Displacement in the nth Second Formula | Complete Notes with Derivation

Displacement during the nth second under constant acceleration explained with formula and illustration.

-  Dr.Sanjaykumar Pawar

 nth Second Displacement Formula | Class 11 Physics Notes

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 Internal Links

• Motion in a Straight Line Notes

• Distance and Displacement Explained

• Speed, Velocity and Acceleration

• Equations of Motion (SUVAT)

• Graphs of Motion (Distance-Time & Velocity-Time)

• Average Velocity Formula

• Relative Motion Notes

• Uniform and Non-Uniform Motion

• Free Fall Motion

• Projectile Motion Basics

• Kinematics Formula Sheet

• Motion Formula Cheat Sheet

• Class 11 Physics Chapter 3 Notes

• JEE Kinematics Questions

• NEET Physics Motion Practice Problems

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FAQ 

Q1. What is displacement in the nth second?
A. It is the displacement covered by an object during only the nth second of its motion.

Q2. What is the formula for displacement in the nth second?
A. The formula is:
Sₙₜₕ = u + (a/2)(2n−1)

Q3. When can this formula be used?
A. It is applicable only when the acceleration is constant.

Q4. Is nth second displacement the same as total displacement?
A. No. Total displacement is measured from the start of motion, whereas nth second displacement is measured only during one specific second.

Q5. Which exams include questions on the nth second formula?
A. This topic is important for Class 11 Physics, JEE Main, JEE Advanced, NEET, and other competitive examinations. 

Displacement in the n^th Second

1. What is Displacement in the n^th Second?

Displacement in the n^th second means the displacement covered during only one particular second.

Examples

• 1st second → 0 s to 1 s
• 2nd second → 1 s to 2 s
• 3rd second → 2 s to 3 s
• 5th second → 4 s to 5 s

It does not mean the total displacement from the starting point.

2. Total Displacement After n Seconds

The displacement after n seconds is given by:

S_n = un + ½ an²

Where:

• u = Initial velocity
• a = Constant acceleration
• n = Time in seconds
• S_n = Total displacement after n seconds

3. Total Displacement After (n−1) Seconds

The displacement up to the end of the previous second is:

S_n−1 = u(n−1) + ½ a(n−1)²

This represents the total displacement before the n^th second begins.

4. Formula for Displacement in the n^th Second

Displacement during only the n^th second is:

S_nth = S_n − S_n−1

This means:

Displacement in one interval = Total displacement at the end − Total displacement at the beginning.

5. Substitute the Values

S_nth = (un + ½an²) − [u(n−1) + ½a(n−1)²]

Now simplify the expression.

6. Final Formula

S_nth = u + (a/2)(2n − 1)

This is the standard formula used to calculate displacement during the n^th second.

Meaning of Symbols

Symbol	Meaning
Snth	Displacement during the nth second
u	Initial velocity
a	Constant acceleration
n	Number of the second

Example

Given:

• u = 5 m/s
• a = 2 m/s²
• Find displacement in the 4th second.

S_4th = 5 + (2/2)(2×4 −1)

= 5 + 7

= 12 m

Answer: The object moves 12 m during the 4th second.

Important Points

• Acceleration must remain constant.
• The formula gives displacement, not total displacement.
• The time interval is always 1 second.
• The n^th second starts at (n−1) s and ends at n s.
• Examples:
  • 3rd second → 2 s to 3 s
  • 6th second → 5 s to 6 s
  • 10th second → 9 s to 10 s

Total Displacement vs Displacement in the n^th Second

Total Displacement	Displacement in nth Second
Measured from 0 s to n s	Measured only during one second
S = ut + ½at²	Snth = u + (a/2)(2n−1)
Gives total position change	Gives position change in one specific second

Quick Revision

1. Find total displacement after n seconds.
2. Find total displacement after (n−1) seconds.
3. Subtract both values.
4. Use the final formula:

S_nth = u + (a/2)(2n−1)

Memory Trick:

n^th second = Total displacement till n − Total displacement till (n−1).

NEET Practice Questions – Displacement in the n^th Second

Formula: S_nth = u + ^a⁄₂(2n − 1)

Type 1: Direct MCQs

Q1 (Easy)
Displacement in the n^th second is:

A. u + a/2(2n+1)      B. u + a/2(2n−1)      C. ut + ½at²      D. un + ½an²

Answer: B

Q2
u = 8 m/s, a = 4 m/s². Find displacement in 3rd second.

Answer: 18 m

Q3
Starts from rest, a = 6 m/s². Displacement in 5th second?

Answer: 27 m

Q4
6th second displacement = 29 m, a = 2 m/s². Find u.

Answer: 18 m/s

Q5 (Hard)
8th second displacement = 40 m, a = 4 m/s². Find u.

Answer: 10 m/s

Type 2: Statement Based

Q6
Statement I: S_nth = S_n − S_(n−1).
Statement II: nth-second displacement equals difference of total displacements.

Answer: Both True.

Q7
Statement I: Formula works for variable acceleration.
Statement II: Constant acceleration is required.

Answer: I False, II True.

Q8
Statement I: Formula gives displacement.
Statement II: Distance and displacement are always equal.

Answer: I True, II False.

Type 3: Assertion & Reason

Q9
Assertion (A): 4th-second displacement = S₄ − S₃.
Reason (R): 4th second is from 3 s to 4 s.

Answer: Both true and R explains A.

Q10
Assertion (A): Formula works for changing acceleration.
Reason (R): Derivation assumes constant acceleration.

Answer: A False, R True.

Q11
Assertion (A): If u = 0, S_nth = (a/2)(2n−1).
Reason (R): Body starts from rest.

Answer: Both true and R explains A.

Type 4: Match the Columns

Column I	Column II
A. 1st second	P. 4–5 s
B. 2nd second	Q. 1–2 s
C. 5th second	R. 0–1 s
D. 8th second	S. 7–8 s

Correct Match: A-R, B-Q, C-P, D-S

Type 5: Diagram Based

Q12

0s ---- 1 ---- 2 ---- 3 ---- 4
4th second = 3 s to 4 s

Answer: 3 s to 4 s

Q13

S
|
|      *
|   *
| *
+---------------> t

Correct equation: S = ut + ½at²

Answer Key

1-B, 2-18 m, 3-27 m, 4-18 m/s, 5-10 m/s, 6-Both True, 7-I False, II True, 8-I True, II False, 9-Both true and R explains A, 10-A False, R True, 11-Both true and R explains A, 12-3 s to 4 s, 13-S = ut + ½at²

Column Match: A-R, B-Q, C-P, D-S

NEET Physics: Motion in One Dimension – Mechanics, Kinematics & Dynamics Notes

 

NEET Physics – Motion in One Dimension

Motion in One Dimension explained with Mechanics, Kinematics, Dynamics, and Frame of Reference for NEET aspirants.

Chapter 1: Basic Concepts of Mechanics (Complete Notes)

-Dr.Sanjaykumar Pawar 

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1. What is Physics?

Physics is the branch of science that studies matter, energy, motion, and forces.

Example:

• Why does a ball fall?

• Why does a bike move?

• Why do planets revolve around the Sun?

All these are studied in Physics.

---

2. What is Mechanics?

Mechanics is the branch of Physics that studies the motion of objects and the forces causing that motion.

Simple Definition

Mechanics = Study of motion + causes of motion.

---

Mechanics is divided into two parts

              Mechanics
             /          \
      Kinematics      Dynamics

---

3. Kinematics

Definition

Kinematics is the branch of mechanics that studies motion without considering the force causing it.

It tells us

• Where the object is

• How fast it moves

• How far it moves

• How long it takes

But NOT why it moves.

Example

A car moves with speed 40 km/h.

Kinematics studies

✔ Speed

✔ Distance

✔ Time

✔ Displacement

It does NOT study engine force.

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4. Dynamics

Definition

Dynamics studies motion along with the forces responsible for the motion.

It answers

Why does the object move?

Example

A football moves because a player kicks it.

Dynamics studies the force of the kick.

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Difference Between Kinematics and Dynamics

Kinematics	Dynamics
Studies motion only	Studies motion + force
Force is ignored	Force is considered
Easier chapter	Uses Newton's Laws
Example: Speed	Example: Force

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5. Frame of Reference

This is one of the most important concepts in NEET.

Definition

A Frame of Reference is a coordinate system attached to an observer from which measurements of position and time are made.

Simply,

It is the point from which an observer watches motion.

---

It contains

(1) Observer

Who is watching?

Example

You

Driver

Passenger

Teacher

---

(2) Space Coordinate

Where is the object?

Usually represented by

x-axis

y-axis

z-axis

---

(3) Time Coordinate

When is the object observed?

Measured using a clock.

---

Example

A train is moving.

Passenger says

"I am at rest."

A person standing on the platform says

"The train is moving."

Both are correct because their frames of reference are different.

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Important Points

Motion depends on

✔ Observer

✔ Position

✔ Time

---

There is NO Absolute Motion

Motion is always relative.

There is nothing called

❌ Absolute Motion

❌ Absolute Rest

Everything depends upon the observer.

---

Example

You are sitting inside a moving bus.

With respect to

Seat → You are at rest.

Road → You are moving.

Both answers are correct.

---

6. Rest and Motion

Rest

An object is at rest if its position does not change with respect to the observer.

Example

Book lying on a table.

---

Motion

An object is in motion if its position changes with time with respect to an observer.

Example

Running car.

Flying bird.

Moving train.

---

Remember

Motion requires

✔ Observer

✔ Position

✔ Time

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7. Point Object (Particle)

Very important for NEET.

Definition

A point object is an object whose size is very small compared to the distance travelled, so its dimensions can be ignored.

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Example

Bike length = 2 m

Distance travelled = 500 km

Compared to 500 km,

2 m is negligible.

So bike is treated as a point object.

---

Another Example

Earth revolves around the Sun.

Earth diameter = 12,742 km

Distance from Sun = 150 million km

Compared to this huge distance,

Earth behaves like a point object.

---

When can an object be treated as a Point Object?

Ignore size when

Distance travelled ≫ Size of object

---

Example

Object	Distance	Point Object?
Car travelling 200 km	Yes	✔
Cricket ball moving 40 m	Yes	✔
Pencil moving 2 cm	No	✖

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8. Position

Position tells where an object is located with respect to a chosen origin.

Example

If a boy stands 5 m to the right of the origin,

Position = +5 m

If he stands 3 m to the left,

Position = −3 m

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9. Origin

Origin is the reference point from which distances are measured.

Usually represented by

O

Example

<------|------|------|------>

     -2    -1    O    +1   +2

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10. Coordinate System

Used to locate objects.

One-dimensional (1D)

Only x-axis

Example

Train on straight track.

---

Two-dimensional (2D)

x-axis and y-axis

Example

Football ground.

---

Three-dimensional (3D)

x-axis

y-axis

z-axis

Example

Flying aeroplane.

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Important NEET Concepts

Observer

Person who measures motion.

---

Coordinate

Shows position.

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Time

Shows when motion occurs.

---

Motion depends on observer.

Always remember

Motion is Relative.

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Frequently Asked NEET Facts

SI Unit of Distance

metre (m)

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SI Unit of Time

second (s)

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SI Unit of Speed

m/s

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SI Unit of Force

Newton (N)

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Previous Year NEET Concepts

Example 1

A passenger sitting inside a moving train is

A) Moving with respect to train

B) At rest with respect to train

C) Moving with respect to seat

D) None

Answer

✅ B

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Example 2

A car moves from Delhi to Jaipur.

The car is treated as

A) Rigid body

B) Point object

C) Fluid

D) None

Answer

✅ B

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Example 3

Motion is

A) Absolute

B) Relative

C) Constant

D) None

Answer

✅ B

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Formula Sheet (This Topic)

There are no numerical formulas in this introductory topic. Focus on understanding the concepts.

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One-Page Revision

Mechanics

• Study of motion and the causes of motion.

Kinematics

• Motion only.

• No force.

Dynamics

• Motion + force.

Frame of Reference

• Observer + coordinate system + time.

Motion

• Position changes with time relative to an observer.

Rest

• Position does not change relative to an observer.

Point Object

• Size is negligible compared to the distance travelled.

Key Rule

• There is no absolute rest or absolute motion; all motion is relative to the observer.

These notes provide a strong conceptual foundation for Motion in One Dimension, a high-weightage topic in the NEET Physics syllabus. 

Internal Links

1. Units and Measurements Notes for NEET
2. Vector and Scalar Quantities
3. Motion in Two Dimensions
4. Laws of Motion
5. Work, Energy and Power
6. System of Particles and Rotational Motion
7. Gravitation Notes
8. Oscillations and SHM
9. Mechanical Properties of Solids
10. Mechanical Properties of Fluids
11. Thermal Physics Complete Notes
12. Waves and Sound Notes
13. Complete NEET Physics Formula Sheet
14. NEET Physics PYQs Chapter-wise
15. Motion in One Dimension MCQs with Answers
16. Motion in One Dimension Numericals for NEET 
17. Motion in One Dimension Revision Notes
18. Class 11 Physics Complete Notes
19. NEET Physics Mock Tests
20. NEET Physics Study Plan

🔥 NEET Physics – Motion in One Dimension

Part 1 : Ultra-Hard NEET Numericals

Instructions

• Total Questions : 25
• Difficulty : NEET Advanced
• Use SI Units.
• Take g = 10 m/s² whenever required.
• Do not use calculator unless instructed.

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Q1.

A car starts from rest and moves with a constant acceleration of 4 m/s² for 15 s. Calculate

1. Final velocity
2. Total displacement
3. Average velocity

Q2.

A train moving at 20 m/s accelerates uniformly to 50 m/s in 15 s. Find

1. Acceleration
2. Distance travelled

Q3.

A body travels

• 40 m East
• 30 m West
• 20 m East

Calculate

1. Total distance
2. Displacement

Q4.

A particle moves with velocity v = 5 + 2t where t is in seconds. Find

1. Velocity after 8 s
2. Displacement in first 8 s

Q5.

A ball is thrown vertically upward with speed 40 m/s. Calculate

1. Maximum height
2. Time to reach highest point
3. Total time of flight

Q6.

A runner completes

• First half distance at 10 m/s
• Second half distance at 15 m/s

Find average speed.

Q7.

A particle covers

• 100 m in North direction
• 100 m in East direction

Calculate

1. Distance
2. Displacement

Q8.

A car moving at 25 m/s stops uniformly in 5 seconds. Calculate

1. Retardation
2. Stopping distance

Q9.

Two cars move towards each other. Car A = 30 m/s Car B = 20 m/s Initially separated by 500 m. Find time to meet.

Q10.

A particle travels

• 2 hours at 40 km/h
• 3 hours at 60 km/h

Find

1. Total distance
2. Average speed

Q11.

A body starts from rest with acceleration 3 m/s² Find displacement in 12 seconds.

Q12.

A train moving at 72 km/h crosses a 300 m platform in 30 s. Find length of train.

Q13.

A cyclist increases speed uniformly from 5 m/s to 15 m/s over a distance of 100 m. Find acceleration.

Q14.

A body moves 10 m East 10 m North 10 m West Find displacement.

Q15.

A particle has velocity v = 20 − 5t Find

1. Time when particle stops
2. Distance travelled till then

Q16.

A stone falls freely from a tower. It reaches ground in 6 s. Find

1. Height of tower
2. Velocity on reaching ground

Q17.

A bus travels 80 km at 40 km/h and 120 km at 60 km/h. Find average speed.

Q18.

A particle moves according to s = 4t² + 3t Find velocity after 5 seconds.

Q19.

Two trains Length = 200 m each Speed = 15 m/s and 25 m/s move in opposite directions. Find time taken to cross each other.

Q20.

A particle covers first one-third distance with speed 10 m/s remaining distance with speed 20 m/s Find average speed.

Q21.

A body starts with velocity 15 m/s Acceleration 2 m/s² Find distance covered in 20 s.

Q22.

A particle moves 50 m East 120 m West 70 m East Calculate displacement.

Q23.

A train moving at 54 km/h crosses a man standing on platform in 12 seconds. Length of train?

Q24.

A particle starts from rest. Acceleration = 5 m/s² Find distance travelled during 4th second.

Q25. ⭐ NEET Challenge

A particle moves according to x = 5 + 4t + 2t² Find

1. Position after 6 s
2. Velocity after 6 s
3. Acceleration
4. Distance covered between 2 s and 6 s

Practice Rule

• Attempt all questions without looking at formulas.
• Write complete steps.
• Time yourself (45 minutes).
• Target Score : 25/25

End of Part 1
Next: Part 2 – Full NEET Mock Test (45 Questions)

🔥 Supplementary Part 2

Ultra-Hard Motion in One Dimension Numericals (Q26–Q50)

Instructions

• Difficulty: NEET Advanced
• Attempt without calculator.
• Take g = 10 m/s² unless otherwise stated.

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Q26

A car accelerates uniformly from 10 m/s to 40 m/s in 12 seconds. Find:

1. Acceleration
2. Distance travelled

Q27

A particle moves with velocity v = 8 + 3t. Find displacement during first 10 seconds.

Q28

A train moving at 90 km/h crosses a pole in 18 s. Find length of train.

Q29

A ball is projected vertically upward with speed 50 m/s. Find:

1. Maximum height
2. Total time of flight

Q30

A body covers:

• 120 m at 6 m/s
• 180 m at 9 m/s

Find average speed.

Q31

Two cars move in the same direction. Car A = 18 m/s Car B = 24 m/s Initial gap = 240 m. Find time taken by B to overtake A.

Q32

A particle starts from rest with acceleration 4 m/s². Find distance travelled during the 8th second.

Q33

A cyclist slows uniformly from 12 m/s to rest in 6 s. Find stopping distance.

Q34

A particle moves:

• 40 m North
• 30 m East
• 40 m South

Find displacement.

Q35

Velocity equation: v = 24 − 4t Find:

1. Time to stop
2. Distance travelled

Q36

A freely falling stone reaches the ground with velocity 60 m/s. Find height of the tower.

Q37

A train of length 180 m crosses a bridge of length 420 m in 24 s. Find speed of train.

Q38

A body moves according to x = 3t² + 2t + 5. Find velocity at t = 8 s.

Q39

A car travels:

• 100 km at 50 km/h
• 150 km at 75 km/h

Find average speed.

Q40

Two trains of lengths 250 m and 150 m move in opposite directions at 18 m/s and 22 m/s. Find crossing time.

Q41

A particle starts from rest. Acceleration = 6 m/s². Find displacement after 15 seconds.

Q42

A ball is thrown downward from a building with velocity 15 m/s. It reaches ground in 5 s. Find height of building.

Q43

A particle covers one-fourth distance at 12 m/s and remaining distance at 24 m/s. Find average speed.

Q44

Position equation: s = 2t³ − 5t² + 8. Find velocity at t = 3 s.

Q45

A train moving at 54 km/h overtakes another moving at 36 km/h. Both trains are 150 m long. Find overtaking time.

Q46

A body starts with velocity 20 m/s and acceleration 3 m/s². Find velocity after travelling 150 m.

Q47

A particle moves:

• 100 m East
• 60 m West
• 20 m East
• 40 m West

Find:

1. Total distance
2. Displacement

Q48

A stone is thrown upward from a cliff of height 80 m with speed 30 m/s. Find time taken to hit the ground.

Q49

A particle has acceleration a = 4 m/s². Initial velocity = 6 m/s. Find displacement in first 12 s.

Q50 ⭐ NEET Challenge

Position equation: x = 10 + 6t + 3t² Find:

1. Position at t = 8 s
2. Velocity at t = 8 s
3. Acceleration
4. Distance travelled between 4 s and 8 s

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Self Evaluation

Score	Performance
23–25	Excellent (NEET Top Rank Level)
18–22	Very Good
12–17	Good, Needs Practice
Below 12	Revise Theory & Formulae

End of Supplementary Part 2
Next: Part 3 – Full NEET Mock Test (45 Questions)

🔥 NEET Physics Mock Test

Motion in One Dimension (Part 3)

Total Questions : 45
Time : 45 Minutes
Marks : 180
Correct : +4
Wrong : −1

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Section A (MCQs)

Q1. SI unit of displacement is

A) m/s
B) m
C) km
D) s

Q2. Distance is a

A) Vector
B) Scalar
C) Tensor
D) None

Q3. Velocity can become zero during upward motion of a projectile at

A) Highest point
B) Lowest point
C) Throughout
D) Never

Q4. Average velocity equals average speed when

A) Circular path
B) Straight line without changing direction
C) Closed path
D) Random motion

Q5. Which quantity may be zero even when distance is not zero?

A) Speed
B) Time
C) Displacement
D) Velocity

Q6. A body starts from rest. Initial velocity is

A) 1 m/s
B) 10 m/s
C) 0 m/s
D) −1 m/s

Q7. Uniform velocity means

A) Constant speed only
B) Constant velocity
C) Variable speed
D) Variable direction

Q8. Unit of acceleration

A) m
B) m/s
C) m/s²
D) km/h

Q9. Graph of uniform motion is

A) Straight line
B) Circle
C) Parabola
D) Hyperbola

Q10. Which is a vector?

A) Speed
B) Distance
C) Velocity
D) Time

Q11. A car moves 20 m East then 20 m West. Displacement is

A)40 m B)20 m C)0 D)10 m

Q12. Average speed is

A)Total Distance/Total Time B)Displacement/Time C)v²-u² D)None

Q13. Motion along straight line is called

A)Rectilinear B)Circular C)Rotational D)Projectile

Q14. Retardation is

A)Positive acceleration B)Negative acceleration C)Zero acceleration D)Infinite acceleration

Q15. Speedometer measures

A)Velocity B)Distance C)Instantaneous Speed D)Average Speed

Q16. Odometer measures

A)Speed B)Distance C)Acceleration D)Velocity

Q17. Motion is always

A)Absolute B)Relative C)Constant D)Uniform

Q18. A freely falling body has acceleration

A)g B)0 C)2g D)Depends on mass

Q19. Unit of velocity

A)m B)m/s C)m/s² D)km

Q20. Equation v=u+at is valid for

A)Variable acceleration B)Constant acceleration C)Circular motion D)Random motion

Q21. Which quantity is never negative?

A)Velocity B)Displacement C)Speed D)Acceleration

Q22. Area under velocity-time graph gives

A)Acceleration B)Distance/Displacement C)Speed D)Force

Q23. Slope of displacement-time graph gives

A)Acceleration B)Velocity C)Force D)Momentum

Q24. Slope of velocity-time graph gives

A)Displacement B)Acceleration C)Time D)Distance

Q25. A body moving with constant velocity has

A)Zero acceleration B)Maximum acceleration C)Infinite acceleration D)Variable acceleration

Q26. Graph of rest on displacement-time graph is

A)Horizontal line B)Vertical line C)Curve D)Circle

Q27. SI unit of time

A)Hour B)Second C)Minute D)Day

Q28. Which is scalar?

A)Velocity B)Acceleration C)Distance D)Displacement

Q29. Initial velocity symbol

A)s B)t C)u D)v

Q30. Final velocity symbol

A)u B)a C)v D)s

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Section B (Assertion-Reason)

Q31. Motion is relative.
Reason: There is no absolute rest.

Q32. Average speed is always less than average velocity.
Reason: Speed is scalar.

Q33. Displacement may be zero while distance is not zero.
Reason: Distance is scalar.

Q34. Constant speed always means constant velocity.
Reason: Direction may change.

Q35. Uniform acceleration means equal change in velocity in equal intervals of time.

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Section C (Numericals)

Q36. A body starts from rest with acceleration 5 m/s². Find velocity after 8 s.

Q37. A train moving at 20 m/s travels for 30 s. Find distance.

Q38. A body moves 60 m East then 80 m West. Find displacement.

Q39. A stone is dropped from height 45 m. Find time to reach ground.

Q40. Find average speed if a car travels 120 km in 2 hours.

Q41. Find displacement using s=ut+½at² where u=10 m/s, a=2 m/s², t=5 s.

Q42. Find acceleration if velocity changes from 10 m/s to 30 m/s in 4 s.

Q43. A train of length 180 m crosses a pole in 12 s. Find speed.

Q44. A ball is thrown upward with 20 m/s. Find maximum height.

Q45. A body moves with velocity equation v=6+2t. Find velocity after 8 s.

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OMR Answer Sheet

Q	Ans	Q	Ans	Q	Ans
1	___	16	___	31	___
2	___	17	___	32	___
3	___	18	___	33	___
4	___	19	___	34	___
5	___	20	___	35	___
6	___	21	___	36	___
7	___	22	___	37	___
8	___	23	___	38	___
9	___	24	___	39	___
10	___	25	___	40	___
11	___	26	___	41	___
12	___	27	___	42	___
13	___	28	___	43	___
14	___	29	___	44	___
15	___	30	___	45	___

End of Part 3 – Full NEET Mock Test
Part 4: Detailed Answer Key & Step-by-Step Solutions

🔥 NEET Physics

Part 4 – Answer Key & Step-by-Step Solutions

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Section A (MCQs)

Q	Answer	Q	Answer	Q	Answer
1	B	11	C	21	C
2	B	12	A	22	B
3	A	13	A	23	B
4	B	14	B	24	B
5	C	15	C	25	A
6	C	16	B	26	A
7	B	17	B	27	B
8	C	18	A	28	C
9	A	19	B	29	C
10	C	20	B	30	C

---

Section B (Assertion–Reason)

Question	Correct Option
31	A (Both true, Reason correctly explains)
32	D (Assertion false, Reason true)
33	B (Both true, Reason not correct explanation)
34	D (Assertion false, Reason true)
35	A (Both true, Reason correctly explains)

---

Section C (Numericals)

Q36

Given u = 0 m/s a = 5 m/s² t = 8 s

v = u + at
v = 0 + 5 × 8

Answer = 40 m/s

Q37

Distance = Speed × Time
= 20 × 30

Answer = 600 m

Q38

East = +60 West = -80

Displacement = 60 - 80 = -20 m

Answer = 20 m West

Q39

s = 45 m u = 0 g =10

45 = 5t²
t² =9
t =3 s

Answer = 3 s

Q40

Average Speed =120/2

Answer =60 km/h

Q41

s = ut + ½at²
=10×5 + ½×2×25
=50+25

Answer =75 m

Q42

a=(30−10)/4
=20/4

Answer =5 m/s²

Q43

Speed =180/12

Answer =15 m/s

Q44

Maximum Height H=u²/2g
=20²/(20)
=20 m

Answer =20 m

Q45

v=6+2t
=6+16
Answer =22 m/s

---

Score Analysis

Marks	Performance
170–180	Outstanding ⭐⭐⭐⭐⭐
150–169	Excellent
120–149	Very Good
90–119	Good
Below 90	Needs Revision

Common Mistakes in NEET Motion in One Dimension

• Confusing distance with displacement.
• Using average velocity instead of average speed.
• Incorrect sign convention (+ and -).
• Using wrong kinematic equation.
• Mixing km/h and m/s units.
• Ignoring direction in vector quantities.
• Calculation mistakes in free-fall problems.

✔ Part 4 Completed

Next: Part 5 – 100 Previous Year Questions (NEET + AIIMS + JIPMER) with Detailed Solutions

🔥 Part 5

Previous Year Question Bank (NEET / AIPMT / AIIMS Style)

This section contains NEET-style questions inspired by previous exam patterns. They are designed for practice and concept building.

---

Q1. A body moves 10 m east and then 10 m west. What is its displacement?

A) 20 m   B) 10 m   C) 0 m   D) 5 m

Answer: C (0 m)

Q2. Which of the following is a scalar quantity?

A) Velocity B) Acceleration C) Distance D) Displacement

Answer: C

Q3. SI unit of acceleration is

A) m B) m/s C) m/s² D) km/h

Answer: C

Q4. A car starts from rest with acceleration 4 m/s². Find velocity after 5 s.

Answer: 20 m/s

Q5. A train moving at 25 m/s crosses a pole in 8 s. Find length of train.

Answer: 200 m

Q6. A body falls freely for 4 s. Find its velocity on reaching the ground.

Answer: 40 m/s

Q7. Average speed is defined as A) Distance / Time B) Displacement / Time C) Velocity / Time D) None

Answer: A

Q8. Which graph represents uniform motion?

Answer: Straight line on displacement–time graph

Q9. A body moves with constant velocity. Its acceleration is

Answer: Zero

Q10. A stone is thrown upward with speed 30 m/s. Find maximum height.

Answer: 45 m

---

🔥 PYQ Practice Set (Without Answers)

1. A body covers 240 m in 30 s. Find average speed.
2. A car accelerates from 10 m/s to 30 m/s in 5 s. Find acceleration.
3. Define displacement with one example.
4. Differentiate between speed and velocity.
5. A particle moves according to x = 4 + 5t + 2t². Find velocity at t = 6 s.
6. A train of length 250 m crosses a bridge of length 350 m in 24 s. Find speed.
7. A freely falling body reaches the ground in 8 s. Find height.
8. State any two differences between distance and displacement.
9. Find stopping distance of a car moving at 30 m/s if retardation is 5 m/s².
10. Find average speed if equal distances are covered at 20 km/h and 30 km/h.

---

🔥 Formula Revision Sheet

• v = u + at
• s = ut + ½at²
• v² = u² + 2as
• Average Speed = Total Distance / Total Time
• Average Velocity = Total Displacement / Total Time
• Distance = Speed × Time
• Acceleration = (v − u) / t

---

🎯 NEET Rapid Revision

• Motion is relative.
• Distance is scalar.
• Displacement is vector.
• Speed is scalar.
• Velocity is vector.
• Acceleration can be positive or negative.
• Retardation = Negative acceleration.
• Slope of displacement–time graph = Velocity.
• Slope of velocity–time graph = Acceleration.
• Area under velocity–time graph = Displacement.

★★★★★ Congratulations! ★★★★★ You have completed the complete practice package for Motion in One Dimension

Recommended Next Chapter: Motion in a Plane (2D Motion & Projectile Motion)

Stopping Distance Formula Explained with Solved Examples | Reaction & Braking Distance

 - Dr Sanjay Kumar Pawar

Stopping Distance: Formula, Reaction Distance, Braking Distance & Questions

Stopping Distance consists of Reaction Distance and Braking Distance. Learn the formula with an easy solved example.

Internal Links

• Motion in a Straight Line
• Equations of Motion
• Uniform and Non-Uniform Motion
• Speed, Velocity and Acceleration
• Distance and Displacement
• Newton's Laws of Motion
• Friction in Physics
• Work, Energy and Power
• Projectile Motion
• Free Fall Equations
• Average Speed Formula
• Relative Motion
• SUVAT Equations
• Kinematics Notes
• Physics Formula Sheet

Stopping Distance (Reaction Distance + Braking Distance)

What is Stopping Distance?

When a driver sees an obstacle, the car does not stop immediately. The car travels some distance before stopping completely.

The stopping distance has two parts:

• Reaction Distance → Distance travelled while the driver reacts.
• Braking Distance → Distance travelled after applying the brakes.

Stopping Distance = Reaction Distance + Braking Distance

S = Sr + Sb

1. Reaction Distance

Sr = u × t

Where

• u = Initial speed (m/s)
• t = Reaction time (seconds)

2. Braking Distance

Use the equation of motion:

v² = u² + 2as

When the car stops:

• v = 0
• a = Retardation

Therefore,

Sb = u² / 2a

Solved Example

Question

A car is moving with velocity 20 m/s. The driver's reaction time is 0.5 s. The maximum retardation is 4 m/s². Find the stopping distance.

Step 1 : Reaction Distance

Sr = u × t

Sr = 20 × 0.5

Sr = 10 m

Step 2 : Braking Distance

Sb = u² / 2a

= 20² / (2 × 4)

= 400 / 8

= 50 m

Step 3 : Stopping Distance

S = Sr + Sb

= 10 + 50

= 60 m

Answer = 60 m

Shortcut Formula

Stopping Distance

S = ut + u² / 2a

Important Points

• Reaction distance depends on reaction time.
• Braking distance depends on speed and braking force.
• Higher speed means longer stopping distance.
• Wet roads increase braking distance.
• Always maintain a safe distance while driving.

Practice Questions

Q1. A car moves at 10 m/s. Reaction time = 1 s Retardation = 5 m/s² Find stopping distance.

Q2. A car moves at 30 m/s. Reaction time = 1 s Retardation = 6 m/s² Find stopping distance.

Q3. A car moves at 15 m/s. Reaction time = 0.4 s Retardation = 3 m/s² Find stopping distance.

Q4. A car moves at 25 m/s. Reaction time = 0.8 s Retardation = 5 m/s² Find stopping distance.

Answers

Question	Answer
Q1	20 m
Q2	105 m
Q3	43.5 m
Q4	82.5 m

Quick Revision

Quantity	Formula
Reaction Distance	Sr = ut
Braking Distance	Sb = u² / 2a
Stopping Distance	S = Sr + Sb

Memory Trick

Think → Brake → Stop

• Think = Reaction Distance
• Brake = Braking Distance
• Stop = Stopping Distance

© Physics Notes for Beginners

Stopping Distance, Reaction Distance & Braking Distance Explained | NEET Physics Notes

- Dr Sanjay Kumar Pawar 

Reaction Distance, Braking Distance and Stopping Distance for NEET 

Stopping distance consists of reaction distance and braking distance, an important concept in NEET Physics.

Internal Links

1. Motion in One Dimension Notes
2. Distance, Displacement and Speed
3. Velocity and Acceleration Explained
4. Equations of Motion (SUVAT Equations)
5. Uniform and Non-uniform Motion
6. Graphs of Motion (Distance-Time & Velocity-Time)
7. Relative Motion Notes
8. Newton's Laws of Motion
9. Friction and Braking Force
10. Work, Energy and Power
11. Circular Motion
12. NEET Physics Formula Sheet
13. Top 100 NEET Physics MCQs
14. Kinematics Practice Questions
15. NEET Physics Revision Notes 

Stopping Distance (Reaction Distance + Braking Distance)

Definition

Stopping distance is the total distance travelled by a vehicle from the moment the driver notices an obstacle until the vehicle comes to complete rest.

Stopping Distance = Reaction Distance + Braking Distance

S = sr + sb

---

1. Reaction Distance (Thinking Distance)

Reaction distance is the distance travelled by the vehicle during the driver's reaction time before applying the brakes.

Reaction Time

Reaction time is the time taken by the driver from observing an obstacle to applying the brakes.

Generally, Reaction Time < 1 second

Formula

sr = u × tr

Where,

• sr = Reaction Distance
• u = Initial Velocity
• tr = Reaction Time

During reaction time:

• Brakes are not applied.
• Velocity remains constant.
• Acceleration = 0

---

2. Braking Distance

Braking distance is the distance travelled after applying the brakes until the vehicle comes to rest.

During braking,

• Vehicle decelerates.
• Final velocity becomes zero.

Formula

v² = u² − 2as

Since

v = 0

Therefore,

sb = u² / 2a

Where

• sb = Braking Distance
• a = Deceleration

---

3. Stopping Distance Formula

S = ut_r + u² / 2a

This is the most important formula for NEET examinations.

---

4. Safe Distance

Safe distance is the minimum distance required between two vehicles to avoid collision.

Safe Distance ≥ Stopping Distance

---

5. Difference Between Reaction Distance and Braking Distance

Reaction Distance	Braking Distance
Before applying brakes	After applying brakes
Velocity remains constant	Velocity decreases
Acceleration = 0	Negative acceleration
Depends on reaction time	Depends on braking force

---

6. Important NEET Points

• Reaction distance depends on speed and reaction time.
• Braking distance depends on speed².
• If speed doubles, braking distance becomes four times.
• Stopping distance = Reaction distance + Braking distance.
• Reaction time is usually less than 1 second.

---

7. Formula Sheet

Reaction Distance = u × tr

Braking Distance = u² / 2a

Stopping Distance = ut_r + u² / 2a

---

8. Practice Questions

Q1. A car moves with speed 20 m/s. Reaction time is 0.5 s. Find the reaction distance.

Answer: 20 × 0.5 = 10 m

Q2. A vehicle moves at 30 m/s. Braking deceleration is 5 m/s². Find braking distance.

Answer: 30² / (2 × 5) = 900 /10 = 90 m

Q3. A car travels at 25 m/s. Reaction time = 0.8 s Deceleration = 5 m/s² Find stopping distance.

Reaction Distance = 25 × 0.8 = 20 m
Braking Distance = 25² /10 = 62.5 m
Stopping Distance = 20 + 62.5
= 82.5 m

---

9. Multiple Choice Questions

1. During reaction time acceleration is
   • A. Positive
   • B. Negative
   • C. Zero ✔
   • D. Infinite
2. Reaction distance depends upon
   • A. Speed
   • B. Reaction time
   • C. Both A and B ✔
   • D. Braking force
3. Braking distance is proportional to
   • A. Speed
   • B. Speed² ✔
   • C. Time
   • D. Mass only
4. Stopping distance equals
   • A. Reaction distance
   • B. Braking distance
   • C. Reaction distance + Braking distance ✔
   • D. None
5. If speed doubles, braking distance becomes
   • A. Double
   • B. Half
   • C. Four Times ✔
   • D. Eight Times

---

10. One Minute Revision

✔ Reaction Distance = Before braking
✔ Braking Distance = After braking
✔ Stopping Distance = Reaction Distance + Braking Distance
✔ Reaction Distance = u × tr
✔ Braking Distance = u² / 2a
✔ Stopping Distance = ut_r + u² / 2a
✔ Braking Distance ∝ Speed²

Prepared for NEET Physics Students

Topic: Reaction Distance • Braking Distance • Stopping Distance

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

---

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

---

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

---

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

---

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

---

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

---

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

---

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

---

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

---

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)

---

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

---

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.

---

---

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

---

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

---

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)

---

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

---

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.

---

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

 - Dr.Sanjaykumar Pawar  

```html

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

```

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