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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NCERT Physics Class 11 Chapter 2 Notes: Instantaneous Velocity & Acceleration

NCERT Physics Class 11 Chapter 2

Continuation: Instantaneous Velocity

“The sixth column lists the difference Δx = x(t₂) − x(t₁)...”

Easy Meaning

• Find displacement between two times:

Δx = x(t₂) − x(t₁)

• This tells how much position changed.

“...the last column gives the ratio Δx/Δt...”

Meaning

Average velocity is:

v_avg = Δx / Δt

• Divide displacement by time interval.

“As we decrease Δt from 2.0 s to 0.010 s, the average velocity approaches 3.84 m/s.”

Important Idea

• Smaller time interval gives more accurate velocity.
• As:

Δt → 0

average velocity becomes instantaneous velocity.

“...which is the value of velocity at t = 4.0 s.”

Final Result

v = 3.84 m/s

“In this manner, we can calculate velocity at each instant...”

Meaning

• Using calculus or graphs:
• velocity can be found at any time.

Graphical Method Limitation

“The graphical method... is not always a convenient method.”

Meaning

Drawing tangents accurately is difficult.

“We must carefully plot the position-time graph...”

Meaning

• For accurate velocity:
• graph must be very precise.

“It is easier to calculate velocity if we have data or an equation.”

Important Point

• Equations are easier than graphs.
• Example:
• If position equation is known, velocity can be calculated directly.

Example 2.1

Given Equation

“The position of an object moving along x-axis is given by”

x = a + bt²

Where:

• a = 8.5 m
• b = 2.5 m/s²

Finding Velocity

“In differential calculus, velocity is”

v = dx / dt

Differentiate Position Equation

Given:

x = a + bt²

Differentiate with respect to time:

v = d/dt (a + bt²) = 2bt

Substitute Value of b

Since:

• b = 2.5

Then:

v = 5.0t

Unit: m/s

Velocity at t = 0 s

“At t = 0 s, v = 0 m/s”

Meaning

Object starts from rest.

Velocity at t = 2 s

Substitute (t = 2):

v = 5 × 2

v = 10 m/s

Average Velocity Between 2 s and 4 s

Formula

v_avg = [x(4.0) − x(2.0)] / (4.0 − 2.0)

Substitute Position Values

Using:

x = a + bt²

At t = 4:

x(4) = 8.5 + 2.5(16)

x(4) = 48.5 m

At t = 2:

x(2) = 8.5 + 2.5(4)

x(2) = 18.5 m

Calculate Average Velocity

v_avg = (48.5 − 18.5) / 2

v_avg = 15 m/s

Uniform Motion

“For uniform motion, velocity is same as average velocity.”

Meaning

• If velocity does not change:
• instantaneous velocity = average velocity

Instantaneous Speed

“Instantaneous speed is the magnitude of velocity.”

Definition

Speed = magnitude (numerical value) of velocity.

Example

Velocity: +24 m/s

or

Velocity: −24 m/s

Both have speed:

24 m/s

because speed has no direction.

Important Difference

Quantity	Direction Needed?
Velocity	Yes
Speed	No

2.3 ACCELERATION

“The velocity of an object changes during motion.”

Meaning

• Objects may:
• speed up
• slow down
• change direction

“How do we describe this change?”

Answer

Using acceleration.

Galileo’s Idea

“Galileo concluded that rate of change of velocity with time is constant...”

Meaning

• In free fall:
• velocity changes uniformly with time.

This led to acceleration concept.

Definition of Acceleration

“Acceleration is the rate of change of velocity with time.”

Main Definition

a = Δv / Δt

Average Acceleration Formula

ā = (v₂ − v₁) / (t₂ − t₁)

Meaning of Symbols

Symbol	Meaning
v₁	initial velocity
v₂	final velocity
t₁	initial time
t₂	final time

SI Unit of Acceleration

m/s²

Meaning:

• velocity changes by meters per second every second.

Graphical Meaning

“On velocity-time graph, acceleration is the slope.”

Important Rule

Slope = Δv / Δt

Summary Table

Concept	Formula
Velocity	v = dx/dt
Average velocity	v = Δx/Δt
Acceleration	a = Δv/Δt
Speed	Magnitude of velocity

One-Line Summary: Acceleration tells how quickly velocity changes with time, and it is equal to the slope of the velocity-time graph.

Instantaneous Velocity and Acceleration Class 11 Physics Explained 

CBSE Class 11 Physics - Motion in a Straight Line

Very Short Answer Questions (1 Mark)

Q1. Define instantaneous velocity.

Ans: Instantaneous velocity is the velocity of an object at a particular instant of time.

Q2. Write the formula for average velocity.

Ans: v = Δx / Δt

Q3. What is the SI unit of velocity?

Ans: m/s

Q4. What is acceleration?

Ans: Acceleration is the rate of change of velocity with time.

Q5. What is the SI unit of acceleration?

Ans: m/s²

Short Answer Questions (2-3 Marks)

Q6. Differentiate between speed and velocity.

Speed	Velocity
Scalar quantity	Vector quantity
Magnitude only	Magnitude and direction
Always positive	May be positive, negative or zero

Q7. Define average acceleration.

Ans: Average acceleration is the change in velocity divided by the time interval.
a = (v₂ − v₁)/(t₂ − t₁)

Q8. Why is instantaneous velocity more accurate than average velocity?

Ans: Instantaneous velocity is calculated over an extremely small time interval, giving the exact velocity at a particular instant.

Long Answer Questions (5 Marks)

Q9. Derive the expression for velocity when position is given by: x = a + bt²

Ans:
Given:
x = a + bt²

Velocity:
v = dx/dt

Differentiating:
v = d(a + bt²)/dt
v = 2bt

Therefore,
v = 2bt

Q10. Calculate velocity at t = 2 s if b = 2.5 m/s².

v = 2bt
v = 2 × 2.5 × 2
v = 10 m/s

Multiple Choice Questions (MCQs)

1. The slope of a velocity-time graph represents:

• A. Speed
• B. Distance
• C. Acceleration
• D. Displacement

Answer: C. Acceleration

2. Speed is:

• A. Vector quantity
• B. Scalar quantity
• C. Tensor quantity
• D. None of these

Answer: B. Scalar quantity

Assertion and Reason

Assertion (A): Acceleration is the rate of change of velocity.
Reason (R): Velocity is a vector quantity.

Answer: Both A and R are true and R correctly explains A.

Fill in the Blanks

1. Speed is the ______ of velocity.
2. The SI unit of acceleration is ______.
3. The slope of a velocity-time graph gives ______.
4. Average velocity equals ______ divided by time interval.

Answers:

1. Magnitude
2. m/s²
3. Acceleration
4. Displacement

Match the Columns

Column A	Column B
Velocity	dx/dt
Acceleration	Δv/Δt
Speed	Magnitude of Velocity
Slope of x-t graph	Velocity

Case Study Questions

A particle moves along the x-axis according to:
x = 8.5 + 2.5t²

1. Find the velocity equation.

v = dx/dt = 5t

2. Find velocity at t = 2 s.

v = 5 × 2 = 10 m/s

3. Find velocity at t = 0 s.

0 m/s

Instantaneous velocity and acceleration explained using position-time and velocity-time graphs for NCERT Class 11 Physics.

Internal Links

Velocity Section

Motion in a Straight Line Notes

Difference Between Speed and Velocity

Graphs in Kinematics Explained

Position-Time Graph Questions

Velocity-Time Graph Notes

Example 2.1 Section

NCERT Class 11 Physics Solved Examples

Numerical Problems on Motion in a Straight Line

Derivatives in Physics Made Easy

Acceleration Section

Uniform and Non-Uniform Motion

Acceleration and Retardation Notes

Free Fall Motion Explained

Galileo's Contributions to Physics

Exam Preparation

Class 11 Physics Important Questions

NEET Physics Kinematics Questions

JEE Motion in One Dimension Problems

CBSE Class 11 Physics Revision Notes

NCERT Class 11 Physics Chapter 2.2 Instantaneous Velocity and Speed Notes

NCERT Physics Class 11 Chapter 2

2.2 Instantaneous Velocity and Speed

“The average velocity tells us how fast an object has been moving over a given time interval...”

Easy Meaning

• Average velocity gives motion over a long interval of time.
• It does NOT tell the exact velocity at a particular moment.

Example

If a car travels:

• 100 m in 10 s

v_avg = Δx / Δt

= 100 / 10 = 10 m/s

But the car may move:

• slowly at first
• faster later

So average velocity hides instant changes.

“...but does not tell us how fast it moves at different instants of time...”

Easy Meaning

• We need velocity at one exact moment.
• Example:
  • velocity exactly at 4 s
  • not from 0 to 10 s

Instantaneous Velocity

“For this, we define instantaneous velocity or simply velocity v at an instant t.”

Definition

Instantaneous velocity = velocity at a particular instant of time.

Example

• Speedometer of a bike shows instantaneous speed.

Mathematical Definition

“The velocity at an instant is defined as the limit of average velocity...”

Idea

Take smaller and smaller time intervals.

• When interval becomes extremely tiny:
• average velocity becomes instantaneous velocity.

Formula of Instantaneous Velocity

v = lim_Δt→0 (Δx / Δt)

Meaning of Symbols

• (v) → instantaneous velocity
• (Δx) → small displacement
• (Δt) → small time interval

Calculus Form

v = dx / dt

Meaning

Velocity = rate of change of position with time.

“...the quantity on the right hand side is the differential coefficient of x with respect to t...”

Easy Meaning

In calculus:

dx / dt

means how quickly position changes with time.

“It is the rate of change of position with respect to time...”

Important Point

• Velocity tells:
• how fast position changes
• and in which direction

Finding Velocity Graphically

“We can use Eq. (2.1a) for obtaining the value of velocity graphically or numerically.”

Two Methods

1. Graph method
2. Numerical/table method

“Suppose we want to obtain graphically the value of velocity at t = 4 s...”

Meaning

We want exact velocity at 4 seconds.

“Let us take Δt = 2 s centered at t = 4 s.”

Meaning

Choose time interval around 4 s:

• from 3 s to 5 s

Average Velocity from Graph

Slope of Line

Slope = Δx / Δt

Meaning

Slope of position-time graph gives velocity.

“The slope of line P₁P₂ gives average velocity over interval 3 s to 5 s.”

Easy Meaning

• Draw line between two points.
• Measure rise/change in position.
• Divide by time interval.

“Now, we decrease the value of Δt from 2 s to 1 s.”

Meaning

Take smaller intervals:

• 3.5 to 4.5
• 3.75 to 4.25
• etc.

“In the limit Δt → 0, the line P₁P₂ becomes tangent to the curve...”

Very Important Concept

• When interval becomes extremely small:
• secant line changes into tangent.

Tangent Slope = Instantaneous Velocity

Tangent Concept

“Velocity at t = 4 s is given by the slope of tangent at point P.”

Final Conclusion

Instantaneous velocity at a point = slope of tangent to position-time graph at that point.

Numerical Method

“It is difficult to show this process graphically...”

Meaning

Exact tangent drawing is hard. So tables and calculations are easier.

“Table 2.1 gives values of Δx/Δt...”

What Table Shows

As:

Δt → 0

The value of:

Δx / Δt

approaches a fixed number.

Observation from Table

Δt gets smaller	Velocity value
2.0	3.92
1.0	3.86
0.5	3.845
0.1	3.8402
0.01	3.8400

Final Instantaneous Velocity

v ≈ 3.84 m/s

Important Concepts Summary

Average Velocity

v_avg = Δx / Δt

• For large interval

Instantaneous Velocity

v = dx / dt

• Velocity at exact instant

Graph Rule

Position-Time Graph

• Slope of secant → average velocity
• Slope of tangent → instantaneous velocity

Easy Real-Life Examples

Situation	Type
Average speed of trip	Average velocity
Car speedometer reading	Instantaneous velocity
Straight line slope	Velocity from graph

One-Line Summary: Instantaneous velocity is the velocity of an object at one exact moment and is equal to the slope of the tangent to the position-time graph at that instant.

Learn Instantaneous Velocity and Speed Class 11 Physics with formulas, graphs, examples, numerical methods, and NCERT explanations. 

Internal Links

Add these naturally throughout the article:

Motion & Kinematics

Motion in a Straight Line Notes

Distance and Displacement Explained

Speed vs Velocity Difference

Average Velocity Formula and Examples

Acceleration in One Dimension

Position-Time Graph Explained

Velocity-Time Graph Explained

Equations of Motion Class 11

Calculus & Concepts

Introduction to Differentiation in Physics

Graphical Interpretation of Motion

Slope of a Graph in Physics

Numerical Methods in Kinematics

Exam Preparation

Class 11 Physics Important Questions

NEET Kinematics Questions

JEE Motion in a Straight Line Problems

NCERT Solutions Motion in a Straight Line

Previous Year Physics Questions 

Instantaneous velocity is equal to the slope of the tangent to the position-time graph at a given instant.