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# Derivation of third equation of Motion

Aug 22, 2022

### Key Concepts

1. Uniformly accelerated/ decelerated motion

2. Equations of motion

3. Third equation of motion

## 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 are called the equations of motion. One or two unknown quantities can be calculated by using the equations of motion if other quantities are known

## 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 at2

v2 = u2 + 2as

### Derivation of 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 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)

=

(OA+DC) x (OD)2OA+DC x (OD)2

=

(u+v) x (t)2(u+v) x (t)2

s =

(u+v)t2(u+v)t2

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 = v2 – u2

v2 = u2 + 2as

#### Problems:

1. 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/s2

The distance covered, s = 1.8 km = 1800 m

On using the third equation of motion we get,

v2 = u2 + 2as

Or, v2 = (20)2 + 2 x 2 x 1800

Or, v2 = 400 + 7200

Or, v2 = 7600

Or, v = 87.17 m/s

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

1. A car moving at a velocity of 5 m/s accelerates to 25 m/s while covering a distance of 1.8 km. Workout 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,

v2 = u2 + 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/s2

Therefore, the acceleration of the car is 1.67 m/s2

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.

1. A bus starts from rest and attains a velocity of 90 km/h in 10 minutes. Find out the acceleration and the distance travelled 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/s2

Therefore, the acceleration is 1/24 m/s2

On using the third equation of motion we get,

v2 = u2 + 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

1. A ball is dropped from the top of a building of height 50 m. Considering the acceleration due to gravity of the Earth to be 10 m/s2, 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/s2

The displacement, s = 80 m

On using the third equation of motion we get,

v2 = u2 + 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

### Summary

1. A moving body is said to be in a uniformly accelerated motion if its
velocity increases by equal amounts in equal intervals of time.

2. Different attributes of a uniformly accelerated motion such as
displacement, velocity and acceleration in a certain time interval can be
related to each other by a set of equations called the equations of motion.

3. These equations of motion can be used to work out one or two unknown
attributes of a uniformly accelerated motion if all other attributes are known.

4. The equations of motion are as follows:

v = u + at

s = ut + 1/2 at2

v2 = u2 + 2as

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