Equation Of Motion And Its Application

In this article we will how we can relate quantities like velocity, time, acceleration and displacement provided the acceleration remains constant. These relations are collectively known as equation of motion. There are three equations of motion. We will try to derive those using the following graph.

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

tanθ = $\frac {xy}{uy}$

tanθ = $\frac {v~-~u}{t}$

We also know that tanθ is nothing but the slope and slope of 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

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

In this case,

Displacement = Area of trapezium (ouxt)

S = $\frac 12$  x sum of parallel sides x height

S = $\frac 12$  x (v + u) x t ———- (2)

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

S = ut + $\frac 12~ at^2$

Where symbols have their usual meaning.

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

S = $\frac 12$  x (v + u) x t

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

S = $\frac 12$  x $(v + u)$ x $\frac{(v-u)}{a}$

S = $\frac {(v^2~-~ u^2)}{2a}$

$v^2$  = $u^2~ +~ 2as$

The above equation represents our third equation of motion. So now that we have seen all the 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.

This was just the introduction to the equations. To learn how to apply these get connected to Byjus.com

Practise This Question

A fruit is dropped from a height of 20 m. Find the velocity with which it would hit the ground.