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.

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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.

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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

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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

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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.

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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

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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

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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

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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