Derive the equation $v^{2}-u^{2}=2 a S$ graphically.

(N/A) Consider the velocity-time graph shown in the figure.
The distance $S$ covered by an object moving with uniform acceleration $a$ is equal to the area under the velocity-time graph,which is the area of the trapezium $OABD$.
Area of trapezium $OABD = \frac{(OA + BD) \times OD}{2}$
From the graph,we have:
$OA = u$ (initial velocity)
$BD = v$ (final velocity)
$OD = t$ (time taken)
Substituting these values,we get:
$S = \frac{(u + v) t}{2}$ --- (Equation $1$)
We know that for uniform acceleration,$v = u + at$,which can be rearranged to find time $t$:
$at = v - u$
$t = \frac{v - u}{a}$ --- (Equation $2$)
Substituting the value of $t$ from Equation $2$ into Equation $1$:
$S = \frac{(u + v)(v - u)}{2a}$
Using the algebraic identity $(a+b)(a-b) = a^2 - b^2$,we get:
$S = \frac{v^2 - u^2}{2a}$
Rearranging the terms,we get the final equation:
$v^2 - u^2 = 2aS$