Train Crossing Bridge Questions and Answers Class 11 Physics | CBSE & NEET Practice

 - Dr.Sanjaykumar Pawar  

Train Crossing Bridge Numerical Questions with Solutions | Class 11 Physics 

Train Crossing Bridge Numerical for CBSE Class 11 Physics with Step-by-Step Solution

Internal Links

1. Motion in a Straight Line Complete Notes
2. Equations of Motion Explained
3. Relative Motion Notes
4. Average Speed and Average Velocity
5. Instantaneous Velocity Notes
6. Acceleration Complete Notes
7. Graphical Analysis of Motion
8. Kinematics Formula Sheet
9. Work, Energy and Power Notes
10. Laws of Motion Complete Notes
11. NEET Physics Chapter-wise MCQs
12. CBSE Class 11 Physics Important Questions
13. Assertion and Reason Questions for Physics
14. Case Study Questions for Class 11 
15. NCERT Solutions for Motion in a Straight Line
16. Previous Year CBSE Physics Questions
17. NEET Physics Practice Tests
18. One-Dimensional Motion Numericals
19. Train Crossing and River Boat Problems
20. Class 11 Physics Revision Notes

CBSE Class 11 Physics

Practice Questions and Answers

Topic: Motion in a Straight Line – Train Crossing Bridge with Constant Acceleration

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Question 1

A 120 m long train crosses a 480 m long bridge with a constant acceleration of 1 m/s². The train enters the bridge with an initial velocity of 20 m/s.

Find the time taken by the train to completely cross the bridge.

Answer

Given:

• Length of Train = 120 m
• Length of Bridge = 480 m
• Initial Velocity (u) = 20 m/s
• Acceleration (a) = 1 m/s²

Step 1: Total Distance Covered

Distance = Length of Train + Length of Bridge

s = 120 + 480 = 600 m

Step 2: Apply Equation of Motion

s = ut + ½at²

600 = 20t + ½(1)t²

600 = 20t + 0.5t²

Multiply by 2

1200 = 40t + t²

t² + 40t − 1200 = 0

Factorisation

(t + 60)(t − 20) = 0

Time Taken = 20 s

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Question 2

A 150 m long train crosses a 450 m long bridge with constant acceleration 2 m/s². The initial velocity of the train is 10 m/s.

Find the time taken to completely cross the bridge.

Answer

Given:

• Length of Train = 150 m
• Length of Bridge = 450 m
• Initial Velocity = 10 m/s
• Acceleration = 2 m/s²

Step 1: Calculate Total Distance

s = 150 + 450 = 600 m

Step 2: Apply Equation

600 = 10t + ½(2)t²

600 = 10t + t²

t² + 10t − 600 = 0

Factorisation

(t + 30)(t − 20) = 0

Time Taken = 20 s

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Question 3

A 200 m long train crosses a 300 m long bridge. It enters the bridge with a speed of 15 m/s and accelerates uniformly at 1 m/s².

Find the time taken by the train to completely cross the bridge.

Answer

Given:

• Length of Train = 200 m
• Length of Bridge = 300 m
• Initial Velocity = 15 m/s
• Acceleration = 1 m/s²

Step 1: Total Distance

s = 200 + 300 = 500 m

Step 2: Equation of Motion

500 = 15t + ½(1)t²

500 = 15t + 0.5t²

Multiply by 2

1000 = 30t + t²

t² + 30t − 1000 = 0

Using quadratic formula,

t = 20 s

Time Taken = 20 s

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Question 4

A 100 m long train completely crosses a 400 m long bridge. The train enters the bridge with an initial speed of 25 m/s and moves with a constant acceleration of 2 m/s².

Find the time taken by the train to completely cross the bridge.

Answer

Given:

• Length of Train = 100 m
• Length of Bridge = 400 m
• Initial Velocity (u) = 25 m/s
• Acceleration (a) = 2 m/s²

Step 1: Calculate Total Distance

Distance = Length of Train + Length of Bridge

s = 100 + 400 = 500 m

Step 2: Apply Equation of Motion

s = ut + ½at²

500 = 25t + ½(2)t²

500 = 25t + t²

Rearranging,

t² + 25t − 500 = 0

Using the quadratic formula,

t = 13.9 s (approximately)

Time Taken = 13.9 s

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Question 5

A 250 m long train crosses a 350 m long bridge. It enters the bridge with an initial speed of 20 m/s and accelerates uniformly at 2 m/s².

Find the time taken by the train to completely cross the bridge.

Answer

Given:

• Length of Train = 250 m
• Length of Bridge = 350 m
• Initial Velocity = 20 m/s
• Acceleration = 2 m/s²

Step 1: Total Distance

s = 250 + 350 = 600 m

Step 2: Apply Equation of Motion

600 = 20t + ½(2)t²

600 = 20t + t²

Rearranging,

t² + 20t − 600 = 0

Using the quadratic formula,

t = 15.6 s (approximately)

Time Taken = 15.6 s

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Question 6 (Assertion & Reason)

Assertion (A):

To completely cross a bridge, a train travels a distance equal to the sum of the length of the train and the length of the bridge.

Reason (R):

The rear end of the train leaves the bridge only after the front end has already crossed the bridge.

Choose the correct option.

1. Both A and R are true, and R is the correct explanation of A.
2. Both A and R are true, but R is not the correct explanation of A.
3. A is true, but R is false.
4. A is false, but R is true.

Answer

Correct Option: A. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Explanation:

When a train completely crosses a bridge, its front end first reaches the other end of the bridge. However, the train is considered completely out of the bridge only when its rear end also leaves the bridge. Therefore, the train travels a distance equal to the sum of its own length and the length of the bridge.

Correct Answer: Option A

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Question 7 (Very Short Answer)

Why do we add the length of the train and the length of the bridge while solving train crossing problems?

Answer

The train completely crosses the bridge only when its rear end leaves the bridge. Therefore, the front end of the train travels a distance equal to the sum of the train's length and the bridge's length.

Distance = Length of Train + Length of Bridge

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Question 8 (Short Answer)

Which equation of motion is generally used when displacement, initial velocity, acceleration, and time are involved in train crossing problems?

Answer

The required equation of motion is:

s = ut + ½at²

Where,

• s = Displacement
• u = Initial Velocity
• a = Acceleration
• t = Time Taken

Equation Used: s = ut + ½at²

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Question 9 (Multiple Choice Question)

A 100 m long train crosses a 500 m long bridge with an initial velocity of 10 m/s and an acceleration of 2 m/s².

The time taken is:

1. 10 s
2. 15 s
3. 20 s
4. 25 s

Answer

Total Distance

100 + 500 = 600 m

Using

600 = 10t + t²

t² + 10t − 600 = 0

(t + 30)(t − 20) = 0

Correct Option: C (20 s)

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Question 10 (Concept-Based Question)

A train takes 20 seconds to completely cross a 500 m long bridge. The train is 100 m long.

Find the total displacement of the front end of the train during crossing.

Answer

The front end of the train travels:

Distance = Length of Train + Length of Bridge

Distance = 100 + 500 = 600 m

Total Displacement = 600 m

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CBSE Exam Tips

• Always calculate the total distance as the length of the train + length of the bridge.
• Choose the correct equation of motion according to the given data.
• Write every step clearly to score full marks in CBSE board examinations.
• Include SI units (m, m/s, s) in every calculation.
• Highlight the final answer inside a box.

---

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Question 7 (Very Short Answer)

Why do we add the length of the train and the length of the bridge while solving train crossing problems?

Answer

The train completely crosses the bridge only when its rear end leaves the bridge. Therefore, the front end of the train travels a distance equal to the sum of the train's length and the bridge's length.

Distance = Length of Train + Length of Bridge

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Question 8 (Short Answer)

Which equation of motion is generally used when displacement, initial velocity, acceleration, and time are involved in train crossing problems?

Answer

The required equation of motion is:

s = ut + ½at²

Where,

• s = Displacement
• u = Initial Velocity
• a = Acceleration
• t = Time Taken

Equation Used: s = ut + ½at²

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Question 9 (Multiple Choice Question)

A 100 m long train crosses a 500 m long bridge with an initial velocity of 10 m/s and an acceleration of 2 m/s².

The time taken is:

1. 10 s
2. 15 s
3. 20 s
4. 25 s

Answer

Total Distance

100 + 500 = 600 m

Using

600 = 10t + t²

t² + 10t − 600 = 0

(t + 30)(t − 20) = 0

Correct Option: C (20 s)

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Question 10 (Concept-Based Question)

A train takes 20 seconds to completely cross a 500 m long bridge. The train is 100 m long.

Find the total displacement of the front end of the train during crossing.

Answer

The front end of the train travels:

Distance = Length of Train + Length of Bridge

Distance = 100 + 500 = 600 m

Total Displacement = 600 m

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CBSE Exam Tips

• Always calculate the total distance as the length of the train + length of the bridge.
• Choose the correct equation of motion according to the given data.
• Write every step clearly to score full marks in CBSE board examinations.
• Include SI units (m, m/s, s) in every calculation.
• Highlight the final answer inside a box.

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NCERT Solutions Class 11 Physics Chapter 2 Exercise 2.15 to 2.18 with Answers

 - Dr.Sanjaykumar Pawar  

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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.

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(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.

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(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.

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(b) Average Speed

The average speed is greatest in the interval where the speed values are highest.

Average Speed is greatest in Interval 3.

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(c) Signs of Velocity and Acceleration

Interval	Velocity	Acceleration
1	Positive	Positive
2	Positive	Negative
3	Positive	Positive

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(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

Difference Between Speed and Velocity

Graphs in Physics Explained

Simple Harmonic Motion Basics

Class 11 Physics Revision Notes

NCERT Solutions Chapter 2 Complete Series

Physics Formula Sheet for Class 11

NEET Physics Kinematics Practice Questions

JEE Motion in a Straight Line Problems

CBSE Class 11 Physics Sample Questions

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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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.

Motion of an Object Under Free Fall | NEET Physics Notes with Graphs & Formulas

CBSE Class 11 Physics (Motion Under Free Fall) – Question Bank with Answers  

Motion of an object under free fall showing acceleration, velocity, and displacement-time graphs.

- Dr.Sanjaykumar Pawar 

A. Multiple Choice Questions (MCQs)

1. A body is said to be in free fall when:

a) It falls with constant velocity
b) It falls under gravity only
c) It falls in vacuum only
d) It moves downward

Answer: b) It falls under gravity only

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2. The acceleration due to gravity near Earth's surface is:

a) 8.9 m/s²
b) 9.8 m/s²
c) 10.8 m/s²
d) 12 m/s²

Answer: b) 9.8 m/s²

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3. The velocity-time graph for a freely falling body is:

a) Horizontal line
b) Parabola
c) Straight line
d) Circle

Answer: c) Straight line

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

a) Velocity
b) Distance
c) Acceleration
d) Momentum

Answer: c) Acceleration

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5. Distance travelled in successive equal intervals of time during free fall follows:

a) 1 : 2 : 3 : 4
b) 2 : 4 : 6 : 8
c) 1 : 3 : 5 : 7
d) 1 : 4 : 9 : 16

Answer: c) 1 : 3 : 5 : 7

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6. Stopping distance of a vehicle is proportional to:

a) Speed
b) Square of speed
c) Cube of speed
d) Inverse of speed

Answer: b) Square of speed

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7. For a freely falling body released from rest:

a) u = g
b) u = 1
c) u = 0
d) u = 10

Answer: c) u = 0

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8. Which equation represents free fall motion?

a) v = u + gt
b) v = u − gt
c) v = u/t
d) v = gt²

Answer: b) v = u − gt

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9. The SI unit of acceleration due to gravity is:

a) m
b) m/s
c) m/s²
d) kg

Answer: c) m/s²

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10. If speed doubles, stopping distance becomes:

a) Double
b) Triple
c) Four times
d) Eight times

Answer: c) Four times

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B. Very Short Answer Questions (1 Mark)

1. Define free fall.

Answer: Motion of a body under the influence of gravity alone is called free fall.

2. What is the value of g near Earth's surface?

Answer: 9.8 m/s².

3. What is the acceleration of a freely falling body?

Answer: g downward.

4. Who proposed the law of odd numbers?

Answer: Galileo Galilei.

5. What is stopping distance?

Answer: Distance travelled by a vehicle after brakes are applied until it comes to rest.

6. What is the slope of a v–t graph?

Answer: Acceleration.

7. What is the shape of displacement-time graph in free fall?

Answer: Parabola.

8. What is the initial velocity of a body released from rest?

Answer: Zero.

---

C. Short Answer Questions (2–3 Marks)

1. Why is acceleration constant during free fall?

Answer: The only force acting on the body is gravity. Near Earth's surface, gravity remains nearly constant. Therefore acceleration remains constant and equal to g.

---

2. Write the equations of motion for free fall.

Answer:

---

3. State Galileo's Law of Odd Numbers.

Answer: The distances covered by a freely falling body during successive equal intervals of time are proportional to odd numbers:

1 : 3 : 5 : 7 : 9 ...

---

4. Why is the displacement-time graph parabolic?

Answer: Displacement in free fall is proportional to the square of time.

s = \frac{1}{2}gt^2

Since displacement depends on , the graph is a parabola.

---

5. Explain why stopping distance increases with speed.

Answer: Stopping distance is given by:

d_s = \frac{v_0^2}{2a}

Hence stopping distance is proportional to the square of velocity. Therefore higher speed results in much larger stopping distance.

---

D. Long Answer Questions (5 Marks)

1. Explain the variation of acceleration, velocity and displacement with time during free fall.

Answer:

(i) Acceleration-Time Graph

• Acceleration remains constant.
• Value = –g.
• Graph is a horizontal straight line.

(ii) Velocity-Time Graph

• Velocity changes uniformly with time.
• Equation:

v = u - gt

• Graph is a straight line with negative slope.

(iii) Displacement-Time Graph

• Displacement increases as square of time.

s = ut - 1/2gt²

• Graph is parabolic.

Thus acceleration is constant, velocity changes uniformly, and displacement changes non-uniformly.

---

2. Prove Galileo's Law of Odd Numbers.

Answer:

For free fall:

s=1/2gt²

After time t:

s_1=1/2gt²

After 2t:

s_2=4s_1

After 3t:

s_3=9s_1

Distance in successive intervals:

First interval:

s_1

Second interval:

s_2-s_1=3s_1

Third interval:

s_3-s_2=5s_1

Fourth interval:

s_4-s_3=7s_1

Hence ratio:

1:3:5:7

Thus proved.

---

E. Assertion and Reason Questions

1.

Assertion (A): A freely falling body has constant acceleration.

Reason (R): Gravity acts uniformly near Earth's surface.

Answer: Both A and R are true, and R is the correct explanation of A.

---

2.

Assertion (A): Stopping distance depends on velocity.

Reason (R): Stopping distance is proportional to velocity squared.

Answer: Both A and R are true, and R is the correct explanation of A.

---

3.

Assertion (A): Velocity-time graph in free fall is a straight line.

Reason (R): Acceleration remains constant.

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

---

4.

Assertion (A): Displacement-time graph in free fall is linear.

Reason (R): Displacement is proportional to time squared.

Answer: Assertion is false but Reason is true.

---

F. Fill in the Blanks

1. Motion under gravity alone is called free fall.

2. The value of acceleration due to gravity is approximately 9.8 m/s².

3. The slope of a velocity-time graph gives acceleration.

4. Galileo's law follows the ratio 1 : 3 : 5 : 7.

5. Stopping distance is proportional to the square of velocity.

6. The SI unit of acceleration is m/s².

7. The displacement-time graph of free fall is a parabola.

8. A body released from rest has initial velocity zero.

---

G. Case Study Questions

Case Study

A ball is dropped from the top of a tower. It falls freely under gravity. The acceleration remains constant throughout the motion. The velocity increases uniformly while displacement increases rapidly with time.

Questions

1. What is the acceleration acting on the ball?

Answer: g = 9.8 m/s² downward.

---

2. Which force acts on the ball during free fall?

Answer: Gravitational force.

---

3. What is the shape of the velocity-time graph?

Answer: Straight line.

---

4. What is the shape of the displacement-time graph?

Answer: Parabola.

---

5. Which law explains distances covered in successive seconds?

Answer: Galileo's Law of Odd Numbers.

---

H. Statement-Based Questions

Statement 1

Acceleration due to gravity remains constant during free fall.

Statement 2

Velocity changes uniformly with time.

a) Both statements are true.
b) Both statements are false.
c) Statement 1 true, Statement 2 false.
d) Statement 1 false, Statement 2 true.

Answer: a) Both statements are true.

---

Statement 1

Stopping distance is proportional to speed.

Statement 2

Stopping distance is proportional to square of speed.

Answer: Statement 1 is false and Statement 2 is true.

---

I. Match the Columns

Column A	Column B
1. Free Fall	a. Gravity only
2. g	b. 9.8 m/s²
3. v–t graph slope	c. Acceleration
4. Galileo	d. Odd number law
5. Stopping Distance	e. Depends on v²

Answers

1 → a

2 → b

3 → c

4 → d

5 → e

---

CBSE Exam Important Questions

1. Define free fall and explain its characteristics.

2. Draw and explain acceleration-time, velocity-time and displacement-time graphs for free fall.

3. State and prove Galileo's Law of Odd Numbers.

4. Derive the formula for stopping distance.

5. Explain why stopping distance increases with speed.

6. Write equations of motion for a freely falling body.

7. Differentiate between velocity and acceleration during free fall.

These questions cover MCQs, competency-based questions, assertion-reason, case study, fill in the blanks, statement-based questions, match the columns, short answers, and long answers as per the latest CBSE Class 11 examination pattern. 

 Internal Links

Motion in a Straight Line Notes

Acceleration and Velocity Concepts

Kinematics Formula Sheet

Free Fall Concepts

Acceleration Due to Gravity Notes

Newton's Law of Universal Gravitation

Projectile Motion Basics

Graph Section

Velocity-Time Graph Explained

Acceleration-Time Graph Problems

Position-Time Graph Interpretation

Galileo Law Section

Motion Under Constant Acceleration

Important NEET Kinematics Questions

NCERT Kinematics Solutions

Stopping Distance Section

Newton's Laws of Motion

Friction Notes Class 11

Braking Force and Retardation Problems

Revision Section

NEET Physics Formula Handbook

Most Important Kinematics Numericals

NEET Physics Previous Year Questions

Motion of an Object Under Free Fall

NEET Physics Easy Notes

1. Free Fall – Basic Idea

• When an object falls under the effect of gravity only, the motion is called free fall.
• Air resistance is neglected in free fall problems.
• Acceleration due to gravity is represented by g.

Near Earth’s surface:

g ≈ 9.8 m/s²

• Direction of gravity is always downward.
• If upward direction is taken positive, then acceleration becomes negative.

Fig. 2.7 : Motion of Object Under Free Fall

(a) Variation of Acceleration with Time

• Acceleration remains constant throughout the motion.
• The graph is a horizontal straight line.
• Value of acceleration is always:

a = -g

Important NEET Point

Constant acceleration means velocity changes uniformly with time.

(b) Variation of Velocity with Time

• Initial velocity for a freely falling body released from rest:

u = 0

• Velocity increases linearly with time.
• Equation of velocity:

v = u - gt

Since (u = 0),

v = -gt

Graph Understanding

• Straight line with negative slope.
• Slope of v–t graph = acceleration = (-g).

Important NEET Concepts

• Velocity becomes more negative with time.
• Body gains speed while falling downward.

(c) Variation of Distance (Position) with Time

• Distance covered in free fall is proportional to square of time.
• Equation of motion:

y = ut - 1/2 gt²

For (u = 0),

y = -1/2 gt²

Graph Understanding

• Graph is a parabola.
• Distance increases rapidly with time.
• Motion is non-uniform because velocity changes continuously.

Galileo’s Law of Odd Numbers

Statement

• “The distances travelled during successive equal intervals of time by a freely falling body are in the ratio of odd numbers.”

Ratio is:

1 : 3 : 5 : 7 : 9 : ...

Proof in Simple Steps

Step 1: Position after Different Times

For free fall:

y = -1/2 gt²

After time (t):

y₁ = 1/2 gt²

After time (2t):

y₂ = 1/2 g(2t)² = 4y₁

After time (3t):

y₃ = 1/2 g(3t)² = 9y₁

After time (4t):

y₄ = 16y₁

Step 2: Distance in Successive Intervals

First Interval

y₁ = 1y₁

Second Interval

y₂ - y₁ = 4y₁ - y₁ = 3y₁

Third Interval

y₃ - y₂ = 9y₁ - 4y₁ = 5y₁

Fourth Interval

y₄ - y₃ = 16y₁ - 9y₁ = 7y₁

Hence ratios become:

1 : 3 : 5 : 7

Important Result for NEET

For a body starting from rest under gravity:

sₙ ∝ (2n - 1)

where:

• (sₙ) = distance travelled in nth second.

Example 2.6 – Stopping Distance of Vehicles

Definition

• Distance travelled by a vehicle before coming to rest after brakes are applied is called stopping distance.

Derivation

Using equation of motion:

v² = u² + 2as

For stopping:

• Final velocity (v = 0)
• Initial velocity (u = v₀)

So,

0 = v₀² + 2adₛ

Therefore,

dₛ = -v₀² / 2a

Conclusions

• Stopping distance is proportional to square of initial velocity.

dₛ ∝ v₀²

NEET Important Points

• If speed doubles → stopping distance becomes 4 times.
• Stronger brakes mean larger retardation and smaller stopping distance.

Quick Revision Formula Sheet

Concept	Formula
Velocity-Time Relation	v = u - gt
Position-Time Relation	s = ut - 1/2 gt²
Velocity-Position Relation	v² = u² - 2gs

One-Line NEET Tricks

• Acceleration due to gravity is constant.
• v–t graph slope gives acceleration.
• Distance in nth second follows odd number rule.
• Stopping distance depends on square of speed.
• Free fall graphs are very important for NEET numericals.

NEET Physics Easy Notes © 2026

Kinematics Equations Explained for NEET | 3 Equations of Motion Notes

KINEMATICS EQUATIONS – EASY NEET NOTES

1. Three Important Equations of Motion

These equations are used for:

• ✅ Straight line motion
• ✅ Constant acceleration

They connect:

• Initial velocity (\(v_0\))
• Final velocity (\(v\))
• Acceleration (\(a\))
• Time (\(t\))
• Displacement (\(x\))

First Equation of Motion

\[ v = v_0 + at \]

Meaning

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.

© Easy NEET Physics Notes – Kinematics Equations

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. 

 INTERNAL LINKS

Motion in a Straight Line Notes

Velocity and Acceleration Explained

Example 3.4 Solution Explained for Beginners

Block and Trolley System NEET Solution

Newton's Laws of Motion Notes

Vector Addition and Subtraction

Important Physics Derivations for NEET

Projectile Motion Notes

Free Fall and Gravity Problems

NCERT Kinematics Solutions

Kinematic Equations of Motion | NEET & Class 11 Physics Notes (Complete Guide)

KINEMATIC EQUATIONS FOR UNIFORMLY ACCELERATED MOTION

NEET Physics Easy Notes

1. Area under v–t Graph = Displacement

Fig. 2.4 Explanation

• Velocity–time graph is a straight horizontal line.
• Horizontal line means velocity is constant.
• Time is shown on x-axis.
• Velocity is shown on y-axis.

The shaded area forms a rectangle.

Rectangle dimensions:

• Height = velocity = (u)
• Base = time = (T)

Area = u × T

Since area under velocity-time graph gives displacement,

x = uT

Important Point

• Area under v–t graph always gives displacement.
• Unit check:

(velocity) × (time) = m/s × s = m

So result is displacement.

2. Real Graphs are Smooth

Book Note Meaning

• Some graphs show sharp corners.
• In reality, velocity and acceleration do not change suddenly.
• Physical quantities change continuously.

NEET Point

• Real motion graphs are smooth curves.
• Sudden jumps are not realistic.

3. Uniformly Accelerated Motion

Uniform acceleration means:

• Acceleration remains constant throughout motion.

Main Quantities

• (u) = initial velocity
• (v) = final velocity
• (a) = acceleration
• (t) = time
• (x) = displacement

4. First Equation of Motion

Relationship between velocity and time:

v = u + at

Meaning

• Final velocity increases uniformly with time.
• Every second velocity changes by (a).

Rearranged Forms

a = (v - u)/t

t = (v - u)/a

NEET Tip

Use when:

• time is involved
• displacement is NOT required

5. Derivation of Second Equation of Motion

Step 1: Area under v–t graph

Displacement = area of rectangle + area of triangle

Rectangle Area

• Height = (u)
• Base = (t)

Rectangle Area = ut

Triangle Area

• Height = (v - u)
• Base = (t)

Triangle Area = 1/2 (v - u)t

Total Displacement

x = ut + 1/2 (v - u)t

From first equation:

v - u = at

Substitute:

x = ut + 1/2 at²

6. Second Equation Meaning

• Object already moves with initial velocity (u).
• Additional displacement comes due to acceleration.

Special Case

If object starts from rest:

u = 0

Then,

x = 1/2 at²

7. Third Equation of Motion

Using:

t = (v - u)/a

and substituting into displacement equation:

We get:

v² = u² + 2ax

8. Third Equation Meaning

• Connects velocity and displacement directly.
• Time is absent.

NEET Tip

Use when:

• time is NOT given
• displacement, velocity and acceleration are given

9. Average Velocity Formula

For constant acceleration:

v̄ = (u + v)/2

Displacement:

x = v̄t

So,

x = (u + v)t / 2

10. Important NEET Formula Sheet

Concept	Formula
First Equation	v = u + at
Second Equation	x = ut + 1/2 at²
Third Equation	v² = u² + 2ax
Average Velocity	v̄ = (u + v)/2

11. Quick Concept Tricks for NEET

If acceleration is zero

a = 0

Then:

v = u

Motion becomes uniform motion.

If object starts from rest

u = 0

Equations become:

v = at

x = 1/2 at²

v² = 2ax

12. Graph-Based NEET Concepts

Velocity-Time Graph

• Slope = acceleration
• Area under graph = displacement

Acceleration-Time Graph

• Area under graph = change in velocity

Displacement-Time Graph

• Slope = velocity

13. Most Important NEET Mistakes

❌ Using km/h instead of m/s

❌ Forgetting sign of acceleration

❌ Using wrong equation

❌ Taking displacement as distance

❌ Forgetting (u = 0) for rest condition

14. One-Line Revision

• Area under v–t graph → displacement
• Slope of v–t graph → acceleration
• Constant acceleration → use equations of motion
• Average velocity in uniform acceleration:

(u + v)/2

NEET Physics Easy Notes © 2026

- Dr.Sanjaykumar Pawar  

CBSE Class 11 Physics: Kinematics (Uniformly Accelerated Motion)

Complete Question Bank with Answers

---

Very Short Answer Questions (1 Mark)

Q1. What is uniform acceleration?
Ans: Uniform acceleration is acceleration that remains constant throughout the motion.

Q2. What is SI unit of acceleration?
Ans: m/s²

Q3. What does slope of velocity-time graph represent?
Ans: Acceleration

Q4. What does area under velocity-time graph represent?
Ans: Displacement

Q5. Write first equation of motion.
Ans: v = u + at

Q6. Write second equation of motion.
Ans: x = ut + 1/2 at²

Q7. Write third equation of motion.
Ans: v² = u² + 2ax

Q8. What is average velocity in uniform acceleration?
Ans: (u + v)/2

Q9. What is acceleration when velocity is constant?
Ans: Zero

Q10. What is initial velocity for a body starting from rest?
Ans: Zero

---

Short Answer Questions (2–3 Marks)

Q1. Why does area under velocity-time graph give displacement?
Ans: Velocity = displacement/time.
Area = velocity × time = m/s × s = m.
Hence, area under v–t graph gives displacement.

Q2. Differentiate between distance and displacement.
Ans: Distance is scalar and total path length.
Displacement is vector and shortest distance between initial and final position.

Q3. Define average velocity.
Ans: Average velocity = total displacement / total time.
For uniform acceleration, (u + v)/2.

Q4. What happens when acceleration is zero?
Ans: Velocity remains constant and motion becomes uniform motion.

---

Derivations (3–5 Marks)

Q1. Derive first equation of motion.
Ans: a = (v - u)/t
v - u = at
v = u + at

Q2. Derive second equation of motion.
Ans: Displacement = area under v–t graph
x = ut + 1/2 (v - u)t
Using v - u = at,
x = ut + 1/2 at²

Q3. Derive third equation of motion.
Ans: v = u + at ⇒ t = (v - u)/a
x = (u + v)/2 × t
x = (u + v)(v - u)/2a
2ax = v² - u²
v² = u² + 2ax

---

Numerical Question

Q1. A body starts from rest and accelerates at 4 m/s² for 5 s. Find final velocity.

Ans:
u = 0, a = 4 m/s², t = 5 s
v = u + at
v = 0 + 4 × 5 = 20 m/s

---

MCQs

Q1. Slope of v–t graph gives:
Ans: Acceleration

Q2. Area under v–t graph gives:
Ans: Displacement

Q3. If acceleration is zero:
Ans: v = u

Q4. Which equation does not contain time?
Ans: v² = u² + 2ax

Q5. Unit of displacement is:
Ans: meter (m)

---

Assertion and Reason

Q1.
Assertion: Area under v–t graph gives displacement.
Reason: Velocity × time gives displacement.
Ans: Both are true and Reason is correct explanation.

Q2.
Assertion: Slope of s–t graph gives acceleration.
Reason: Slope of s–t graph gives velocity.
Ans: Assertion false, Reason true.

---

Fill in the Blanks

1. Slope of v–t graph gives ________.
Answer: acceleration

2. Area under v–t graph gives ________.
Answer: displacement

3. SI unit of velocity is ________.
Answer: m/s

4. A body at rest has initial velocity ________.
Answer: zero

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Match the Column

1. Slope of v–t graph → Acceleration
2. Area under v–t graph → Displacement
3. Slope of s–t graph → Velocity
4. Area under a–t graph → Change in velocity

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Case Study

Case: A car starts with velocity 10 m/s and accelerates at 2 m/s² for 5 s.

Q1. Final velocity?
v = 10 + 2×5 = 20 m/s

Q2. Displacement?
x = 10×5 + 1/2×2×25 = 75 m

Q3. Average velocity?
(u + v)/2 = (10 + 20)/2 = 15 m/s

Q4. Type of motion?
Uniformly accelerated motion

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Conclusion

This question bank covers MCQs, numericals, derivations, assertion-reason, fill in blanks, and case-based questions for CBSE Class 11 Physics exam preparation.

Kinematics Made Easy: All 3 Equations of Motion Explained for NEET & JEE 

Kinematic Equations of Motion with Velocity-Time Graph and Key Formulas Explained

🔗 Internal Links 

Add these as internal linking keywords:

Motion in a Straight Line Notes

Velocity-Time Graph Explanation

Derivation of v = u + at

Second Equation of Motion Proof

Third Equation of Motion Formula

NEET Physics Formula Sheet

Important Physics Graphs for NEET

Uniform Acceleration Problems

Class 11 Physics Chapter 3 Notes

Kinematics MCQs Practice Set