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

-  Dr.Sanjaykumar Pawar  

 

Internal Links

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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