**Introduction:**

In a uniformly accelerated motion, the velocity of the body increases or decreases by equal amounts in equal intervals of time. For such kinds of motion, the various attributes of motion such as the displacement, velocity, and acceleration for a certain time interval can be related by a set of equations. This set of equations is called the equations of motion. One or two unknown quantities can be calculated by using the equations of motion if other quantities are known.

**Explanation:**

### Equations of Motion:

Let us consider the following physical quantities by the symbols given below for a body in a uniformly accelerated or decelerated motion.

- Initial velocity = u
- Final velocity = v
- Acceleration = a
- Time = t
- Displacement = s

The equations of motion connecting the above-mentioned physical quantities are as follows:

**v = u + at**

**s = ut + ½ at ^{2}**

**v ^{2} = u^{2} + 2as**

## Derivation of the Third Equation of Motion:

All the equations of motion can be derived from the V-t graph. Consider the V-t graph of a body in a uniformly accelerated motion shown below. The v-t graph (AC) represents the motion of a body in a uniformly accelerated motion with a non-zero initial velocity. The constructions CD (perpendicular to the **x-axis**) and **AB** (perpendicular to **CD**) are drawn. Now, **OACD** forms a trapezium which is made of a triangle **ACB** and a rectangle **OABD**.

The above **v-t graph **represents the motion of a body with a non-zero initial velocity **u** (given by the length **OA** of the graph). The body increases its velocity to **v **(given by the length **CD**) in time **t** (given by the length **OD**). Let **u **be the displacement of the body in time **t** and the velocity changes at a uniform rate **a**.

### Third Equation of Motion:

From the graph, it is seen that,

Displacements = Area (trapezium OACD)

=

=

s =

2s = t (u + v) ————-(1)

From the first equation of motion, we have,

v = u + at

Or, t = (v – u)/a ————-(2)

On substituting equation (2) in equation (1) we get,

2s = (v + u) (v – u)/a

2as = v^{2} – u^{2}

**v ^{2} = u^{2} + 2as**

#### Problems:

**A body moving at a speed of 72 km/h accelerates uniformly at the rate of 2 m/s to cover 1.8 km. Calculate the final velocity of the body. How long did it take for the body to cover that distance?**

**Solution:**

Given that,

The initial velocity, u = 72 km/h = 72 x (5/18) = 20 m/s

The acceleration, a = 2 m/s^{2}

The distance covered, s = 1.8 km = 1800 m

On using the third equation of motion we get,

**v ^{2} = u^{2} + 2as**

Or, v^{2} = (20)^{2} + 2 x 2 x 1800

Or, v^{2} = 400 + 7200

Or, v^{2} = 7600

Or, v = 87.17 m/s

Therefore, the final velocity of the car is **87.17 m/s**.

**2**. **A car moving at a velocity of 5 m/s accelerates to 25 m/s while covering a distance of 1.8 km. Work out the acceleration of the car and the time it takes to cover that distance.**

**Solution:**

Given that,

The initial velocity, u = 5 m/s

The final velocity, v = 25 m/s

The distance covered, s = 1.8 km = 1800 m

On using the third equation of motion we get,

**v ^{2} = u^{2} + 2as**

Or, (25)^{2} = (5)^{2} + 2 x a x 1800

Or, 625 = 25 + 3600a

Or, 3600 a = 625 – 25

Or, 3600 a = 600

Or, a = 600/ 3600 = 1.67 m/s^{2}

Therefore, the acceleration of the car is **1.67 m/s ^{2}**.

On using the first equation of motion we get,

**v = u + at**

Or, 25 = 5 + (1/6) x t

Or, t/6 = 20

Or, t = 20 x 6

Or, t = 120 s

Therefore, the time taken by the car is 120 s.

**3**. **A bus starts from rest and attains a velocity of 90 km/h in 10 minutes. Find out the acceleration and the distance traveled by the bus to attain this velocity.**

**Solution:**

Given that,

The initial velocity, u = 0

The final velocity, v = 90 km/h = 90 x (5/18) = 25 m/s

Time taken, t = 10 minutes = 60 x 10 = 600 s

On using the first equation of motion we get,

**v = u + at**

Or, 25 = 0 + a x 600

Or, a = 25/600

Or, a = 1/24 m/s^{2}

Therefore, the acceleration is **1/24 m/s ^{2}**.

On using the third equation of motion we get,

**v ^{2} = u^{2} + 2as**

Or, (25)^{2} = 0 + 2 x (1/24) x s

Or, 625 = s/12

Or, s = 625 x 12

Or, s = 7500 m

Or, s = 7.5 km

Therefore, the distance covered by the bus is **7.5 km**.

4. **A ball is dropped from the top of a building of a **height of 50 m. Considering the acceleration due to the **gravity of the Earth to be 10 m/s ^{2}, find out the velocity of the ball right before it touches the ground and the time it takes to hit the ground.**

**Solution:**

Given that,

The initial velocity, u = 0

The acceleration, a = 10 m/s^{2}

The displacement, s = 80 m

On using the third equation of motion we get,

**v ^{2} = u^{2} + 2as**

Or, (v)^{2} = 0 + 2 x 10 x 80

Or, (v)^{2} = 1600

Or, v = sqrt (1600)

Or, v = 40 m/s

Therefore, the velocity of the ball before hitting the ground is **40 m/s**.

On using the first equation of motion we get,

**v = u + at**

Or, 40 = 0 + 10 x t

Or, t = 40/10

Or, t = 4 s

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