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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
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.
Step-by-step solutions for NCERT Physics Examples 4.2 and 4.3 demonstrating Newton's Second Law and force calculations.
Physics Examples 4.2 and 4.3
Physics Solved Examples
Example 4.2
Question:
A bullet of mass 0.04 kg moving with a speed of 90 m/s enters a heavy wooden block and is stopped after travelling a distance of 60 cm. What is the average resistive force exerted by the block on the bullet?
Given:
Mass of bullet, m = 0.04 kg
Initial velocity, u = 90 m/s
Final velocity, v = 0 m/s (bullet stops)
Distance travelled, s = 60 cm = 0.6 m
To Find:
Average resistive force exerted by the block on the bullet.
Step 1: Use the equation of motion
v² = u² + 2as
Substitute the values:
0² = 90² + 2(a)(0.6)
0 = 8100 + 1.2a
1.2a = -8100
a = -8100 / 1.2
a = -6750 m/s²
The negative sign indicates retardation (deceleration).
Step 2: Apply Newton's Second Law
F = ma
F = 0.04 × 6750
F = 270 N
Final Answer: Average resistive force = 270 N
Example 4.3
Question:
The motion of a particle of mass m is described by the equation:
y = ut + ½gt².
Find the force acting on the particle.
Given:
y = ut + ½gt²
To Find:
Force acting on the particle.
Step 1: Find Velocity
Velocity is the rate of change of displacement.
v = dy/dt
Differentiate y = ut + ½gt²
v = u + gt
Step 2: Find Acceleration
Acceleration is the rate of change of velocity.
a = dv/dt
Differentiate v = u + gt
a = g
Step 3: Apply Newton's Second Law
F = ma
Since a = g,
F = mg
Final Answer: F = mg
Therefore, the particle moves under the influence of gravitational force.
