CBSE Class 11 Physics Vectors Notes, MCQs, Questions & Answers | NEET Preparation

 - Dr.Sanjaykumar Pawar  

Vectors Class 11 Physics Notes, MCQs, Assertion Reason, Case Study & CBSE Questions

Complete Class 11 Physics Vectors Notes with formulas, diagrams, MCQs, assertion-reason, case studies and CBSE exam questions.

Internal Links

• Class 11 Physics Units and Measurements
• Motion in a Straight Line Notes
• Motion in a Plane
• Projectile Motion
• Laws of Motion
• Work, Energy and Power
• System of Particles
• Circular Motion
• Kinematics Formula Sheet
• Physics Formula Handbook
• NEET Physics Notes
• CBSE Class 11 Physics MCQs
• Class 11 Physics Previous Year Questions
• Class 11 Physics Sample Papers
• NCERT Solutions for Class 11 Physics
• Important Physics Derivations
• Physics Practical Experiments
• Physics Revision Notes
• Physics Chapter-wise Question Bank
• CBSE Class 11 Study Material

NEET Physics Chapter : Vectors

Vectors are one of the most important topics in NEET Physics. Almost every chapter uses vectors. Therefore understanding vectors properly makes Mechanics very easy.

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1. Equality of Vectors

Two vectors are called equal if

• Magnitude is same.
• Direction is same.

Position does NOT matter. Even if vectors are shifted parallel, they are still equal.

Equal vectors ⇒ Same Magnitude + Same Direction

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2. Addition of Vectors

Vector addition means combining two vectors to get one resultant vector.

Triangle Law

Place the tail of second vector at the head of first vector. Join the starting point to the final point. That gives resultant.

Remember: Head to Tail Rule

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3. Parallelogram Law

If two vectors start from the same point, complete a parallelogram. Diagonal gives resultant vector.

Resultant = Diagonal of Parallelogram

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4. Magnitude of Resultant

Suppose

First Vector = a

Second Vector = b

Angle between them = θ

R = √(a² + b² + 2ab cosθ)

This is one of the MOST IMPORTANT formulas for NEET. Learn it perfectly.

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5. Direction of Resultant

tanα = (b sinθ)/(a + b cosθ)

α = angle made by resultant with first vector.

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6. Special Cases

Angle	Magnitude
0°	a+b
180°	|a-b|
90°	√(a²+b²)

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7. Example

Question: Two vectors have equal magnitude A. Angle between them is θ. Find resultant.

Solution

R = √(A²+A²+2A²cosθ)

= √(2A²(1+cosθ))

Using 1+cosθ=2cos²(θ/2)

R = 2A cos(θ/2)

Resultant = 2A cos(θ/2)

Direction:

α = θ/2

The resultant bisects the angle between two equal vectors.

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8. Memory Tricks

✔ Triangle Rule → Head to Tail

✔ Parallelogram Rule → Diagonal

✔ Equal Vectors → Same Magnitude + Same Direction

✔ 90° → Pythagoras

✔ 180° → Subtraction

✔ 0° → Addition

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9. NEET Important Points

• Magnitude is always positive.
• Direction decides vector.
• Vectors obey triangle law.
• Resultant depends on angle.
• Equal vectors can have different positions.
• Parallelogram law is frequently asked in NEET.

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10. Practice Questions

1. Define equal vectors.
2. State triangle law.
3. State parallelogram law.
4. Write magnitude formula.
5. Write direction formula.
6. Find resultant when angle is 90°.
7. Find resultant when angle is 180°.
8. Two vectors 10 N each make 60°. Find resultant.
9. Two vectors 5 N each make 120°. Find resultant.
10. Why does the resultant bisect equal vectors?

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Summary

• Equal vectors → Same magnitude + same direction
• Triangle Law → Head to Tail
• Parallelogram Law → Diagonal
• Magnitude → √(a²+b²+2abcosθ)
• Direction → tanα=(bsinθ)/(a+bcosθ)
• Equal vectors → Resultant = 2Acos(θ/2)
• Direction = θ/2

CBSE Class 11 Physics

Chapter : Vectors Question Bank

1. Multiple Choice Questions (MCQs)

1. A vector quantity has

1. Only magnitude
2. Only direction
3. Magnitude and direction
4. None

Answer: C

2. Equal vectors have

1. Equal magnitude only
2. Equal direction only
3. Equal magnitude and direction
4. Different directions

Answer: C

3. The diagonal of a parallelogram represents

1. Difference of vectors
2. Resultant vector
3. Unit vector
4. Zero vector

Answer: B

2. Very Short Answer Questions (1 Mark)

Q1. Define a vector.

A quantity having both magnitude and direction is called a vector.

Q2. Give one example of a vector.

Velocity.

Q3. What is a zero vector?

A vector whose magnitude is zero.

3. Short Answer Questions (2-3 Marks)

Q1. Define equal vectors.

Two vectors having equal magnitude and same direction are called equal vectors.

Q2. State the triangle law of vector addition.

If two vectors are represented by two sides of a triangle taken in order, the third side taken in opposite order represents the resultant.

4. Long Answer Questions (5 Marks)

Q1. Explain the parallelogram law of vector addition with diagram.

If two vectors acting simultaneously are represented by two adjacent sides of a parallelogram, then the diagonal passing through the common point represents the resultant vector. Magnitude: R = √(A² + B² + 2AB cosθ) Direction: tanα = (B sinθ)/(A + B cosθ)

5. Assertion and Reason

Assertion: Equal vectors may have different positions.

Reason: A vector depends only on magnitude and direction.

Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.

Assertion: Scalar quantities have direction.

Reason: Scalars possess only magnitude.

Answer: Assertion is false. Reason is true.

6. Fill in the Blanks

Question	Answer
A vector has ______ and ______.	Magnitude, Direction
The diagonal of a parallelogram gives the ______.	Resultant
A quantity having only magnitude is called ______.	Scalar
The SI unit of displacement is ______.	metre

7. Match the Columns

Column A	Column B
Velocity	Vector
Mass	Scalar
Acceleration	Vector
Time	Scalar

Answers Velocity → Vector Mass → Scalar Acceleration → Vector Time → Scalar

8. Statement Based Questions

Statement I: The resultant of two equal vectors bisects the angle between them.

Statement II: The magnitudes of both vectors are equal.

Both statements are true. Statement II explains Statement I.

9. Case Study Questions

Rahul pushes a box with force 20 N towards east. Aman pushes the same box with force 20 N making an angle of 60°. Answer the following.

1. Which law is used?
2. Write the magnitude formula.
3. If angle becomes 180°, what happens?

1. Parallelogram law.
2. R = √(A²+B²+2ABcosθ)
3. Resultant = |A−B|

10. Numerical Questions

Two vectors of magnitude 5 N each make an angle of 60°. Find the resultant.

R = √(25+25+50×0.5) = √75 = 8.66 N

11. HOTS Questions

Why can two vectors be equal even if they are drawn at different places?

Because a vector depends only on magnitude and direction, not on position.

12. Competency Based Questions

A boat is moving across a river. Which physical quantities should be treated as vectors?

Velocity, displacement and acceleration.

13. One Word Questions

Question	Answer
Quantity having direction	Vector
Quantity having only magnitude	Scalar
Magnitude zero vector	Zero Vector
Vector of magnitude one	Unit Vector

14. Important CBSE Questions

1. Define vector.
2. State triangle law.
3. State parallelogram law.
4. Define equal vectors.
5. What is a unit vector?
6. What is a null vector?
7. Derive the magnitude formula.
8. Derive the direction formula.
9. Differentiate scalar and vector.
10. Give five examples each of scalars and vectors.

Conservation of Mechanical Energy Class 11 Notes, MCQs, Questions & Answers | CBSE & NEET

-  Dr Sanjay Kumar Pawar 

Conservation of Mechanical Energy Class 11 Physics Notes PDF | CBSE & NEET 

Conservation of Mechanical Energy explained with a falling ball showing the conversion of potential energy into kinetic energy.

Internal Links

Link this page to related Class 11 Physics topics to improve SEO and user navigation:

1. Work, Energy and Power Class 11 Notes
2. Work-Energy Theorem Explained
3. Potential Energy Class 11 Notes
4. Kinetic Energy Formula and Examples
5. Conservative and Non-Conservative Forces
6. Gravitational Potential Energy Notes
7. Free Fall Motion Class 11
8. Laws of Motion Class 11
9. Newton's Laws of Motion Notes
10. Circular Motion Class 11 Notes
11. System of Particles and Rotational Motion
12. Gravitation Class 11 Notes
13. Mechanical Properties of Solids
14. Complete Class 11 Physics Notes Index 
15. Class 11 Physics MCQs with Answers
16. CBSE Class 11 Physics Important Questions
17. NEET Physics Chapter-wise Notes
18. NCERT Solutions for Class 11 Physics
19. Class 11 Physics Formula Sheet
20. Previous Year CBSE Class 11 Physics Questions

Chapter 5.8
Conservation of Mechanical Energy

Definition:
Mechanical Energy is the sum of Kinetic Energy (KE) and Potential Energy (PE).

Mechanical Energy = KE + PE

1. Work-Energy Theorem

Suppose a body moves through a small distance Δx under the action of force F. According to the Work-Energy Theorem,

ΔK = F(x) Δx

This means the work done by a force changes the kinetic energy of the body.

• Positive work increases kinetic energy.
• Negative work decreases kinetic energy.

2. Conservative Force

If the force is conservative, then potential energy can be defined.

−ΔU = F(x) Δx

The negative sign shows that whenever potential energy decreases, kinetic energy increases.

Example:
A falling stone loses potential energy and gains kinetic energy.

3. Combining the Equations

From the two equations:

ΔK = −ΔU

Therefore,

ΔK + ΔU = 0

or

Δ(K + U) = 0

4. Conservation of Mechanical Energy

Since Δ(K+U)=0, the total mechanical energy never changes.

K + U = Constant

This is called the Law of Conservation of Mechanical Energy.

5. Equation Between Two Positions

Ki + Ui = Kf + Uf

The total mechanical energy before motion equals the total mechanical energy after motion.

6. Conservative Force - Important Properties

• Potential energy can be defined.
• Work depends only on initial and final positions.
• Work does not depend on the path.
• Work done in a closed path is zero.
• Mechanical energy remains conserved.

7. Example - Falling Ball

A ball of mass m is dropped from height H. Initially the velocity is zero.

At Height H

PE = mgH

KE = 0

EH = mgH

At Height h

PE = mgh

KE = ½mv²h

Eh = mgh + ½mv²h

At Ground Level

PE = 0

KE = ½mv²f

E0 = ½mv²f

8. Conservation of Energy

EH = Eh = E0

Since only gravity acts on the body, mechanical energy remains constant.

mgH = mgh + ½mv²h = ½mv²f

9. Final Velocity

Using conservation of energy,

mgH = ½mv²f

After simplifying,

vf = √(2gH)

10. Velocity at Height h

mgH = mgh + ½mv²h

Therefore,

vh² = 2g(H − h)

11. Energy Conversion

Position	Potential Energy	Kinetic Energy
Top	Maximum	Zero
Middle	Decreasing	Increasing
Ground	Zero	Maximum

12. Important Points for NEET

• Mechanical Energy = KE + PE
• Gravity is a conservative force.
• Spring force is also conservative.
• Mechanical energy remains constant if only conservative forces act.
• Work done by a conservative force depends only on the initial and final positions.
• Work done in a closed path is zero.
• At the highest point, PE is maximum and KE is zero.
• At the ground, KE is maximum and PE is zero.
• Potential energy converts into kinetic energy during free fall.

13. Formula Sheet

Mechanical Energy = KE + PE

ΔK + ΔU = 0

K + U = Constant

Ki + Ui = Kf + Uf

PE = mgh

KE = ½mv²

vf = √(2gH)

vh² = 2g(H − h)

Work done in a Closed Path = 0

Prepared for NEET Physics Students • Chapter 5.8 - Conservation of Mechanical Energy

Conservation of Mechanical Energy

What is Mechanical Energy?

Mechanical Energy is the sum of Kinetic Energy (KE) and Potential Energy (PE).

Mechanical Energy = KE + PE

Work-Energy Theorem

When a force acts on an object, its kinetic energy changes.

ΔKE = Work Done

For conservative forces, Potential Energy decreases when Kinetic Energy increases.

ΔKE + ΔPE = 0

Energy Conversion During Falling

Top
PE Maximum
KE Zero

Middle
PE ↓
KE ↑

Ground
PE Zero
KE Maximum

Visual Falling Ball

⬇

As the ball falls, Potential Energy continuously converts into Kinetic Energy.

Example

Position	Potential Energy	Kinetic Energy	Total Energy
Top	100 J	0 J	100 J
Middle	60 J	40 J	100 J
Ground	0 J	100 J	100 J

Important Formulae

PE = mgh

KE = ½mv²

KE + PE = Constant

vf = √(2gH)

vh² = 2g(H − h)

NEET Remember:

• Gravity is a conservative force.
• Total Mechanical Energy remains constant if only conservative forces act.
• At the highest point: PE is maximum and KE is zero.
• At the ground: KE is maximum and PE is zero.
• Potential Energy converts into Kinetic Energy during falling.

CBSE Class 11 Physics

Chapter 5.8 - Conservation of Mechanical Energy

Question Bank with Answers

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Part A - Multiple Choice Questions

1. Mechanical energy is the sum of

1. Potential energy and Heat energy
2. Kinetic energy and Potential energy
3. Heat energy and Electrical energy
4. Sound energy and Potential energy

Answer: B

2. Mechanical energy remains constant when

1. Friction acts
2. Air resistance acts
3. Only conservative forces act
4. External force acts

Answer: C

3. Which one is a conservative force?

1. Friction
2. Gravity
3. Air resistance
4. Viscous force

Answer: B

4. Work done by a conservative force depends on

1. Path followed
2. Distance travelled
3. Initial and final positions only
4. Speed

Answer: C

5. Work done in a closed path by gravity is

1. Positive
2. Negative
3. Zero
4. Infinite

Answer: C

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Part B - Very Short Answer Questions

1. Define mechanical energy.

Answer: Mechanical energy is the sum of kinetic energy and potential energy.

2. Write the formula of mechanical energy.

Answer: E = KE + PE

3. Name one conservative force.

Answer: Gravitational force.

4. Write the formula of kinetic energy.

Answer: KE = ½mv²

5. Write the formula of potential energy.

Answer: PE = mgh

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Part C - Short Answer Questions

1. What is conservation of mechanical energy?

Answer:
When only conservative forces act on a body, the total mechanical energy (kinetic energy + potential energy) remains constant throughout the motion.

2. What is a conservative force?

Answer:
A conservative force is a force whose work depends only on the initial and final positions and not on the path followed. Examples:

• Gravity
• Spring force

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Part D - Long Answer Questions

1. State the law of conservation of mechanical energy.

Answer:
If only conservative forces act on a body, its total mechanical energy remains constant. Mechanical Energy = Kinetic Energy + Potential Energy Initial Energy Ki + Ui Final Energy Kf + Uf Therefore, Ki + Ui = Kf + Uf Example: A freely falling body loses potential energy and gains kinetic energy. The total mechanical energy remains constant.

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Part E - Assertion and Reason

Assertion: Mechanical energy remains constant when only conservative forces act.

Reason: Gravity is a conservative force.

Answer: Both Assertion and Reason are true, and Reason is the correct explanation.

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Part F - Fill in the Blanks

1. Mechanical energy is the sum of ______ and ______.
2. Gravity is a ______ force.
3. Potential energy converts into ______ during free fall.
4. Work done in a closed path is ______.
5. Mechanical energy remains ______ when only conservative forces act.

Answers:

1. Kinetic energy, Potential energy
2. Conservative
3. Kinetic energy
4. Zero
5. Constant

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Part G - Match the Columns

Column A	Column B
Gravity	Conservative force
Friction	Non-conservative force
Potential Energy	mgh
Kinetic Energy	½mv²

Matching Answers 1 → Conservative force 2 → Non-conservative force 3 → mgh 4 → ½mv²

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Part H - Case Study

A ball of mass 2 kg is dropped from a height of 20 m. Ignore air resistance.

Q1. Which energy is maximum at the top?

Answer: Potential Energy

Q2. Which energy is maximum at the ground?

Answer: Kinetic Energy

Q3. Calculate total mechanical energy at the top. Take g = 10 m/s².

PE = mgh = 2 × 10 × 20 = 400 J Answer = 400 Joule

Q4. Is mechanical energy conserved?

Yes. Only gravity acts.

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Important Formulae

• E = KE + PE
• KE = ½mv²
• PE = mgh
• Ki + Ui = Kf + Uf
• Δ(KE + PE) = 0
• v = √(2gH)
• v² = 2g(H − h)
• Work in closed path = 0

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End of CBSE Class 11 Physics Question Bank

Retardation (Negative Acceleration) Practice Questions with Answers | CBSE Class 9 & NEET Physics Notes

 Dr Sanjay Kumar Pawar

Retardation (Negative Acceleration) Notes, MCQs & Numericals | CBSE & NEET 

Retardation (Negative Acceleration) explained with formulas, solved examples, and practice questions for CBSE Class 9 and NEET.

Internal Links

1. Motion in One Dimension Notes

2. Distance and Displacement Explained

3. Speed, Velocity and Acceleration Notes

4. Uniform and Non-Uniform Motion

5. Equations of Motion with Derivations

6. Graphs of Motion (Distance-Time & Velocity-Time)

7. Newton's Laws of Motion

8. Force and Inertia Notes

9. Work, Energy and Power

10. Gravitation Notes

11. Units and Measurements

12. Scalars and Vectors

13. Physics Formula Sheet for Class 9

14. CBSE Class 9 Science Chapter-wise Notes

15. NEET Foundation Physics Practice Questions

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Retardation (Negative Acceleration)

CBSE | NCERT | NEET Foundation

Very Short Answer Questions (1 Mark)

1. What is retardation?
   Retardation is the decrease in velocity of a moving body with time. It is also called negative acceleration.
2. What is another name for negative acceleration?
   Deceleration or Retardation.
3. What is the SI unit of retardation?
   m/s²
4. Can acceleration be negative?
   Yes. Negative acceleration is called retardation.
5. A vehicle slows down. What is the sign of acceleration?
   Negative (−)
6. What is the final velocity of a body that comes to rest?
   0 m/s
7. Which equation of motion is used when time is not given?
   v² = u² + 2as
8. What happens to the speed during retardation?
   The speed decreases.
9. What is the SI unit of velocity?
   m/s
10. What is the SI unit of displacement?
    metre (m)

Short Answer Questions (2 Marks)

Q1. Differentiate between Acceleration and Retardation

Acceleration	Retardation
Speed increases	Speed decreases
Positive	Negative
Velocity increases	Velocity decreases

Q2. Write the three equations of motion.

v = u + at

s = ut + ½at²

v² = u² + 2as

Q3. Write the equations for retardation.

v = u − at

s = ut − ½at²

v² = u² − 2as

Solved Numerical Questions (3 Marks)

Question 1

A bike moves with an initial velocity of 20 m/s. It stops after 5 seconds. Find its retardation.

Given
u = 20 m/s
v = 0 m/s
t = 5 s

v = u + at

0 = 20 + 5a

5a = -20

a = -4 m/s²

Retardation = 4 m/s²

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Question 2

A car moving at 30 m/s comes to rest in 6 seconds. Find the retardation.

Given
u = 30 m/s
v = 0 m/s
t = 6 s

v = u + at

0 = 30 + 6a

6a = -30

a = -5 m/s²

Retardation = 5 m/s²

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Question 3

A train slows down from 40 m/s to 20 m/s in 10 seconds. Find the acceleration.

Given
u = 40 m/s
v = 20 m/s
t = 10 s

v = u + at

20 = 40 + 10a

10a = -20

a = -2 m/s²

Retardation = 2 m/s²

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Question 4

A bus moving at 25 m/s stops after travelling 125 m. Find the retardation.

Given
u = 25 m/s
v = 0 m/s
s = 125 m

v² = u² + 2as

0 = 25² + 2 × a × 125

0 = 625 + 250a

250a = -625

a = -2.5 m/s²

Retardation = 2.5 m/s²

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Question 5

A car moving at 15 m/s stops after travelling 45 m. Find the retardation.

Given
u = 15 m/s
v = 0 m/s
s = 45 m

v² = u² + 2as

0 = 15² + 2 × a × 45

0 = 225 + 90a

90a = -225

a = -2.5 m/s²

Retardation = 2.5 m/s²

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Question 6

A scooter slows from 18 m/s to 6 m/s in 4 seconds. Find the retardation.

a = (6 − 18) / 4

a = -3 m/s²

Retardation = 3 m/s²

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Question 7

A train moving at 72 m/s stops in 12 seconds. Find the retardation.

a = (0 − 72) / 12

a = -6 m/s²

Retardation = 6 m/s²

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Question 8

A bus slows from 50 m/s to 20 m/s in 10 seconds. Find the retardation.

a = (20 − 50) / 10

a = -3 m/s²

Retardation = 3 m/s²

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Question 9

A car moving at 10 m/s stops in 2 seconds. Find the retardation.

a = (0 − 10) / 2

a = -5 m/s²

Retardation = 5 m/s²

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Question 10

A truck moving at 36 m/s comes to rest in 9 seconds. Find the retardation.

a = (0 − 36) / 9

a = -4 m/s²

Retardation = 4 m/s²

Multiple Choice Questions (MCQs)

1. Retardation is also called

A. Positive Acceleration
B. Uniform Motion
C. Negative Acceleration
D. Velocity

Answer: C. Negative Acceleration

2. SI unit of acceleration is

A. metre
B. second
C. m/s²
D. km/h

Answer: C. m/s²

3. When brakes are applied, acceleration becomes

A. Positive
B. Zero
C. Negative
D. Infinite

Answer: C. Negative

4. Which equation is used when time is not given?

A. v = u + at
B. s = ut
C. v² = u² + 2as
D. a = v/t

Answer: C. v² = u² + 2as

5. Final velocity of a body at rest is

A. 5 m/s
B. 10 m/s
C. 0 m/s
D. 100 m/s

Answer: C. 0 m/s

6. Retardation decreases

A. Time
B. Velocity
C. Distance
D. Mass

Answer: B. Velocity

7. Initial velocity is represented by

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

Answer: C. u

8. Final velocity is represented by

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

Answer: D. v

9. Time is represented by

A. a
B. t
C. s
D. u

Answer: B. t

10. Displacement is represented by

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

Answer: A. s

Higher Order Thinking Questions (HOTS)

Q1. Why is retardation called negative acceleration?

Retardation acts opposite to the direction of motion and decreases the velocity of a moving body. Therefore, its value is negative.

Q2. Can a body move forward while having negative acceleration?

Yes. If a moving body slows down while still moving forward, it has negative acceleration (retardation).

Assertion and Reason

Question 1

Assertion (A): Retardation decreases the speed of a moving body.

Reason (R): Retardation acts opposite to the direction of motion.

Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Question 2

Assertion (A): A car moving at constant speed has retardation.

Reason (R): Constant speed means acceleration is zero.

Assertion is false, but Reason is true.

Case Study Question

A car is moving with an initial velocity of 60 m/s. The driver applies brakes and the car stops after travelling 180 m.

Questions

1. What is the initial velocity?
2. What is the final velocity?
3. Which equation of motion will be used?
4. Find the retardation.

Answers

Initial Velocity = 60 m/s

Final Velocity = 0 m/s

Formula Used:

v² = u² + 2as

0 = 60² + 2 × a × 180

0 = 3600 + 360a

360a = -3600

a = -10 m/s²

Retardation = 10 m/s²

Quick Revision

v = u + at

s = ut + ½at²

v² = u² + 2as

For Retardation

v = u − at

s = ut − ½at²

v² = u² − 2as

Exam Tips

• Always write the SI unit in the final answer.
• Write: Given → Formula → Substitution → Calculation → Final Answer.
• Use v² = u² + 2as when time is not given.
• Negative acceleration means retardation.
• If the body stops, final velocity (v) = 0.

CBSE & NEET Foundation Physics Notes

Practice Daily • Revise Formulas • Solve Numericals • Score Full Marks

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.

---

Standard of Comparison (Large and Small Physical Quantities)

 Standard of Comparison Explained | CBSE Class 11 Physics | NEET Notes 

Standard of Comparison in Physics with real-life examples for NEET and CBSE students.

- Dr. Sanjaykumar Pawar 

Q. 1.4 Explain this statement clearly:

“To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison.”

Answer (CBSE Topper Style)

A dimensional physical quantity has both magnitude and unit. Whether it is called large or small depends on what it is being compared with.

Therefore, simply saying that a quantity is "large" or "small" has no meaning unless a standard or reference quantity is mentioned.

Example:

• A distance of 10 m is large compared to the size of a classroom but very small compared to the distance between two cities.

Hence, every statement describing a dimensional quantity as large or small should include a suitable standard of comparison.

---

Reframing the given statements

(a) Atoms are very small objects.

Reframed Statement:
Atoms are very small compared to ordinary objects, such as a grain of sand or a pinhead.

---

(b) A jet plane moves with great speed.

Reframed Statement:
A jet plane moves with great speed compared to road vehicles, but its speed is much smaller than the speed of light.

---

(c) The mass of Jupiter is very large.

Reframed Statement:
The mass of Jupiter is very large compared to the mass of the Earth (or any other planet like Mars).

---

(d) The air inside this room contains a large number of molecules.

Reframed Statement:
The air inside this room contains a very large number of molecules compared to the number of people or other visible objects present in the room.

---

(e) A proton is much more massive than an electron.

Answer:
This statement is already meaningful because it clearly specifies the standard of comparison (electron).

---

(f) The speed of sound is much smaller than the speed of light.

Answer:
This statement is already meaningful because it compares the speed of sound with the speed of light, providing a clear standard of comparison.

---

Final Answer (Exam Points)

• A quantity can be called large or small only with respect to a specified standard or reference.
• Without comparison, such statements are meaningless.

Corrected Statements:

(a) Atoms are very small compared to ordinary objects.
(b) A jet plane moves with great speed compared to road vehicles.
(c) The mass of Jupiter is very large compared to the mass of the Earth.
(d) The air inside this room contains a very large number of molecules compared to the number of people in the room.
(e) No change required, as the comparison is with an electron.
(f) No change required, as the comparison is with the speed of light. 

NEET Master Learning Structure (NMLS)

हर टॉपिक के लिए एक यूनिवर्सल फॉर्मेट

---

1. Topic Name

Chapter: Units and Measurements

Topic: Standard of Comparison (Large & Small Quantities)

---

2. One-Line Definition (Exam Definition)

Rule:

किसी भी भौतिक राशि (Physical Quantity) को "बड़ा" या "छोटा" तभी कहा जा सकता है जब उसके साथ तुलना (Comparison) का मानक (Standard) दिया गया हो।

---

3. Why? (Concept Building)

याद रखने वाला नियम

👉 बड़ा या छोटा = हमेशा तुलना

यदि तुलना नहीं है तो कथन अधूरा है।

---

4. Visual Flow Chart

Physical Quantity
        │
        ▼
Large / Small ?
        │
        ▼
Compared With?
      /      \
    Yes      No
    │         │
Meaningful  Meaningless

---

5. Memory Trick (Mnemonic)

Mnemonic

"LSC"

L → Large

S → Small

C → Comparison

Large or Small ⇒ Comparison Must

---

दूसरा आसान Mnemonic

"No Comparison = No Conclusion"

(NC = NC)

---

6. Quick Table

Statement	Comparison Given?	Correct / Incorrect
Atom is small	❌ No	Incorrect
Atom is smaller than a grain of sand	✅ Yes	Correct
Jet is fast	❌ No	Incorrect
Jet is faster than a car	✅ Yes	Correct
Proton is heavier than electron	✅ Yes	Correct
Speed of sound is less than speed of light	✅ Yes	Correct

---

7. Comparison Table

Quantity	Compared With	Final Statement
Atom	Everyday objects	Very small
Jet Plane	Car	Very fast
Jupiter	Earth	Very massive
Air Molecules	Humans	Very large number
Proton	Electron	Much heavier
Sound	Light	Much slower

---

8. Mind Map

             Large / Small
                    │
        ┌───────────┼───────────┐
        │           │           │
     Heavy       Fast       Long
        │           │           │
        └───────────┼───────────┘
                    │
          Need Comparison
                    │
           Otherwise Wrong

---

9. Common Mistakes

❌ Atom is very small.

✔ Atom is very small compared to everyday objects.

---

❌ Car is very fast.

✔ Car is faster than a bicycle.

---

❌ Jupiter is very massive.

✔ Jupiter is much more massive than Earth.

---

10. Examiner's Favourite Questions

Type-1

Explain why comparison is necessary.

---

Type-2

Rewrite the statement correctly.

---

Type-3

Which statement is meaningful?

---

Type-4

Assertion-Reason

---

Type-5

Choose the incorrect statement.

---

11. NEET Trick

Whenever you read these words

Large

Small

Heavy

Light

Fast

Slow

High

Low

Long

Short

Immediately ask

"Compared to What?"

यदि उत्तर मिल गया

✅ Statement Correct

यदि उत्तर नहीं मिला

❌ Statement Incorrect

---

12. PYQ Thinking Method

Question पढ़ते ही

STEP-1

Physical Quantity पहचानो

↓

STEP-2

Large/Small लिखा है?

↓

STEP-3

Comparison दिया है?

↓

Yes → Correct Statement

No → Wrong Statement

---

13. 10-Second Revision

✔ Large = Relative

✔ Small = Relative

✔ Heavy = Relative

✔ Fast = Relative

✔ Comparison Must

✔ No Comparison = No Meaning

---

14. Golden Rule (100% Exam Point)

यदि किसी वाक्य में Large, Small, Heavy, Light, Fast, Slow, High या Low लिखा हो, तो तुरंत जाँचें कि तुलना (Comparison) दी गई है या नहीं।

यदि तुलना नहीं दी गई है, तो कथन वैज्ञानिक रूप से अधूरा (Meaningless) माना जाएगा।

---

15. One-Line Formula

Large/Small + Comparison = Meaningful Statement

Large/Small − Comparison = Meaningless Statement

NEET Practice Question Bank

Chapter: Units and Measurements

Topic: Standard of Comparison (Large & Small Quantities)

---

Section A: Direct MCQs (Single Correct Option)

Easy Level

Q1.

Which of the following statements is scientifically meaningful?

A. A mountain is very high.

B. An atom is very small.

C. A proton is more massive than an electron.

D. A river is very long.

✅ Answer: C

Explanation: Only option C provides a standard of comparison.

---

Q2.

Which statement is incomplete?

A. Earth is larger than the Moon.

B. Light travels faster than sound.

C. Jupiter is more massive than Earth.

D. A train is very fast.

✅ Answer: D

---

Moderate Level

Q3.

Which statement correctly follows the principle of comparison?

A. Molecules are tiny.

B. A jet plane is fast compared to a bicycle.

C. A room contains many molecules.

D. The Sun is hot.

✅ Answer: B

---

Q4.

Which statement is NOT meaningful?

A. A proton is heavier than an electron.

B. Sound travels slower than light.

C. Mercury is smaller than Earth.

D. Iron is very heavy.

✅ Answer: D

---

Hard Level

Q5.

Which option correctly identifies all meaningful statements?

1. Atom is very small.

2. Proton is heavier than electron.

3. Speed of light is greater than speed of sound.

4. Jupiter is very massive.

Options

A. 1 and 4

B. 2 and 3

C. 1,2,3

D. All

✅ Answer: B

---

Section B: Statement-Based Questions

Q6.

Statement I

Every physical quantity described as "large" or "small" requires a comparison.

Statement II

Without a standard of comparison, such statements are scientifically incomplete.

A. Both statements are true and II explains I.

B. Both true but II does not explain I.

C. I true II false.

D. I false II true.

✅ Answer: A

---

Q7.

Statement I

The statement "An atom is very small" is scientifically complete.

Statement II

A meaningful statement must include a standard of comparison.

A. TT

B. TF

C. FT

D. FF

✅ Answer: C

---

Section C: Assertion & Reason

Q8.

Assertion (A)

"The speed of sound is much smaller than the speed of light."

Reason (R)

The statement specifies a standard for comparison.

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

B. Both true but R is not the explanation.

C. A true R false.

D. A false R true.

✅ Answer: A

---

Q9.

Assertion

"A jet plane moves with great speed."

Reason

The statement gives the comparison standard.

A. TT Correct Explanation

B. TT Wrong Explanation

C. TF

D. FT

✅ Answer: C

---

Section D: Match the Columns

Column I

A. Atom

B. Proton

C. Sound

D. Jupiter

Column II

1. Compared with Earth

2. Compared with Electron

3. Compared with Light

4. Compared with Everyday Objects

Correct Matching

A → 4

B → 2

C → 3

D → 1

Answer

A-4

B-2

C-3

D-1

---

Section E: Diagram-Based Question

Flow Chart

          Statement
               │
               ▼
      Large / Small ?
               │
        ┌──────┴──────┐
        │             │
Comparison Given?   No Comparison
        │             │
        ▼             ▼
Meaningful      Meaningless

Question

The statement

"Gold is heavy."

lies in which branch?

A. Meaningful

B. Meaningless

C. Dimensionless

D. Scalar

✅ Answer: B

---

Section F: Higher Order Thinking (HOTS)

Q10.

A student writes

"The Earth is very large."

The teacher marks it incorrect.

Which correction is most appropriate?

A. Earth is large compared to the Moon.

B. Earth is always large.

C. Earth is absolutely large.

D. Earth has a large radius.

✅ Answer: A

---

Q11.

Which one of the following is an example of a relative statement?

A. Length is measured in metre.

B. Speed of light is constant.

C. The Himalayas are higher than the Aravalli Hills.

D. Mass is a fundamental quantity.

✅ Answer: C

---

Section G: NEET PYQ-Type Mixed MCQs

Q12.

Identify the meaningful statements.

1. Air contains many molecules.

2. Jupiter is more massive than Earth.

3. Electron is lighter than proton.

4. Mountain is very high.

Options

A. 1,2

B. 2,3

C. 1,3

D. All

✅ Answer: B

---

Q13.

Which of the following words usually require a comparison to become scientifically meaningful?

1. Large

2. Small

3. Fast

4. Heavy

Options

A. Only 1

B. 1 and 2

C. 1,2,3

D. All

✅ Answer: D

---

Section H: Very Hard NEET Conceptual Question

Q14.

Which statement best explains why "The Sun is very large" is scientifically incomplete?

A. The Sun is not large.

B. Size has no unit.

C. No comparison standard has been specified.

D. The Sun's mass is unknown.

✅ Answer: C

---

Section I: NEET Trap Question

Q15.

Which one is correctly written?

A. Atom is tiny.

B. Jupiter is huge.

C. Proton is much heavier than electron.

D. Car is fast.

✅ Answer: C

---

Final Revision Trick

Whenever you read these words in a NEET question:

✔ Large

✔ Small

✔ Heavy

✔ Light

✔ Fast

✔ Slow

✔ High

✔ Low

Immediately ask:

👉 "Compared to What?"

If comparison is given → Correct

If comparison is missing → Incorrect or Incomplete

 Internal Links

From this article link to

✔ Physical Quantities

✔ Fundamental and Derived Quantities

✔ SI Units

✔ International System of Units

✔ Dimensions

✔ Dimensional Formula

✔ Dimensional Analysis

✔ Significant Figures

✔ Errors in Measurement

✔ Scientific Notation

✔ Order of Magnitude

✔ Accuracy and Precision

✔ Vernier Calipers

✔ Screw Gauge

✔ NCERT Class 11 Physics Chapter 2 Notes

✔ NEET Physics MCQ Collection

✔ Physics Formula Sheet

✔ Previous Year Questions 

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.

---

(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