Equation of Motion and Its Application

As we know, motion is the phenomenon in which an object changes its position. Motion is represented in terms of displacement, distance, velocity, acceleration, speed, and time. Let us know the equations of motion, and applications of the equations of motion.

 

Equations of Motion

The motion of an object moving at uniform acceleration can be described with the help of three equations, namely

(i) v = u + at

(ii) v2 – u2 = 2as

(iii) s = ut + (1/2)at2

where u is the initial velocity, v is the final velocity, t is the time, a is the acceleration and s is the displacement.

Derivation of Velocity-Time Relation by Graphical Method

Velocity – Time Graph

A body starts with some initial non-zero velocity at A and goes to B  with constant acceleration a.

From the graph BD = v (final velocity) – DC = u (initial velocity)…………..(eq 1).

BD = BC – DC……………..(eq 2).

$\begin{array}{l}�������������������=�����=\frac{��}{��}����=��=�(���������������ℎ������)\end{array}$

.

$Therefore$ BD = at………………….(eq 3).

Substitute everything we get: at = v – u.

Rearrange to get v = u + at.

Derivation of Position-Time Relation by Graphical Method

Velocity – Time Graph

A body starts with some initial non-zero velocity at A and goes to B  with constant acceleration a

Area under the graph gives Displacement as follows:

$\begin{array}{l}�(\mathrm{\Delta }���)+�(����)=\frac{1}{2}(��\times ��)+(��\times ��)\dots .���(1)\end{array}$

OA = u , OC = t and BD = at

Substituting in (eq 1) we get $s=$

$\begin{array}{l}��+\frac{1}{2}�{�}^2\end{array}$

Derivation of Position-Velocity Relation by Graphical Method

Velocity – Time Graph

A body starts with some initial non-zero velocity at A and goes to B  with constant acceleration a

Displacement covered will be the area under the curve which is the trapezium OABC.

We know the area of trapezium is 

$\begin{array}{l}�=\frac{(��+��)}{2\ast ��}\end{array}$

OA = u and BC = v and OC = t

$\begin{array}{l}�ℎ�������,�=\frac{(�+�)}{2\ast �}\dots \dots \dots \dots \dots (��1)\end{array}$

$\begin{array}{l}�����������ℎ���=\frac{(�-�)}{�}\dots \dots \dots \dots \dots ..(��2)\end{array}$

Substitute (eq 2) in (eq 1) and arrange to get

v2−u2=2a

Introduction to Equations Of Motion

In this article, we will learn how we can relate quantities like velocity, time, acceleration and displacement provided the acceleration remains constant. These relations are collectively known as the equation of motion. There are three equations of motion. There are three ways to derive the equation of motion and here we are going to derive with the help of a graph.

First Equation of Motion

First equation of motion relates velocity, time and acceleration. Now in ∆uxy,

$\begin{array}{l}����=\frac{��}{��}\end{array}$

 

$\begin{array}{l}����=\frac{�-�}{�}\end{array}$

 
We also know that tanθ is nothing but the slope, and slope of the v – t graph represents acceleration.

⇒ v = u + at ———– (1)

This is the first equation of motion where,

v = final velocity

u = initial velocity

a = acceleration

t = time taken

Second Equation of Motion

Now coming to the second equation of motion, it relates displacement, velocity, acceleration and time. The area under the v – t graph represents the displacement of the body.

In this case,

Displacement = Area of the trapezium (ouxt)

$\begin{array}{l}�=\frac{1}{2}\times \text{ sum of parallel sides }\times ℎ���ℎ�\end{array}$

 

$\begin{array}{l}�=\frac{1}{2}\times \text{ (v + u) }\times \text{ t ——— (2)}\end{array}$

 
We can substitute v in terms of others and get the final equation as:

$\begin{array}{l}�=��+\frac{1}{2}�{�}^2\end{array}$

 
Where symbols have their usual meaning.

Third Equation of Motion

The third equation of motion relates to velocity, displacement, and acceleration. Using the same equation (2),

$\begin{array}{l}�=\frac{1}{2}\times \text{ (v + u) }\times \text{ t}\end{array}$

 
Using equation (1), if we replace t, we get,

$\begin{array}{l}�=\frac{1}{2}\times (�+�)\times \frac{(�-�)}{�}\end{array}$

 

$\begin{array}{l}�=\frac{({�}^2-{�}^2)}{2�}\end{array}$

 

$\begin{array}{l}{�}^2={�}^2+2��\end{array}$

 
The above equation represents our third equation of motion.

Applications of Equations of Motion

So now that we have seen all three equations of motion, we can use them to solve kinematic problems. We just have to identify what all parameters are given and then choose the appropriate equation and solve for the required parameter.

The equations of motion are also used in the calculation of optical properties.