**Introduction:**

In a uniformly accelerated motion in a straight line, 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/ 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 + ** **at ^{2}**

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

#### Derivation:

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 **AO** of the graph). The body increases its velocity to **v **(given by the length **CD**) in time **t** (given by the length **OD**). Let **s** be the displacement of the body in time **t**, and the velocity changes at a uniform rate **a**.

### The First Equation of Motion:

From the graph, it is seen that,

CD = CB + BD

= CB + OA (as OABD is a rectangle)

Or, CB = CD – OA

CB = v – u ——- (1)

Now, the acceleration is given by,

a = ——- (2)

Here, the change in velocity is given by **v – u**, which is CB in time **t**.

Therefore, a =

Or, CB = at ——- (3)

Inserting equation (3) into equation (1) we get,

**v – u = at**

This is the first equation of motion. It is also called the **velocity-time relation**.

#### Problems:

**A car starts from rest and acquires a velocity of 20 m/s in 4 seconds. Find its acceleration.**

**Solution:**

Given that,

The initial velocity, u = 0

The final velocity, v = 20 m/s

Time taken, t = 4 s

On using the first equation of motion we get,

**v = u + at**

Or, a = (v – u)/t

Or, a = (20 – 0)/4

Or, a = 20 / 4

Or, a = 5 m/s^{2}

Therefore, the acceleration of the car is **5 m/s ^{2}**.

2. **A vehicle moving at a speed of 5 m/s accelerates at a rate of 3 m/s ^{2} for 25 seconds. Calculate the final velocity of the vehicle.**

**Solution:**

Given that,

The initial velocity, u = 5 m/s

The acceleration, a = 3 m/s^{2}

Time taken, t = 25 s

On using the first equation of motion we get,

**v = u + at**

Or, v = 5 + 3 × 25

Or, v = 5 + 75

Or, v = 80 m/s

Therefore, the final velocity of the vehicle is 80 m/s.

**3**. **An athlete starts running and keeps increasing his speed by 0.5 m/s every second. What is the final speed that he reaches after 24 seconds? What distance does he cover meanwhile?**

**Solution:**

Given that,

The initial velocity, u = 0

The acceleration, a = 0.5 m/s^{2}

Time taken, t = 24 s

On using the first equation of motion we get,

**v = u + at**

Or, v = 0 + 0.5 x 24

Or, v = 12 m/s

Therefore, the final velocity of the athlete is 12 m/s.

On using the second equation of motion we get,

**s = ut +** **at ^{2}**

Or, s = 0 + (0.5) (24)^{2}

Or, s = 144 m

Therefore, the distance covered by the athlete is 144 m.

**4**. **The driver of a car moving at a speed of 25 m/s applies the brakes on seeing a cat crossing the road at some distance. The car stops after 10 s. What is the acceleration of the car? How far did it go after the brakes were applied?**

**Solution:**

Given that,

The initial velocity, u = 25 m/s

The final velocity, v = 0

Time taken, t = 10 s

On using the first equation of motion we get,

**v = u + at**

Or, a = (v – u)/t

Or, a = (0 – 25)/10

Or, a = – 25 / 10

Or, a = – 2.5 m/s^{2}

Therefore, the acceleration of the car is **– 2.5 m/s ^{2}**.

On using the second equation of motion we get,

**s = ut +** **at ^{2}**

Or, s = 25 × 10 + (– 2.5) (10)^{2}

Or, s = 250 – 125

Or, s = 125 m

Therefore, the distance covered by the car is **125 m**.

**5**. **A ball is thrown upwards with a speed of 32 m/s. How high would the ball reach? How long would the ball take to reach the maximum height?**

**Solution:**

Given that,

The initial velocity, u = 32 m/s

The final velocity, v = 0

The acceleration, a = g = 9.8 m/s^{2}

At the maximum height, the velocity of the ball becomes zero.

On using the first equation of motion we get,

**v = u + at**

Or, 0 = 32 + 9.8 × t

Or, t = 32 / 9.8

Or, t = 3.265 s

Therefore, the time taken by the ball to reach the maximum height is **3.265 s**.

On using the second equation of motion we get,

**s = ut +** **at ^{2}**

Or, s = 32 x 3.265 + (9.8) (3.265)^{2}

Or, s = 104.48 – 52.235

Or, s = 52.245 m

Therefore, the maximum distance covered by the ball is **52.245 m**.

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