Derive the following equation of motion by the graphical method : v^{2} = u^{2} + 2as, where the symbols have their usual meanings.

#### Solution

In the given figure, the distance travelled (s) by a body in time (t) is given by the area of the figure OABC which is a trapezium.

Distance travelld = Area of the trapezium OABC

So, Area of trapezium OABC,

= `"(Sum of parallel sides)(Height)"/2`

=`"(OA+CB)(OC)"/2`

Now, (OA + CB) = u + v and (OC) = t.

Putting these values in the above relation, we get:

`s = ((u+v)/2)t` ....(1)

Eliminate t from the above equation. This can be done by obtaining the value of t from the first equation of motion.

v = u + at

So,

`t = "v-u"/a`

Now, put this value of t in equation (1), we get:

`s = (((u+v)(v-u))/(2a))`

On further simplification,

2as = v^{2} – u^{2}

Finally the third equation of motion.

`v^2 = u^2 + 2as`

where

(s) - Displacement

(u) - Initial velocity

(a) - Acceleration

(v) - Final velocity

(t) - Time taken