Uniformly Accelerated Motion Class 11 Physics Notes | NEET & JEE MCQs

 - Dr.Sanjaykumar Pawar  

Uniformly Accelerated Motion (1-D) Physics Notes, Formulas & NEET Questions 

Uniformly Accelerated Motion (1-D): Equations of Motion, Graphs and NEET Physics Concepts

Internal Links

1. Motion in One Dimension Foundation

 Motion in One Dimension Class 11 Physics Notes

Before introducing uniformly accelerated motion, link readers to the basic concepts of displacement, velocity and speed.

2. Instantaneous Velocity & Acceleration

Instantaneous Velocity and Acceleration Explained

NCERT Physics Class 11 Chapter 2: Instantaneous Velocity & Acceleration

Under "Important Terms" section after explaining acceleration.

3. Vectors in Physics

Vectors Class 11 Physics Notes

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

  explaining displacement as a vector quantity.

4. Newton's Laws of Motion

Newton's Laws of Motion Class 11 Physics

Block and Trolley System NEET Solution | Acceleration & Tension Explained

acceleration concepts because force causes acceleration.

5. Free Fall Motion

Free Fall and Acceleration Due to Gravity

NEET Tips section where free fall problems are mentioned.

6. Graphical Motion Analysis

Velocity-Time and Acceleration-Time Graphs

 diagram-based questions.

Uniform acceleration formula

Equations of motion derivation

Motion in one dimension notes

Class 11 Physics chapter 2 notes

NEET physics motion questions

JEE physics kinematics notes

Velocity time graph questions

Acceleration numericals with solutions

CBSE Physics important questions

Physics revision notes for NEET

FAQ Schema Questions

Q1. What is uniformly accelerated motion?

Uniformly accelerated motion is motion in which acceleration remains constant with time.

Q2. What are the three equations of motion?

The three equations are v = u + at, s = ut + ½at² and v² = u² + 2as.

Q3. What does the slope of a velocity-time graph represent?

The slope of a velocity-time graph represents acceleration.

Q4. What does the area under a velocity-time graph represent?

The area represents displacement.

Q5. Is uniformly accelerated motion important for NEET and JEE?

Yes, it is a fundamental topic used in many mechanics problems.

Uniformly Accelerated Motion (1-D)

1. Meaning of Uniformly Accelerated Motion

Uniformly accelerated motion means an object is moving in a straight line and its acceleration remains constant with time.

• The velocity changes by the same amount in equal intervals of time.
• The motion takes place in one dimension.

Example: A bike increases its speed by 5 m/s every second.

Important Terms

1. Initial Velocity (u)

Initial velocity is the velocity of an object at the starting time. It is represented by u.

At starting time:
t = 0, velocity = u

2. Final Velocity (v)

Final velocity is the velocity of an object after a certain time. It is represented by v.

3. Acceleration (a)

Acceleration is the rate of change of velocity.

a = (v - u) / t

Unit of acceleration = m/s²

4. Displacement (s)

Displacement is the distance travelled by an object in a particular direction.

Unit = metre (m)

First Equation of Motion

v = u + at

Derivation:

Acceleration:

a = dv/dt

Rearranging:

a dt = dv

Integrating:

∫a dt = ∫dv

a(t - 0) = v - u

at = v - u

v = u + at

Final velocity = Initial velocity + Change in velocity

Second Equation of Motion

s = ut + 1/2 at²

Derivation:

Velocity:

v = ds/dt

Therefore:

ds = v dt

Using:
v = u + at

ds = (u + at)dt

After integration:

s = ut + 1/2 at²

Displacement = Distance due to initial velocity + Distance due to acceleration

Third Equation of Motion

v² = u² + 2as

Derivation:

From first equation:

v = u + at

Rearranging:

t = (v - u)/a

Using second equation:

s = ut + 1/2 at²

After simplification:

v² = u² + 2as

This equation is useful when time is not given.

Three Equations of Motion Summary

1. Velocity Equation

v = u + at

Used to find final velocity when time is given.

2. Displacement Equation

s = ut + 1/2 at²

Used to find displacement when time is given.

3. Time Independent Equation

v² = u² + 2as

Used when time is not given.

Easy Memory Trick

• V-U-AT: v = u + at
• S-U-T-A-T: s = ut + 1/2 at²
• V-U-AS: v² = u² + 2as

Symbols at a Glance

Symbol	Meaning	Unit
u	Initial Velocity	m/s
v	Final Velocity	m/s
a	Acceleration	m/s²
t	Time	second
s	Displacement	metre

Conclusion

The three equations of motion are used to solve problems involving constant acceleration in one-dimensional motion.

NEET Physics Practice Questions

Chapter: Uniformly Accelerated Motion (1-D)

NEET Question Types

• Direct MCQs (Single Correct Option)
• Statement Based Questions
• Assertion and Reason
• Match the Columns
• Diagram Based / Graphical Questions

PART 1: Direct MCQs (Single Correct Option)

Q1. A car starts from rest and accelerates uniformly at 4 m/s². Its velocity after 5 seconds will be:

A) 10 m/s
B) 20 m/s
C) 25 m/s
D) 40 m/s

Solution:
u = 0
a = 4 m/s²
t = 5 s

v = u + at
v = 0 + 4 × 5
v = 20 m/s

Answer: B) 20 m/s

Q2. The SI unit of acceleration is:

A) m/s
B) m²/s
C) m/s²
D) km/h

Answer: C) m/s²

Q3. A body moving with velocity 20 m/s is brought to rest in 5 seconds. The acceleration is:

A) +4 m/s²
B) -4 m/s²
C) +5 m/s²
D) -5 m/s²

a = (v-u)/t
a = (0-20)/5
a = -4 m/s²

Answer: B) -4 m/s²

Q4. The velocity-time graph for uniformly accelerated motion is:

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

Answer: A) Straight line

Q5. A particle starts from rest and travels 100 m in 5 seconds with uniform acceleration. Find acceleration.

A) 4 m/s²
B) 6 m/s²
C) 8 m/s²
D) 10 m/s²

s = ut + 1/2 at²
100 = 0 + 1/2 × a × 25
a = 8 m/s²

Answer: C) 8 m/s²

PART 2: Statement Based Questions

Options:
A) Both statements true and II explains I
B) Both true but II does not explain I
C) I true, II false
D) I false, II true

Q6.

Statement I: Velocity-time graph for uniformly accelerated motion is a straight line.

Statement II: Acceleration is constant in uniformly accelerated motion.

Answer: A

Q7.

Statement I: Displacement can be zero even if distance travelled is not zero.

Statement II: Displacement depends only on initial and final position.

Answer: A

Q8.

Statement I: Acceleration due to gravity is constant near Earth's surface.

Statement II: Value of g is approximately 9.8 m/s².

Answer: A

PART 3: Assertion and Reason

Options:
A) Both true and R explains A
B) Both true but R does not explain A
C) A true, R false
D) A false, R true

Q9.

Assertion: A body moving with constant velocity has zero acceleration.

Reason: Acceleration is the rate of change of velocity.

Answer: A

Q10.

Assertion: Area under velocity-time graph gives displacement.

Reason: Velocity is displacement divided by time.

Answer: A

Q11.

Assertion: A body can have zero velocity and non-zero acceleration.

Reason: At highest point of upward motion, velocity is zero but acceleration acts downward.

Answer: A

PART 4: Match the Columns

Column I	Column II
Velocity	m/s
Acceleration	m/s²
Displacement	m
Time	second

Answer: Velocity-m/s, Acceleration-m/s², Displacement-m, Time-second

PART 5: Diagram Based / Graphical Questions

Q14. The slope of velocity-time graph represents:

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

Answer: C) Acceleration

Q15. Area under velocity-time graph represents:

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

Answer: B) Displacement

Q16. Velocity-time graph:

v
|
|        /
|       /
|      /
|_____/________ t

The particle has:

A) Constant velocity
B) Constant acceleration
C) Zero acceleration
D) Variable acceleration

Answer: B) Constant acceleration

Hard Numerical Practice

Q17. A train moving at 72 km/h stops in 10 seconds. Find retardation.

72 km/h = 20 m/s a = (0-20)/10 a = -2 m/s² Retardation = 2 m/s²

Q18. A particle has initial velocity 5 m/s and acceleration 2 m/s². Find distance in 10 seconds.

s = ut + 1/2at² s = 5×10 + 1/2×2×100 s = 150 m Answer: 150 m

NEET Formula Revision

v = u + at

s = ut + 1/2 at²

v² = u² + 2as

s = ((u+v)/2)t

NEET Tips

• Practice velocity-time graphs.
• Remember sign convention.
• Understand distance and displacement difference.
• Use correct equation according to given data.
• Practice free fall problems.

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

---

 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

---

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

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

---

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

---

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

---

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

---

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

---

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

---

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

---

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

---

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

---

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

---

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

Kinematics Equations Explained for NEET | 3 Equations of Motion Notes

KINEMATICS EQUATIONS – EASY NEET NOTES

1. Three Important Equations of Motion

These equations are used for:

• ✅ Straight line motion
• ✅ Constant acceleration

They connect:

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

First Equation of Motion

\[ v = v_0 + at \]

Meaning

Final velocity = Initial velocity + increase in velocity due to acceleration.

Use

• Time is given
• Need to find velocity

Second Equation of Motion

\[ x = v_0 t + \frac12 at^2 \]

Meaning

Displacement depends on:

• Initial velocity
• Time
• Acceleration

Important Point:
The term \(at^2\) shows the effect of acceleration.

Third Equation of Motion

\[ v^2 = v_0^2 + 2ax \]

Important Feature:
This equation has NO time term.

Use

• Time is not given
• Need relation between velocity and displacement

2. General Form of Equations

Earlier we assumed:

\[ x_0 = 0 \]

Meaning particle starts from origin.

If particle starts from another position (\(x_0\)), equations become:

Modified Second Equation

\[ x = x_0 + v_0 t + \frac12 at^2 \]

Meaning

Final position =

• Initial position
• Displacement due to initial velocity
• Displacement due to acceleration

Modified Third Equation

\[ v^2 = v_0^2 + 2a(x-x_0) \]

Important Concept:
\((x-x_0)\) represents displacement.

3. Derivation Using Calculus

Definition of Acceleration

Acceleration is rate of change of velocity.

\[ a=\frac{dv}{dt} \]

Rearranging:

\[ dv = a\,dt \]

Integrating Both Sides

\[ \int_{v_0}^{v} dv = \int_0^t a\,dt \]

Since acceleration is constant:

\[ v-v_0 = at \]

Therefore:

\[ v=v_0+at \]

This gives first equation of motion.

4. Derivation of Second Equation

Velocity:

\[ v=\frac{dx}{dt} \]

So,

\[ dx=v\,dt \]

Substitute:

\[ v=v_0+at \]

Then,

\[ dx=(v_0+at)dt \]

Integrating:

\[ x-x_0=v_0 t+\frac12 at^2 \]

Hence,

\[ x=x_0+v_0 t+\frac12 at^2 \]

5. Derivation of Third Equation

We write:

\[ a=\frac{dv}{dt} \]

Using chain rule:

\[ a=\frac{dv}{dx}\frac{dx}{dt} \]

But,

\[ \frac{dx}{dt}=v \]

So,

\[ a=v\frac{dv}{dx} \]

Rearranging:

\[ v\,dv=a\,dx \]

Integrating both sides:

\[ \int_{v_0}^{v} v\,dv = \int_{x_0}^{x} a\,dx \]

After integration:

\[ \frac{v^2-v_0^2}{2}=a(x-x_0) \]

Finally:

\[ v^2=v_0^2+2a(x-x_0) \]

6. Advantage of Calculus Method

✅ This method can also be used for non-uniform acceleration.

Normal equations work only for constant acceleration.

7. Example 2.3 – Ball Thrown Vertically Upward

Given

• Initial velocity: \(v_0=20\,m/s\)
• Acceleration due to gravity: \(a=-10\,m/s^2\)
• At highest point: \(v=0\)

Finding Maximum Height

Using:

\[ v^2=v_0^2+2a(y-y_0) \]

Substitute values:

\[ 0=(20)^2+2(-10)(y-y_0) \]

\[ 0=400-20(y-y_0) \]

\[ 20(y-y_0)=400 \]

\[ y-y_0=20\,m \]

Answer:
Maximum height reached = 20 m

8. Important NEET Sign Convention

Upward direction positive.

• Upward velocity → positive
• Gravity → negative

\[ a=-g \]

9. Important NEET Concepts

At Highest Point

Velocity becomes zero temporarily.

\[ v=0 \]

But acceleration is still:

\[ a=-g \]

10. Quick Formula Revision

Formula	Use
\(v=v_0+at\)	Velocity-time relation
\(x=v_0t+\frac12at^2\)	Displacement-time relation
\(v^2=v_0^2+2ax\)	Velocity-displacement relation
\(a=\frac{dv}{dt}\)	Definition of acceleration
\(v=\frac{dx}{dt}\)	Definition of velocity

11. Most Important NEET Tips

✅ Use equations only for constant acceleration.

✅ Check sign convention carefully.

✅ For upward motion: \[ a=-g \]
✅ At top point: \[ v=0 \] NOT acceleration zero.

✅ Third equation is most useful when time is absent.

© Easy NEET Physics Notes – Kinematics Equations

CBSE Class 11 Physics: Kinematics Equations of Motion Questions and Answers

Multiple Choice Questions (MCQs)

Question: Which equation is known as the first equation of motion?

Answer: v = u + at

Question: Which equation of motion does not contain time?

Answer: v² = u² + 2as

Question: What is the velocity of a body at the highest point of vertical upward motion?

Answer: Zero.

Question: What is the SI unit of acceleration?

Answer: Metre per second square (m/s²).

Question: Under which condition can equations of motion be applied?

Answer: When acceleration remains constant.

---

Very Short Answer Questions

Question: Define acceleration.

Answer: Acceleration is the rate of change of velocity with respect to time.

Question: Write the SI unit of displacement.

Answer: Metre (m).

Question: Write the third equation of motion.

Answer: v² = u² + 2as

Question: What is the acceleration due to gravity near the Earth's surface?

Answer: Approximately 9.8 m/s² downward.

Question: What happens to velocity at the highest point of upward motion?

Answer: Velocity becomes zero momentarily.

---

Short Answer Questions

Question: Write all three equations of motion.

Answer:

First Equation:

v = u + at

Second Equation:

s = ut + ½at²

Third Equation:

v² = u² + 2as

Question: Why is the third equation of motion useful?

Answer: The third equation is useful when time is not given. It directly relates velocity, displacement, and acceleration.

Question: Explain the sign convention used in vertical upward motion.

Answer:

• Upward direction is taken as positive.
• Upward velocity is positive.
• Acceleration due to gravity is negative.
• Downward displacement is negative.

Question: A car starts from rest and accelerates uniformly at 2 m/s² for 5 seconds. Find its final velocity.

Answer:

Given:

Initial velocity, u = 0

Acceleration, a = 2 m/s²

Time, t = 5 s

Using v = u + at

v = 0 + (2 × 5)

v = 10 m/s

Final Velocity = 10 m/s

---

Long Answer Questions

Question: Derive the first equation of motion.

Answer:

Acceleration is defined as:

a = (v − u)/t

Rearranging:

at = v − u

Therefore,

v = u + at

This is called the first equation of motion.

Question: Derive the second equation of motion.

Answer:

Average velocity = (u + v)/2

Displacement:

s = (u + v)t/2

Using the first equation:

v = u + at

Substituting:

s = [u + (u + at)]t/2

s = (2u + at)t/2

s = ut + ½at²

This is the second equation of motion.

Question: A ball is thrown vertically upward with a speed of 20 m/s. Calculate the maximum height reached.

Answer:

Given:

u = 20 m/s

v = 0

a = -10 m/s²

Using:

v² = u² + 2as

0 = (20)² + 2(-10)s

0 = 400 - 20s

20s = 400

s = 20 m

Maximum height reached = 20 m

---

Assertion and Reason Questions

Assertion: At the highest point of upward motion, velocity becomes zero.

Reason: Acceleration due to gravity becomes zero.

Answer: Assertion is true but Reason is false.

Assertion: The third equation of motion is useful when time is absent.

Reason: It does not contain time.

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

Assertion: Equations of motion can be used for variable acceleration.

Reason: These equations are derived assuming constant acceleration.

Answer: Assertion is false but Reason is true.

---

Fill in the Blanks

Question: The SI unit of velocity is ________.

Answer: m/s

Question: The acceleration due to gravity is approximately ________.

Answer: 9.8 m/s²

Question: The equation v = u + at is called the ________ equation of motion.

Answer: First

Question: At the highest point of upward motion, velocity becomes ________.

Answer: Zero

Question: The third equation of motion does not contain ________.

Answer: Time

---

Case Study Based Questions

A student throws a ball vertically upward with an initial velocity of 20 m/s. The ball rises upward, reaches maximum height, and then falls back to the ground. Assume g = 10 m/s².

Question: What is the velocity at the highest point?

Answer: 0 m/s

Question: What is the acceleration at the highest point?

Answer: 10 m/s² downward.

Question: Which equation can be used to find the maximum height?

Answer: v² = u² + 2as

Question: Calculate the maximum height.

Answer: 20 m

---

Match the Following

Column A	Column B
First Equation of Motion	v = u + at
Second Equation of Motion	s = ut + ½at²
Third Equation of Motion	v² = u² + 2as
Acceleration	Rate of change of velocity

---

Important Numerical Problems

Question: A body starts from rest and accelerates uniformly at 4 m/s² for 5 seconds. Find the final velocity.

Answer:

v = u + at

v = 0 + (4 × 5)

v = 20 m/s

Final Velocity = 20 m/s

Question: A car moving with a velocity of 10 m/s accelerates at 2 m/s² for 5 seconds. Find the displacement.

Answer:

s = ut + ½at²

s = (10 × 5) + ½(2)(25)

s = 50 + 25

s = 75 m

Displacement = 75 m

Three equations of motion used in kinematics for constant acceleration explained with formulas and applications. 

 INTERNAL LINKS

Motion in a Straight Line Notes

Velocity and Acceleration Explained

Example 3.4 Solution Explained for Beginners

Block and Trolley System NEET Solution

Newton's Laws of Motion Notes

Vector Addition and Subtraction

Important Physics Derivations for NEET

Projectile Motion Notes

Free Fall and Gravity Problems

NCERT Kinematics Solutions

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

KINEMATIC EQUATIONS FOR UNIFORMLY ACCELERATED MOTION

NEET Physics Easy Notes

1. Area under v–t Graph = Displacement

Fig. 2.4 Explanation

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

The shaded area forms a rectangle.

Rectangle dimensions:

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

Area = u × T

Since area under velocity-time graph gives displacement,

x = uT

Important Point

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

(velocity) × (time) = m/s × s = m

So result is displacement.

2. Real Graphs are Smooth

Book Note Meaning

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

NEET Point

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

3. Uniformly Accelerated Motion

Uniform acceleration means:

• Acceleration remains constant throughout motion.

Main Quantities

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

4. First Equation of Motion

Relationship between velocity and time:

v = u + at

Meaning

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

Rearranged Forms

a = (v - u)/t

t = (v - u)/a

NEET Tip

Use when:

• time is involved
• displacement is NOT required

5. Derivation of Second Equation of Motion

Step 1: Area under v–t graph

Displacement = area of rectangle + area of triangle

Rectangle Area

• Height = (u)
• Base = (t)

Rectangle Area = ut

Triangle Area

• Height = (v - u)
• Base = (t)

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

Total Displacement

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

From first equation:

v - u = at

Substitute:

x = ut + 1/2 at²

6. Second Equation Meaning

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

Special Case

If object starts from rest:

u = 0

Then,

x = 1/2 at²

7. Third Equation of Motion

Using:

t = (v - u)/a

and substituting into displacement equation:

We get:

v² = u² + 2ax

8. Third Equation Meaning

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

NEET Tip

Use when:

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

9. Average Velocity Formula

For constant acceleration:

v̄ = (u + v)/2

Displacement:

x = v̄t

So,

x = (u + v)t / 2

10. Important NEET Formula Sheet

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

11. Quick Concept Tricks for NEET

If acceleration is zero

a = 0

Then:

v = u

Motion becomes uniform motion.

If object starts from rest

u = 0

Equations become:

v = at

x = 1/2 at²

v² = 2ax

12. Graph-Based NEET Concepts

Velocity-Time Graph

• Slope = acceleration
• Area under graph = displacement

Acceleration-Time Graph

• Area under graph = change in velocity

Displacement-Time Graph

• Slope = velocity

13. Most Important NEET Mistakes

❌ Using km/h instead of m/s

❌ Forgetting sign of acceleration

❌ Using wrong equation

❌ Taking displacement as distance

❌ Forgetting (u = 0) for rest condition

14. One-Line Revision

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

(u + v)/2

NEET Physics Easy Notes © 2026

- Dr.Sanjaykumar Pawar  

CBSE Class 11 Physics: Kinematics (Uniformly Accelerated Motion)

Complete Question Bank with Answers

---

Very Short Answer Questions (1 Mark)

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

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

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

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

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

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

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

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

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

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

---

Short Answer Questions (2–3 Marks)

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

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

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

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

---

Derivations (3–5 Marks)

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

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

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

---

Numerical Question

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

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

---

MCQs

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

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

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

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

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

---

Assertion and Reason

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

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

---

Fill in the Blanks

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

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

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

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

---

Match the Column

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

---

Case Study

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

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

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

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

Q4. Type of motion?
Uniformly accelerated motion

---

Conclusion

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

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

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

🔗 Internal Links 

Add these as internal linking keywords:

Motion in a Straight Line Notes

Velocity-Time Graph Explanation

Derivation of v = u + at

Second Equation of Motion Proof

Third Equation of Motion Formula

NEET Physics Formula Sheet

Important Physics Graphs for NEET

Uniform Acceleration Problems

Class 11 Physics Chapter 3 Notes

Kinematics MCQs Practice Set

Class 11 Physics Motion in a Straight Line Notes and Questions

 

NCERT Physics Class 11 Chapter 2 Easy Line-by-Line Notes 

- Dr.Sanjaykumar pawar

Area Under Velocity-Time Graph

---

Fig. 2.4 Explanation

“Area under v–t curve equals displacement...”

Very Important Concept

On a velocity-time graph:

Displacement=\text{Area under v-t graph}

---

“The v–t curve is a straight line parallel to the time axis...”

Meaning

• Velocity remains constant.

• Object moves with uniform velocity.

---

“Area under it between t = 0 and t = T is the area of rectangle...”

Graph Shape

Rectangle:

• Height = velocity (u)

• Base = time (T)

---

Rectangle Area

Area=u\times T

Since:

• area under graph = displacement

Therefore:

x=uT

---

“How come area equals distance?”

Dimensional Understanding

Velocity × Time:

[
(m/s)\times s = m
]

So result becomes displacement.

---

Important Note About Graphs

“x–t, v–t and a–t graphs shown have sharp kinks...”

Meaning

Some graphs in textbooks have sharp corners.

But in real life:

• motion changes smoothly.

---

“Acceleration and velocity cannot change abruptly...”

Important Physical Meaning

Objects cannot instantly jump from:

• slow to very fast

• or stop suddenly

Changes are continuous.

---

2.4 Kinematic Equations for Uniformly Accelerated Motion

---

“For uniformly accelerated motion...”

Meaning

Acceleration remains constant.

Example:

• freely falling object

---

Variables Used

Symbol	Meaning
(x)	displacement
(t)	time
(v_0)	initial velocity
(v)	final velocity
(a)	acceleration

---

First Equation of Motion

“Equation already obtained gives relation between final and initial velocities...”

v=v_0+at

Meaning

Final velocity:

initial velocity + change due to acceleration

---

Graphical Representation

“This relation is graphically represented in Fig. 2.5.”

Meaning

Velocity-time graph becomes a straight line.

Reason:

• acceleration is constant.

---

Area Under Graph

“Area between instants 0 and t = area of triangle + rectangle”

Total displacement:

=
rectangle area + triangle area

---

---

---

Second Equation of Motion

Meaning

Displacement depends on:

• initial velocity

• acceleration

• time

---

Average Velocity Form

Meaning

For constant acceleration:

Average velocity

mean of initial and final velocities.

---

Important Condition

“Constant acceleration only”

This formula works only when:

• acceleration remains constant.

---

Third Equation of Motion

---

Meaning of Third Equation

This equation connects:

• velocity

• displacement

• acceleration

without using time.

---

Three Main Equations of Motion

---

---

Graph Concepts Summary

Graph	Slope Gives	Area Gives
Position-Time	Velocity	—
Velocity-Time	Acceleration	Displacement

---

Real-Life Examples

---

One-Line Summary

For uniformly accelerated motion, displacement and velocity can be calculated using three important equations derived from the velocity-time graph. 

KINEMATIC EQUATIONS FOR UNIFORMLY ACCELERATED MOTION

│

├── Velocity-Time Graph

│   ├── Area under graph = displacement

│   ├── Constant velocity graph

│   │   └── Horizontal straight line

│   └── Uniform acceleration graph

│       └── Sloping straight line

│

├── Area Under v-t Graph

│   ├── Rectangle area

│   │     displacement = velocity × time

│   ├── Formula

│   │     x = uT

│   └── Unit check

│         (m/s) × s = m

│

├── Important Physical Idea

│   ├── Velocity changes continuously

│   ├── Acceleration changes continuously

│   └── Real motion graphs are smooth

│

├── Uniformly Accelerated Motion

│   ├── Constant acceleration

│   └── Variables

│       ├── x → displacement

│       ├── t → time

│       ├── v₀ → initial velocity

│       ├── v → final velocity

│       └── a → acceleration

│

├── First Equation of Motion

│   ├── Formula

│   │     v = v₀ + at

│   ├── Gives final velocity

│   └── Used when time is known

│

├── Displacement from Graph

│   ├── Total area

│   │     = rectangle + triangle

│   │

│   ├── Rectangle area

│   │     = v₀t

│   │

│   ├── Triangle area

│   │     = 1/2 (v - v₀)t

│   │

│   └── Total displacement

│         x = v₀t + 1/2 (v - v₀)t

│

├── Second Equation of Motion

│   ├── Using

│   │     v - v₀ = at

│   ├── Formula

│   │     x = v₀t + 1/2 at²

│   └── Used for displacement

│

├── Average Velocity

│   ├── Formula

│   │     v_avg = (v + v₀)/2

│   ├── Works only for constant acceleration

│   └── Displacement form

│         x = [(v + v₀)/2] t

│

├── Third Equation of Motion

│   ├── Formula

│   │     v² = v₀² + 2ax

│   ├── Time not required

│   └── Relates

│       ├── velocity

│       ├── displacement

│       └── acceleration

│

├── Three Main Equations

│   │

│   ├── First

│   │     v = v₀ + at

│   │

│   ├── Second

│   │     x = v₀t + 1/2 at²

│   │

│   └── Third

│         v² = v₀² + 2ax

│

├── Graph Rules

│   ├── Position-Time Graph

│   │   └── Slope = velocity

│   │

│   └── Velocity-Time Graph

│       ├── Slope = acceleration

│       └── Area = displacement

│

├── Real-Life Examples

│   ├── Accelerating car

│   ├── Falling object

│   ├── Braking vehicle

│   └── Train gaining speed

│

└── Key Ideas

    ├── Constant acceleration simplifies motion

    ├── Area under v-t graph gives displacement

    └── Equations of motion describe straight-line motion 

Internal Links

1. Class 11 Physics Units and Measurements Notes
2. Class 11 Physics Laws of Motion Notes
3. Class 11 Physics Work Energy and Power Questions
4. Kinematics Formula Sheet PDF
5. CBSE Class 11 Physics Important Numericals
6. Class 11 Physics Chapter Wise MCQs
7. Motion in a Plane Complete Notes
8. Physics Graphs and Derivations Guide
9. NCERT Solutions for Class 11 Physics
10. Physics Assertion Reason Questions Collection

CBSE Class 11 Physics Question Bank

Chapter: Motion in a Straight Line

1. Multiple Choice Questions (MCQs)

1. Motion is defined as:

• A. Change in mass
• B. Change in position with time
• C. Change in force
• D. Change in shape

Answer: B. Change in position with time

2. SI unit of acceleration is:

• A. m/s
• B. m/s²
• C. km/h
• D. m²/s

Answer: B. m/s²

3. Area under velocity-time graph gives:

• A. Velocity
• B. Acceleration
• C. Displacement
• D. Speed

Answer: C. Displacement

2. Very Short Answer Questions

1. Define motion.

Motion is the change in position with time.

2. Define acceleration.

Acceleration is the rate of change of velocity with time.

3. Write SI unit of velocity.

m/s

3. Short Answer Questions

1. Differentiate between speed and velocity.

Speed	Velocity
Scalar quantity	Vector quantity
No direction	Has direction
Only magnitude	Magnitude and direction

2. Write formula of average velocity.

v = Δx / Δt

Average velocity is displacement divided by time.

4. Long Answer Questions

1. Derive second equation of motion.

v = u + at

s = ut + 1/2 at²

The second equation of motion is obtained from velocity-time graph by calculating area under the graph.

5. Assertion and Reason Questions

Assertion: Slope of velocity-time graph gives acceleration.
Reason: Acceleration is rate of change of velocity with time.

Both Assertion and Reason are true and Reason correctly explains Assertion.

Assertion: Speed can be negative.
Reason: Speed is scalar quantity.

Assertion is false but Reason is true.

6. Fill in the Blanks

1. Motion is change in ______ with time.

position

2. SI unit of acceleration is ______.

m/s²

3. Area under velocity-time graph gives ______.

displacement

7. Statement Based Questions

Statement I: Velocity has direction.
Statement II: Speed has no direction.

Both statements are true.

Statement I: Uniform motion means constant velocity.
Statement II: Acceleration is zero in uniform motion.

Both statements are true.

8. Match the Columns

Column A	Column B
Slope of x-t graph	Velocity
Slope of v-t graph	Acceleration
Area under v-t graph	Displacement
Constant velocity	Zero acceleration

9. Case Study Questions

A car starts from rest and accelerates uniformly at 2 m/s² for 5 seconds.

1. Find final velocity.
2. Find displacement.

v = u + at

v = 0 + (2)(5) = 10 m/s

s = ut + 1/2 at²

s = 0 + 1/2(2)(25) = 25 m

Final velocity = 10 m/s
Displacement = 25 m

10. Important Formulas

v = u + at

s = ut + 1/2 at²

v² = u² + 2as

a = (v - u)/t

Class 11 Physics Motion in a Straight Line notes, formulas, graphs, and important CBSE exam questions.