NCERT Class 11 Physics Chapter 2 Exercise 2.8 to 2.14 Solutions (CBSE 2026)

-  Dr.Sanjaykumar Pawar  

 

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1. NCERT Class 11 Physics Chapter 2 Motion in a Straight Line Notes

2. NCERT Class 11 Physics Exercise 2.1 to 2.7 Solutions

3. NCERT Class 11 Physics Chapter 3 Motion in a Plane Solutions

4. Important Class 11 Physics Formulas PDF

5. CBSE Class 11 Physics Previous Year Questions

6. Motion in a Straight Line MCQs with Answers

7. Speed, Velocity and Acceleration Notes

8. Class 11 Physics Revision Notes

9. NCERT Exemplar Class 11 Physics Solutions

10. Complete Class 11 Physics Study Material

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NCERT Solutions Class 11 Physics

Chapter 2: Motion in a Straight Line

Exercise 2.8 to 2.14

Question 2.8

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.

Given:

• Height = 90 m
• g = 9.8 m/s²
• Loss of speed after every collision = 10%

Step 1: Time taken to reach the floor

h = ½gt²

90 = ½ × 9.8 × t²

t² = 180/9.8

t = 4.29 s

Step 2: Speed just before collision

v = gt

v = 9.8 × 4.29

v = 42 m/s

Step 3: Speed after collision

v' = 0.9 × 42

v' = 37.8 m/s

Step 4: Time taken to move upward

t = v'/g

t = 37.8/9.8

t = 3.86 s

Graph Description:

• Speed increases uniformly from 0 to 42 m/s during first 4.29 s.
• At collision speed suddenly decreases to 37.8 m/s.
• Speed decreases linearly to zero while moving upward.
• Then increases again while falling downward.

Speed-Time graph consists of straight line segments with sudden vertical drops at collisions.

Question 2.9

Explain clearly, with examples, the distinction between: (a) Magnitude of displacement and total path length (b) Magnitude of average velocity and average speed.

(a) Magnitude of Displacement and Total Path Length

Magnitude of Displacement	Total Path Length
Shortest distance between initial and final positions.	Actual distance travelled.
Depends only on initial and final positions.	Depends on actual path followed.
Always less than or equal to path length.	Always greater than or equal to displacement.

Example:

Particle moves 3 m east and 4 m north. Displacement = √(3² + 4²) = 5 m Path Length = 3 + 4 = 7 m

(b) Magnitude of Average Velocity and Average Speed

Average Velocity = Displacement / Time

Average Speed = Total Path Length / Time

Since total path length ≥ displacement,

Average Speed ≥ Magnitude of Average Velocity

Equality occurs only when motion is along a straight line without changing direction.

Question 2.10

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. Find: (a) Magnitude of average velocity (b) Average speed

(a) Magnitude of Average Velocity

Net displacement = 0 Average Velocity = 0 / Total Time = 0

Magnitude of Average Velocity = 0 km h⁻¹

(b) Average Speed

Time to market = 2.5/5 = 0.5 h

Time to return = 2.5/7.5 = 0.333 h

Total distance = 5 km Total time = 0.5 + 0.333 = 0.833 h

Average Speed = 5 / 0.833 = 6 km h⁻¹

Average Speed = 6 km h⁻¹

Question 2.11

Why is there no distinction between instantaneous speed and magnitude of instantaneous velocity?

Instantaneous Speed = Magnitude of Instantaneous Velocity

Velocity has both magnitude and direction, whereas speed is only magnitude. At any instant, speed is simply the magnitude of velocity.

Instantaneous Speed = |Instantaneous Velocity|

Question 2.12

Look at the graphs (a) to (d) carefully and state which of these cannot represent one-dimensional motion of a particle.

• Graph (a): Impossible because one instant corresponds to more than one position.
• Graph (b): Impossible because one instant corresponds to more than one velocity.
• Graph (c): Impossible because speed cannot be negative.
• Graph (d): Impossible because total path length can never decrease.

All graphs (a), (b), (c) and (d) cannot represent one-dimensional motion.

Question 2.13

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?

No. The graph is an x-t graph and not the actual trajectory of the particle.

For t < 0, x remains constant, showing that the particle is at rest.

For t > 0, x increases with time and velocity increases continuously.

The graph represents variation of position with time and not the actual path of motion.

Question 2.14

A policeman moving at 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?

Step 1: Convert speeds into m/s

Policeman = 30 × 5/18 = 8.33 m/s

Thief = 192 × 5/18 = 53.33 m/s

Step 2: Bullet speed relative to ground

Bullet speed = 150 + 8.33 = 158.33 m/s

Step 3: Relative speed of bullet with respect to thief

158.33 − 53.33 = 105 m/s

Speed of bullet relative to thief's car = 105 m/s

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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.

 Internal Links

CBSE Class 11 Physics Chapter 1 Physical World NCERT Solutions

CBSE Class 11 Physics Chapter 2 Motion in a Straight Line Notes

CBSE Class 11 Physics Chapter 3 Motion in a Plane Solutions

Class 11 Physics Important Formula Sheet PDF

CBSE Class 11 Physics MCQs Chapter 2

Motion in a Straight Line Revision Notes

NCERT Class 11 Physics All Chapters Solutions

CBSE Class 11 Physics Sample Questions with Answers

Kinematics Complete Study Guide for Class 11

Previous Year CBSE Class 11 Physics Important Questions

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.