Derive the following equations for a uniformly accelerated motion:
$(i)$ $v = u + at$
$(ii)$ $S = ut + \frac{1}{2}at^2$
$(iii)$ $v^2 - u^2 = 2aS$, where the symbols have their usual meanings.

(N/A) Suppose the initial velocity of a body is $u$ and it is moving with uniform acceleration $a$ for time $t$. Let the final velocity be $v$ and the distance covered be $S$.
$(i)$ Acceleration is defined as the rate of change of velocity:
$a = \frac{v - u}{t}$
$at = v - u$
$v = u + at$
$(ii)$ The average velocity for uniform acceleration is given by:
$\bar{v} = \frac{u + v}{2} \quad \dots(1)$
Also, average velocity is total displacement divided by time:
$\bar{v} = \frac{S}{t} \quad \dots(2)$
Equating $(1)$ and $(2)$:
$\frac{u + v}{2} = \frac{S}{t} \implies S = \left( \frac{u + v}{2} \right) t \quad \dots(3)$
Substituting $v = u + at$ into equation $(3)$:
$S = \left( \frac{u + (u + at)}{2} \right) t$
$S = \left( \frac{2u + at}{2} \right) t$
$S = ut + \frac{1}{2}at^2$
$(iii)$ From equation $(3)$, $S = \left( \frac{u + v}{2} \right) t$. From $v = u + at$, we get $t = \frac{v - u}{a}$.
Substituting $t$ in equation $(3)$:
$S = \left( \frac{u + v}{2} \right) \left( \frac{v - u}{a} \right)$
$S = \frac{v^2 - u^2}{2a}$
$v^2 - u^2 = 2aS$