motion consists of three parts: walking to the office, staying at the office, and returning home.
- From 9:00 am to the time she reaches the office:
Distance = 2.5 km
Speed = 5 km/h
Time taken = Distance / Speed
= 2.5 / 5 = 0.5 hours = 30 minutes
So, she reaches office at 9:30 am - From 9:30 am to 5:00 pm:
She stays in the office, so position remains constant at 2.5 km. - Return journey:
Distance = 2.5 km
Speed = 25 km/h
Time taken = 2.5 / 25 = 0.1 hours = 6 minutes
So, she returns home at 5:06 pm
The graph is currently unavailable but will be added soon.
Q2.4 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 -
Each full cycle consists of:
Total net displacement in one full cycle = 5 − 3 = 2 m
Total time per cycle = 5 + 3 = 8 s
Let us track his position cycle by cycle:
Now, in the 5th forward phase:
Final Answer: The drunkard falls into the pit at t = 37 s
The graph is currently unavailable but will be added soon.
Each full cycle consists of:
- 5 steps forward = 5 m in 5 s
- 3 steps backward = 3 m in 3 s
Total net displacement in one full cycle = 5 − 3 = 2 m
Total time per cycle = 5 + 3 = 8 s
Let us track his position cycle by cycle:
- After 1st cycle (8 s): 2 m
- After 2nd cycle (16 s): 4 m
- After 3rd cycle (24 s): 6 m
- After 4th cycle (32 s): 8 m
Now, in the 5th forward phase:
- 33 s → 9 m
- 34 s → 10 m
- 35 s → 11 m
- 36 s → 12 m
- 37 s → 13 m ← Drunkard falls into the pit
Final Answer: The drunkard falls into the pit at t = 37 s
The graph is currently unavailable but will be added soon.
Q2.5 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 -
Initial speed, u = 126 km/h = 35 m/s
Final speed, v = 0 m/s (As car comes to rest)
Distance, s = 200 m
Using v2 = u2 + 2as
⇒ 0 = (35)2 + 2 × a × 200
⇒ a = –1225 / 400
⇒ a = –3.06 m/s2
Using v = u + at
⇒ 0 = 35 + (–3.06)t
⇒ t = 35 / 3.06 = 11.44 s
Retardation = 3.06 m/s2, Time taken = 11.44 s
Initial speed, u = 126 km/h = 35 m/s
Final speed, v = 0 m/s (As car comes to rest)
Distance, s = 200 m
Using v2 = u2 + 2as
⇒ 0 = (35)2 + 2 × a × 200
⇒ a = –1225 / 400
⇒ a = –3.06 m/s2
Using v = u + at
⇒ 0 = 35 + (–3.06)t
⇒ t = 35 / 3.06 = 11.44 s
Retardation = 3.06 m/s2, Time taken = 11.44 s
Q2.6 A player throws a ball upwards with an initial speed of 29.4 m s–1.
(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s–2 and neglect air resistance).
(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s–2 and neglect air resistance).
Answer -
- The acceleration is always vertically downward, even during the upward motion of the ball.
- At the highest point:
velocity = 0 m/s
Acceleration = 9.8 m/s2 (downward). - During upward motion:
position = negative
velocity = negative
acceleration = positive.
During downward motion:
position = positive
velocity = positive
acceleration = positive. - Maximum height:
= (u2) / (2g)
= (29.4)2 / (2 × 9.8)
= 864.36 / 19.6
= 44.1 m
Time to reach top:
= u / g
= 29.4 / 9.8 = 3 s
Total time of flight = 2 × 3 = 6 s
Q2.7 Read each statement below carefully and state with reasons and examples, if it is true or false:
(a) A particle in one-dimensional motion with zero speed at an instant may have non-zero acceleration at that instant.
(b) A particle with zero speed may have non-zero velocity.
(c) A particle with constant speed must have zero acceleration.
(d) A particle with positive value of acceleration must be speeding up.
(a) A particle in one-dimensional motion with zero speed at an instant may have non-zero acceleration at that instant.
(b) A particle with zero speed may have non-zero velocity.
(c) A particle with constant speed must have zero acceleration.
(d) A particle with positive value of acceleration must be speeding up.
Answer -
- True — for example, at the highest point in vertical upward motion, speed is zero but acceleration due to gravity is still acting downward.
- False — speed is the magnitude of velocity, so if speed is zero, velocity must also be zero.
- False — in uniform circular motion, speed remains constant but direction changes, so acceleration exists.
- False — if velocity is negative and acceleration is positive, the object may actually be slowing down.
Q2.8 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 - Sorry, the answer for this question is not available yet but will be added soon.
Q2.9 Explain clearly, with examples, the distinction between:
(a) Magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval.
(b) Magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one-dimensional motion only].
(a) Magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval.
(b) Magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one-dimensional motion only].
Answer -
- The magnitude of displacement is the shortest distance between initial and final positions, while the total path length is the actual distance travelled.
Example: If a particle moves 5 m forward and then 3 m backward:
Displacement = 5 − 3 = 2 m
Total path length = 5 + 3 = 8 m
The total path length is always greater than or equal to the displacement.
Equality holds when the particle moves in a straight line without changing direction. - Average velocity = displacement / time
Average speed = total path length / time
Using the same example, if time taken is 4 s:
Average velocity = 2 / 4 = 0.5 m/s
Average speed = 8 / 4 = 2 m/s
So, average speed is greater than or equal to the magnitude of average velocity.
Equality occurs only when the motion is along a straight line in one direction.
Q2.10 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
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ?
[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ?
[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]
Answer -
Distance = 2.5 km
Speed of going = 5 km/h
Speed of returning = 7.5 km/h
Time taken for going
= Distance/Going Speed
= 2.5/5 h
= 0.5 h
= 0.5 × 60 min
= 30 min
Time taken for returning
= Distance/Returning Speed
= 2.5/7.5 h
= 1/3 h
= 1/3 × 60 min
= 20 min
(i) 0 to 30 min:
Time = 30 min = 30/60 h = 0.5 h
Total Distance travelled = 2.5 km
Displacement = 2.5 km
a. Magnitude of average velocity
= Displacement/Time
= 2.5/0.5
= 5 km/h
b. Average speed
= Total Distance travelled/Time Taken
= 2.5/0.5
= 5 km/h
(ii) 0 to 50 min:
Time = 50 min = 50/60 h = 5/6 h
Total Distance travelled
= 2.5 + 2.5 km
= 5 km
Displacement = 0 km
a. Magnitude of average velocity
= Displacement/Time
= 0/ (5/6)
= 0 km/h
b. Average speed
= Total Distance travelled/Time Taken
= 5/ (5/6) km/h
= 6 km/h
(iii) 0 to 40 min:
Time = 40 min = 40/60 h = 2/3 h
The man travels towards market for the first 30 min, and travels towards home for the last 10 mins.
Distance travelled in first 30 min
= 2.5 km
Distance travelled in last 10 min
= Speed × Time
= 7.5 × (10/60)
= 1.25
Total Distance travelled
= 2.5 + 1.25 km
= 3.75
Displacement
= 2.5 - 1.25 km
= 1.25 km
a. Magnitude of average velocity
= Displacement/Time
= 1.25/ (2/3)
= 1.875 km/h
b. Average speed
= Total Distance travelled/Time Taken
= 3.75/ (2/3) km/h
= 5.625 km/h
Distance = 2.5 km
Speed of going = 5 km/h
Speed of returning = 7.5 km/h
Time taken for going
= Distance/Going Speed
= 2.5/5 h
= 0.5 h
= 0.5 × 60 min
= 30 min
Time taken for returning
= Distance/Returning Speed
= 2.5/7.5 h
= 1/3 h
= 1/3 × 60 min
= 20 min
(i) 0 to 30 min:
Time = 30 min = 30/60 h = 0.5 h
Total Distance travelled = 2.5 km
Displacement = 2.5 km
a. Magnitude of average velocity
= Displacement/Time
= 2.5/0.5
= 5 km/h
b. Average speed
= Total Distance travelled/Time Taken
= 2.5/0.5
= 5 km/h
(ii) 0 to 50 min:
Time = 50 min = 50/60 h = 5/6 h
Total Distance travelled
= 2.5 + 2.5 km
= 5 km
Displacement = 0 km
a. Magnitude of average velocity
= Displacement/Time
= 0/ (5/6)
= 0 km/h
b. Average speed
= Total Distance travelled/Time Taken
= 5/ (5/6) km/h
= 6 km/h
(iii) 0 to 40 min:
Time = 40 min = 40/60 h = 2/3 h
The man travels towards market for the first 30 min, and travels towards home for the last 10 mins.
Distance travelled in first 30 min
= 2.5 km
Distance travelled in last 10 min
= Speed × Time
= 7.5 × (10/60)
= 1.25
Total Distance travelled
= 2.5 + 1.25 km
= 3.75
Displacement
= 2.5 - 1.25 km
= 1.25 km
a. Magnitude of average velocity
= Displacement/Time
= 1.25/ (2/3)
= 1.875 km/h
b. Average speed
= Total Distance travelled/Time Taken
= 3.75/ (2/3) km/h
= 5.625 km/h
Q2.11 In Exercises 2.9 and 2.10, 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 - In case of instantaneous velocity, the time interval is extremely small.
In such a short time, the body cannot change its direction of motion.
So, the motion can be considered uniform in a straight line during that instant.
For motion in a straight line, distance and displacement become equal.
That means, instantaneous speed (which is distance/time) becomes equal to the magnitude of instantaneous velocity (which is displacement/time).
Hence, instantaneous speed is always equal to the magnitude of instantaneous velocity.
In such a short time, the body cannot change its direction of motion.
So, the motion can be considered uniform in a straight line during that instant.
For motion in a straight line, distance and displacement become equal.
That means, instantaneous speed (which is distance/time) becomes equal to the magnitude of instantaneous velocity (which is displacement/time).
Hence, instantaneous speed is always equal to the magnitude of instantaneous velocity.
Q2.12 Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
Answer - Sorry, the answer for this question is not available yet but will be added soon.
Q2.13 Figure 2.11 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 - Sorry, the answer for this question is not available yet but will be added soon.
Q2.14 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 - Speed of police van = 30 km/h
Speed of thief's car = 192 km/h
Muzzle speed of bullet (w.r.t. police van) = 150 m/s
First, convert speeds into m/s:
Speed of police van
= 30 × (1000/3600)
= 8.33 m/s
Speed of thief's car
= 192 × (1000/3600)
= 53.33 m/s
Speed of bullet w.r.t. ground
= Muzzle speed + Speed of police van
= 150 + 8.33
= 158.33 m/s
Relative speed of bullet w.r.t. thief's car
= Speed of bullet – Speed of thief's car
= 158.33 – 53.33
= 105 m/s
So, the bullet hits the thief's car with a speed of 105 m s–1 (relative to the car).
Speed of thief's car = 192 km/h
Muzzle speed of bullet (w.r.t. police van) = 150 m/s
First, convert speeds into m/s:
Speed of police van
= 30 × (1000/3600)
= 8.33 m/s
Speed of thief's car
= 192 × (1000/3600)
= 53.33 m/s
Speed of bullet w.r.t. ground
= Muzzle speed + Speed of police van
= 150 + 8.33
= 158.33 m/s
Relative speed of bullet w.r.t. thief's car
= Speed of bullet – Speed of thief's car
= 158.33 – 53.33
= 105 m/s
So, the bullet hits the thief's car with a speed of 105 m s–1 (relative to the car).
Q2.15 Suggest a suitable physical situation for each of the following graphs (Fig 2.12):
Answer - Sorry, the answer for this question is not available yet but will be added soon.
Q2.16 Figure 2.13 gives the x–t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter 13). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, –1.2 s.
Answer - Sorry, the answer for this question is not available yet but will be added soon.
Q2.17 Figure 2.14 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 - Sorry, the answer for this question is not available yet but will be added soon.
Q2.18 Figure 2.15 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 - Sorry, the answer for this question is not available yet but will be added soon.