CBSE Class 11 Physics Chapter 2 Motion in a Straight Line Exercise 2.1 to 2.7 Solutions

-  Dr.Sanjaykumar Pawar  

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?

1. A railway carriage moving without jerks between two stations.
2. A monkey sitting on top of a man cycling smoothly on a circular track.
3. A spinning cricket ball that turns sharply on hitting the ground.
4. A tumbling beaker that has slipped off the edge of a table.

A body can be treated as a point object if its size is negligible compared to the distance travelled.

• (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.