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NCERT Class 11 Physics Chapter 2 Exercise 2.15 to 2.18 Solutions
NCERT Solutions
Class 11 Physics
Chapter 2: Motion in a Straight Line
Exercise 2.15 to 2.18
Question 2.15
Suggest a suitable physical situation for each of the following graphs (Fig. 2.12).
(a) Position-Time Graph (x–t)
Observation:
The particle remains at rest initially.
Then moves away from the origin with constant velocity.
Returns towards the origin with constant velocity.
Finally comes to rest again.
Suitable Physical Situation:
A person standing on a straight road starts walking uniformly away from the origin,
reaches a point A, turns back, crosses the origin at B and finally stops at another point.
The graph represents a person moving away from the origin, returning back and finally coming to rest.
(b) Velocity-Time Graph (v–t)
Observation:
Velocity decreases linearly.
Changes sign repeatedly.
Magnitude decreases after every interval.
Suitable Physical Situation:
A ball thrown vertically upward repeatedly strikes the ground and rebounds.
Each collision reduces its speed due to loss of energy.
The graph represents the motion of a bouncing ball with decreasing speed after each rebound.
(c) Acceleration-Time Graph (a–t)
Observation:
Acceleration is zero for most of the time.
A sudden positive acceleration occurs for a short interval.
Suitable Physical Situation:
A vehicle moving uniformly receives a sudden push or the accelerator is pressed for a short duration.
The graph represents a body receiving acceleration for a short time and then moving uniformly again.
Question 2.16
Figure 2.13 gives the x–t plot of a particle executing one-dimensional simple harmonic motion.
Give the signs of position (x), velocity (v) and acceleration (a) at:
(i) t = 0.3 s
(ii) t = 1.2 s
(iii) t = –1.2 s
Principle Used:
Acceleration in SHM:
a = -ω²x
Velocity is given by the slope of the x-t graph.
(i) At t = 0.3 s
x is negative.
Slope is negative.
Acceleration is positive.
x < 0, v < 0, a > 0
(ii) At t = 1.2 s
x is positive.
Slope is positive.
Acceleration is negative.
x > 0, v > 0, a < 0
(iii) At t = –1.2 s
x is positive.
Slope is positive.
Acceleration is negative.
x > 0, v > 0, a < 0
Time
Position (x)
Velocity (v)
Acceleration (a)
0.3 s
Negative
Negative
Positive
1.2 s
Positive
Positive
Negative
–1.2 s
Positive
Positive
Negative
Question 2.17
Figure 2.14 gives the x–t plot of a particle in one-dimensional motion.
Three different equal intervals of time are shown.
In which interval is the average speed greatest and in which is it least?
Give the sign of average velocity for each interval.
Concept:
Average Velocity = Change in Position / Time
Average Speed = Distance Travelled / Time
Since all intervals have equal duration,
the interval with the greatest change in position will have the greatest average speed.
Interval 1
Position increases slowly.
Average velocity is positive.
Interval 2
Position increases rapidly.
Average velocity is positive.
Interval 3
Position decreases rapidly.
Average velocity is negative.
Greatest Average Speed = Interval 3
Least Average Speed = Interval 1
Interval
Sign of Average Velocity
1
Positive
2
Positive
3
Negative
Question 2.18
Figure 2.15 gives a speed-time graph of a particle moving in a constant direction.
Three equal intervals of time are shown.
(a) In which interval is the average acceleration greatest in magnitude?
(b) In which interval is the average speed greatest?
(c) Choosing the positive direction as direction of motion, give signs of velocity and acceleration in the three intervals.
(d) What are the accelerations at points A, B, C and D?
(a) Average Acceleration
Average Acceleration = Change in Speed / Time
Since all intervals have equal duration,
the third interval shows the largest change in speed.
Greatest Average Acceleration occurs in Interval 3.
(b) Average Speed
The average speed is greatest in the interval where the speed values are highest.
Average Speed is greatest in Interval 3.
(c) Signs of Velocity and Acceleration
Interval
Velocity
Acceleration
1
Positive
Positive
2
Positive
Negative
3
Positive
Positive
(d) Accelerations at A, B, C and D
Acceleration is equal to the slope of the speed-time graph.
At A → Positive slope → aA > 0
At B → Slope = 0 → aB = 0
At C → Slope = 0 → aC = 0
At D → Slope = 0 → aD = 0
aA > 0, aB = 0, aC = 0, aD = 0
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NCERT Class 11 Physics Chapter 2 Motion in a Straight Line Exercise 2.15 to 2.18 solved with graphs, SHM concepts, velocity and acceleration analysis.
Internal Links
Class 11 Physics Motion in a Straight Line Notes
Instantaneous Velocity and Acceleration Notes
NCERT Physics Chapter 2 Important Questions
Numerical Problems on Motion in a Straight Line
Kinematics MCQs for NEET and JEE
Previous Year Questions on Motion in One Dimension
NCERT Class 11 Physics Chapter 2 Motion in a Straight Line Notes
NCERT Class 11 Physics Exercise 2.1 to 2.7 Solutions
NCERT Class 11 Physics Chapter 3 Motion in a Plane Solutions
Important Class 11 Physics Formulas PDF
CBSE Class 11 Physics Previous Year Questions
Motion in a Straight Line MCQs with Answers
Speed, Velocity and Acceleration Notes
Class 11 Physics Revision Notes
NCERT Exemplar Class 11 Physics Solutions
Complete Class 11 Physics Study Material
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NCERT Class 11 Physics - Motion in a Straight Line - Exercise 2.8 to 2.14 Solutions
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
Final velocity = Initial velocity + increase in velocity due to acceleration.
Use
Time is given
Need to find velocity
Second Equation of Motion
\[
x = v_0 t + \frac12 at^2
\]
Meaning
Displacement depends on:
Initial velocity
Time
Acceleration
Important Point:
The term \(at^2\) shows the effect of acceleration.
Third Equation of Motion
\[
v^2 = v_0^2 + 2ax
\]
Important Feature:
This equation has NO time term.
Use
Time is not given
Need relation between velocity and displacement
2. General Form of Equations
Earlier we assumed:
\[
x_0 = 0
\]
Meaning particle starts from origin.
If particle starts from another position (\(x_0\)), equations become:
Modified Second Equation
\[
x = x_0 + v_0 t + \frac12 at^2
\]
Meaning
Final position =
Initial position
Displacement due to initial velocity
Displacement due to acceleration
Modified Third Equation
\[
v^2 = v_0^2 + 2a(x-x_0)
\]
Important Concept:
\((x-x_0)\) represents displacement.
3. Derivation Using Calculus
Definition of Acceleration
Acceleration is rate of change of velocity.
\[
a=\frac{dv}{dt}
\]
Rearranging:
\[
dv = a\,dt
\]
Integrating Both Sides
\[
\int_{v_0}^{v} dv = \int_0^t a\,dt
\]
Since acceleration is constant:
\[
v-v_0 = at
\]
Therefore:
\[
v=v_0+at
\]
This gives first equation of motion.
4. Derivation of Second Equation
Velocity:
\[
v=\frac{dx}{dt}
\]
So,
\[
dx=v\,dt
\]
Substitute:
\[
v=v_0+at
\]
Then,
\[
dx=(v_0+at)dt
\]
Integrating:
\[
x-x_0=v_0 t+\frac12 at^2
\]
Hence,
\[
x=x_0+v_0 t+\frac12 at^2
\]
5. Derivation of Third Equation
We write:
\[
a=\frac{dv}{dt}
\]
Using chain rule:
\[
a=\frac{dv}{dx}\frac{dx}{dt}
\]
But,
\[
\frac{dx}{dt}=v
\]
So,
\[
a=v\frac{dv}{dx}
\]
Rearranging:
\[
v\,dv=a\,dx
\]
Integrating both sides:
\[
\int_{v_0}^{v} v\,dv
=
\int_{x_0}^{x} a\,dx
\]
After integration:
\[
\frac{v^2-v_0^2}{2}=a(x-x_0)
\]
Finally:
\[
v^2=v_0^2+2a(x-x_0)
\]
6. Advantage of Calculus Method
✅ This method can also be used for non-uniform acceleration.
Normal equations work only for constant acceleration.
7. Example 2.3 – Ball Thrown Vertically Upward
Given
Initial velocity: \(v_0=20\,m/s\)
Acceleration due to gravity: \(a=-10\,m/s^2\)
At highest point: \(v=0\)
Finding Maximum Height
Using:
\[
v^2=v_0^2+2a(y-y_0)
\]
Substitute values:
\[
0=(20)^2+2(-10)(y-y_0)
\]
\[
0=400-20(y-y_0)
\]
\[
20(y-y_0)=400
\]
\[
y-y_0=20\,m
\]
Answer:
Maximum height reached = 20 m
8. Important NEET Sign Convention
Upward direction positive.
Upward velocity → positive
Gravity → negative
\[
a=-g
\]
9. Important NEET Concepts
At Highest Point
Velocity becomes zero temporarily.
\[
v=0
\]
But acceleration is still:
\[
a=-g
\]
10. Quick Formula Revision
Formula
Use
\(v=v_0+at\)
Velocity-time relation
\(x=v_0t+\frac12at^2\)
Displacement-time relation
\(v^2=v_0^2+2ax\)
Velocity-displacement relation
\(a=\frac{dv}{dt}\)
Definition of acceleration
\(v=\frac{dx}{dt}\)
Definition of velocity
11. Most Important NEET Tips
✅ Use equations only for constant acceleration.
✅ Check sign convention carefully.
✅ For upward motion:
\[
a=-g
\]
✅ At top point:
\[
v=0
\]
NOT acceleration zero.
✅ Third equation is most useful when time is absent.
CBSE Class 11 Physics: Kinematics Equations of Motion Questions and Answers
Multiple Choice Questions (MCQs)
Question: Which equation is known as the first equation of motion?
Answer: v = u + at
Question: Which equation of motion does not contain time?
Answer: v² = u² + 2as
Question: What is the velocity of a body at the highest point of vertical upward motion?
Answer: Zero.
Question: What is the SI unit of acceleration?
Answer: Metre per second square (m/s²).
Question: Under which condition can equations of motion be applied?
Answer: When acceleration remains constant.
Very Short Answer Questions
Question: Define acceleration.
Answer: Acceleration is the rate of change of velocity with respect to time.
Question: Write the SI unit of displacement.
Answer: Metre (m).
Question: Write the third equation of motion.
Answer: v² = u² + 2as
Question: What is the acceleration due to gravity near the Earth's surface?
Answer: Approximately 9.8 m/s² downward.
Question: What happens to velocity at the highest point of upward motion?
Answer: Velocity becomes zero momentarily.
Short Answer Questions
Question: Write all three equations of motion.
Answer:
First Equation:
v = u + at
Second Equation:
s = ut + ½at²
Third Equation:
v² = u² + 2as
Question: Why is the third equation of motion useful?
Answer: The third equation is useful when time is not given. It directly relates velocity, displacement, and acceleration.
Question: Explain the sign convention used in vertical upward motion.
Answer:
Upward direction is taken as positive.
Upward velocity is positive.
Acceleration due to gravity is negative.
Downward displacement is negative.
Question: A car starts from rest and accelerates uniformly at 2 m/s² for 5 seconds. Find its final velocity.
Answer:
Given:
Initial velocity, u = 0
Acceleration, a = 2 m/s²
Time, t = 5 s
Using v = u + at
v = 0 + (2 × 5)
v = 10 m/s
Final Velocity = 10 m/s
Long Answer Questions
Question: Derive the first equation of motion.
Answer:
Acceleration is defined as:
a = (v − u)/t
Rearranging:
at = v − u
Therefore,
v = u + at
This is called the first equation of motion.
Question: Derive the second equation of motion.
Answer:
Average velocity = (u + v)/2
Displacement:
s = (u + v)t/2
Using the first equation:
v = u + at
Substituting:
s = [u + (u + at)]t/2
s = (2u + at)t/2
s = ut + ½at²
This is the second equation of motion.
Question: A ball is thrown vertically upward with a speed of 20 m/s. Calculate the maximum height reached.
Answer:
Given:
u = 20 m/s
v = 0
a = -10 m/s²
Using:
v² = u² + 2as
0 = (20)² + 2(-10)s
0 = 400 - 20s
20s = 400
s = 20 m
Maximum height reached = 20 m
Assertion and Reason Questions
Assertion: At the highest point of upward motion, velocity becomes zero.
Reason: Acceleration due to gravity becomes zero.
Answer: Assertion is true but Reason is false.
Assertion: The third equation of motion is useful when time is absent.
Reason: It does not contain time.
Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Assertion: Equations of motion can be used for variable acceleration.
Reason: These equations are derived assuming constant acceleration.
Answer: Assertion is false but Reason is true.
Fill in the Blanks
Question: The SI unit of velocity is ________.
Answer: m/s
Question: The acceleration due to gravity is approximately ________.
Answer: 9.8 m/s²
Question: The equation v = u + at is called the ________ equation of motion.
Answer: First
Question: At the highest point of upward motion, velocity becomes ________.
Answer: Zero
Question: The third equation of motion does not contain ________.
Answer: Time
Case Study Based Questions
A student throws a ball vertically upward with an initial velocity of 20 m/s. The ball rises upward, reaches maximum height, and then falls back to the ground. Assume g = 10 m/s².
Question: What is the velocity at the highest point?
Answer: 0 m/s
Question: What is the acceleration at the highest point?
Answer: 10 m/s² downward.
Question: Which equation can be used to find the maximum height?
Answer: v² = u² + 2as
Question: Calculate the maximum height.
Answer: 20 m
Match the Following
Column A
Column B
First Equation of Motion
v = u + at
Second Equation of Motion
s = ut + ½at²
Third Equation of Motion
v² = u² + 2as
Acceleration
Rate of change of velocity
Important Numerical Problems
Question: A body starts from rest and accelerates uniformly at 4 m/s² for 5 seconds. Find the final velocity.
Answer:
v = u + at
v = 0 + (4 × 5)
v = 20 m/s
Final Velocity = 20 m/s
Question: A car moving with a velocity of 10 m/s accelerates at 2 m/s² for 5 seconds. Find the displacement.
Answer:
s = ut + ½at²
s = (10 × 5) + ½(2)(25)
s = 50 + 25
s = 75 m
Displacement = 75 m
Three equations of motion used in kinematics for constant acceleration explained with formulas and applications.
Acceleration tells us how quickly velocity changes with time.
If speed changes → acceleration exists.
If direction changes → acceleration exists.
If both change → acceleration exists.
Acceleration is a vector quantity because it has both magnitude and direction.
2. Average Acceleration
Average acceleration is defined as:
Average Acceleration = Change in Velocity / Time Interval
a = Δv / Δt
Where:
a = average acceleration
Δv = change in velocity
Δt = time interval
3. Velocity Components in x-y Plane
In two-dimensional motion, velocity has two components:
vx → velocity along x-axis
vy → velocity along y-axis
Δv = Δvx i + Δvy j
Therefore acceleration becomes:
a = (Δvx / Δt)i + (Δvy / Δt)j
Where:
i = unit vector along x-axis
j = unit vector along y-axis
4. Components of Acceleration
a = ax i + ay j
Where:
ax = acceleration along x-axis
ay = acceleration along y-axis
5. Instantaneous Acceleration
Instantaneous acceleration means acceleration at a particular instant of time.
It is obtained when the time interval becomes extremely small.
a = lim (Δt → 0) (Δv / Δt)
6. Component Form of Instantaneous Acceleration
ax = dvx / dt
ay = dvy / dt
Meaning:
ax = rate of change of velocity along x-axis
ay = rate of change of velocity along y-axis
7. Graphical Understanding of Acceleration
Suppose an object moves from point P to another point after a small time interval Δt.
The velocity changes from v to another value.
The change in velocity is called Δv.
The direction of acceleration is same as the direction of Δv.
As Δt becomes smaller:
Average acceleration approaches instantaneous acceleration.
The direction becomes more accurate.
8. Velocity Components
If velocity makes angle θ with x-axis:
vx = v cos θ
vy = v sin θ
Where:
vx = horizontal component
vy = vertical component
9. SI Unit of Acceleration
m/s²
Read as: metre per second square
10. Dimensional Formula
[M⁰L¹T⁻²]
11. Important NEET Points
Acceleration depends on change in velocity.
Constant velocity means acceleration is zero.
Negative acceleration is called retardation or deceleration.
In circular motion, acceleration exists even if speed is constant because direction changes continuously.
12. Formula Summary Table
Concept
Formula
Average Acceleration
a = Δv / Δt
Instantaneous Acceleration
a = dv / dt
x-component
ax = dvx / dt
y-component
ay = dvy / dt
Velocity Components
vx = v cos θ, vy = v sin θ
Quick Revision
Acceleration = Rate of change of velocity.
It is a vector quantity.
SI unit = m/s².
Velocity change can be due to speed or direction change.
Average acceleration uses finite time interval.
Instantaneous acceleration uses very small time interval.
Class 11 Physics - Acceleration Questions and Answers
Class 11 Physics - Acceleration Questions and Answers
1. Multiple Choice Questions (MCQs)
Q1. Acceleration is defined as:
A. Change in displacement
B. Change in speed
C. Change in velocity per unit time
D. Distance travelled per unit time
Answer: C. Change in velocity per unit time
Q2. SI unit of acceleration is:
A. m/s
B. m/s²
C. m²/s
D. km/h
Answer: B. m/s²
Q3. Which of the following is a vector quantity?
A. Distance
B. Speed
C. Time
D. Acceleration
Answer: D. Acceleration
2. Very Short Answer Questions
Q1. Define acceleration.
Answer: Acceleration is the rate of change of velocity with time.
Q2. Write the SI unit of acceleration.
Answer: m/s²
Q3. Is acceleration a scalar or vector quantity?
Answer: Vector quantity.
3. Short Answer Questions
Q1. Differentiate between average acceleration and instantaneous acceleration.
Average Acceleration
Instantaneous Acceleration
Calculated over a finite time interval
Calculated at a particular instant
a = Δv / Δt
a = dv / dt
Gives average change
Gives exact change
Q2. Why is acceleration a vector quantity?
Answer: Acceleration depends on change in velocity. Since velocity has both magnitude and direction, acceleration is also a vector quantity.
4. Long Answer Questions
Q1. Define average acceleration and derive its formula.
Average acceleration is the change in velocity divided by time interval.
If initial velocity = u
Final velocity = v
Time taken = Δt
Change in velocity:
Δv = v - u
Therefore,
a = Δv / Δt
Answer: Average acceleration is equal to change in velocity divided by time interval.
Q2. Explain instantaneous acceleration.
Instantaneous acceleration is acceleration at a particular instant of time.
It is obtained when the time interval becomes extremely small.
a = dv / dt
Answer: Instantaneous acceleration gives exact acceleration at any instant.
5. Assertion and Reason Questions
Q1.
Assertion (A): Acceleration is a vector quantity.
Reason (R): Acceleration depends on change in velocity.
A. Both A and R are true and R is correct explanation of A
B. Both A and R are true but R is not correct explanation
C. A is true but R is false
D. A is false but R is true
Answer: A
6. Fill in the Blanks
1. Acceleration is the rate of change of _______.
Answer: velocity
2. SI unit of acceleration is _______.
Answer: m/s²
3. Negative acceleration is called _______.
Answer: retardation
7. Case Study Questions
A car moves along a straight road. Its velocity changes from 10 m/s to 30 m/s in 5 seconds.
Q1. What is the change in velocity?
Answer: 20 m/s
Q2. Calculate acceleration.
a = (v - u)/t
a = (30 - 10)/5 = 4 m/s²
Answer: 4 m/s²
8. Statement Based Questions
Q1. Acceleration can exist without change in speed.
Answer: True
Q2. A body moving in circular path has acceleration.
Answer: True
Q3. Velocity and acceleration always act in same direction.
Answer: False
9. Match the Columns
Column A
Column B
1. Acceleration
a. m/s²
2. Velocity
b. Vector quantity
3. Retardation
c. Negative acceleration
4. SI unit of acceleration
d. Rate of change of displacement
Answers:
1 → b
2 → d
3 → c
4 → a
10. Important Formula Questions
Q1. Write formula for average acceleration.
a = Δv / Δt
Q2. Write formula for instantaneous acceleration.
a = dv / dt
Q3. Write acceleration components in x and y directions.
