Derive following equations for a uniformly accelerated motion

$(i)$ $v=u+a t$

$(ii)$ $S=u t+1 / 2 a t^{2}$

$(iii)$ $v^{2}-u^{2}=2 a S,$ where symbols have their usual meaning.

Suppose, the initial velocity of a body is $u$ and it is moving with uniform acceleration ' $a^{\prime}$ for time $t$. Let the final velocity be $v$ and the distance covered be S. Then, we have

$(i)$ Acceleration $=\frac{\text { Change in velocity }}{\text { Time elapsed }}$

or $a=\frac{v-u}{t} \text { or } v=u+a t$

$(ii)$ The average velocity is given by

$\bar{v}=\frac{u+v}{2} \quad \ldots(1) \quad$ Also, $\bar{v}=\frac{S}{t} \quad \ldots(2)$

From equations $(1)$ and $(2),$ we have

$\frac{u+v}{2}=\frac{ S }{t}$ or $S =\frac{u+v}{2} \times t$ $\ldots(3)$

But $v=u+a t,$ substituting in the above equation, we have

$S=\frac{u+(u+a t)}{2} \times t$

Rewriting, we have $S=u t+1 / 2 a t^{2}$

$(iii)$ From $(3)$, we have $S=\frac{u+v}{2} \times t$

Also, $v=u+$ at or $t=\frac{v-u}{a}$

Substituting in the above equation, we have

$S =\frac{u+v}{2} \times \frac{v-u}{a}$

Rewriting, we have $v^{2}-u^{2}=2 a S$