- Dr.Sanjaykumar Pawar
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| CBSE Class 11 Physics Chapter 2 Motion in a Straight Line NCERT Exercise 2.1 to 2.7 Solved Answers. |
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NCERT Class 11 Physics
Chapter 2: Motion in a Straight Line
Exercise Solutions
Question 2.1
In which of the following examples of motion can the body be considered approximately a point object?
- A railway carriage moving without jerks between two stations.
- A monkey sitting on top of a man cycling smoothly on a circular track.
- A spinning cricket ball that turns sharply on hitting the ground.
- A tumbling beaker that has slipped off the edge of a table.
- (a) Yes. The distance between stations is much greater than the size of the carriage.
- (b) Yes. The radius of the circular track is much larger than the dimensions of the monkey-man system.
- (c) No. Rotation and spin are important.
- (d) No. The tumbling motion depends on the size and shape of the beaker.
Final Answer: (a) and (b)
Question 2.2
The position-time (x-t) graphs for two children A and B returning from school to their homes are shown in Fig. 2.9. Choose the correct entries in the brackets.
(a) (A/B) lives closer to the school than (B/A).
Since OP is less than OQ, A lives closer to the school.
Answer: A lives closer to the school than B.
(b) (A/B) starts from the school earlier than (B/A).
A starts at t = 0 while B starts later.
Answer: A starts earlier than B.
(c) (A/B) walks faster than (B/A).
The slope of an x-t graph represents speed. B has a steeper graph.
Answer: B walks faster than A.
(d) A and B reach home at the (same/different) time.
Both graphs terminate at the same value of time.
Answer: Same time.
(e) (A/B) overtakes (B/A) on the road (once/twice).
The graphs intersect only once and after that B remains ahead.
Answer: B overtakes A once.
Question 2.3
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. Plot the x-t graph of her motion.
Given:
- Distance = 2.5 km
- Walking speed = 5 km h⁻¹
- Auto speed = 25 km h⁻¹
Time taken to reach office:
t = Distance / Speed = 2.5 / 5 = 0.5 hour = 30 minutes
She reaches the office at 9:30 am.
Time taken to return:
t = 2.5 / 25 = 0.1 hour = 6 minutes
She reaches home at 5:06 pm.
| Time | Position (km) |
|---|---|
| 9:00 am | 0 |
| 9:30 am | 2.5 |
| 5:00 pm | 2.5 |
| 5:06 pm | 0 |
The x-t graph consists of an increasing straight line, a horizontal line, and a steep decreasing straight line.
Question 2.4
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward repeatedly. Each step is 1 m long and takes 1 s. Determine how long he takes to fall in a pit 13 m away.
Net displacement in one cycle = 5 − 3 = 2 m
Time taken in one cycle = 8 s
After 4 cycles:
Displacement = 4 × 2 = 8 m
Time = 4 × 8 = 32 s
Then he moves forward:
| Time (s) | Position (m) |
|---|---|
| 33 | 9 |
| 34 | 10 |
| 35 | 11 |
| 36 | 12 |
| 37 | 13 |
Therefore, the drunkard falls into the pit after 37 seconds.
Question 2.5
A car moving along a straight highway with speed 126 km h⁻¹ is brought to a stop within a distance of 200 m. What is the retardation and how long does it take to stop?
Initial speed, u = 126 km h⁻¹ = 35 m s⁻¹
Final speed, v = 0
Distance, s = 200 m
Using:
v² = u² + 2as
0 = (35)² + 2(a)(200)
a = −3.06 m s⁻²
Retardation = 3.06 m s⁻²
Using:
v = u + at
0 = 35 − 3.06t
t = 11.4 s
Retardation = 3.06 m s⁻²
Time taken = 11.4 s
Question 2.6
A player throws a ball upwards with an initial speed of 29.4 m s⁻¹.
(a) What is the direction of acceleration during upward motion?
Acceleration is always vertically downward due to gravity.
(b) What are the velocity and acceleration at the highest point?
Velocity = 0
Acceleration = 9.8 m s⁻² downward
v = 0, a = 9.8 m s⁻² downward
(c) Sign convention:
- During upward motion: Position positive, Velocity negative, Acceleration positive.
- During downward motion: Position positive, Velocity positive, Acceleration positive.
(d) Maximum height and total time of flight
Using:
v² = u² − 2gh
0 = (29.4)² − 2(9.8)h
h = 44.1 m
Time to rise:
t = u/g = 29.4/9.8 = 3 s
Total time:
T = 2 × 3 = 6 s
Maximum height = 44.1 m
Total time = 6 s
Question 2.7
State whether the following statements are true or false with reasons.
(a) A particle with zero speed may have non-zero acceleration.
True. Example: Ball at highest point of vertical motion.
(b) A particle with zero speed may have non-zero velocity.
False. Velocity is zero when speed is zero.
(c) A particle moving with constant speed must have zero acceleration.
False. Uniform circular motion has non-zero acceleration.
(d) A particle with positive acceleration must be speeding up.
False. Positive acceleration can reduce speed if velocity is negative.
