### Theory:

Position-time relation:
Assume an object covering a distance ($$s$$) in time ($$t$$) in an uniform acceleration ($$a$$). The area enclosed within $$OABC$$ gives the distance travelled by the object. Velocity time graph

Distance ($$s$$) travelled by the object,

Substituting $$OA$$ $$=$$ $$u$$, $$OC$$ $$=$$ $$AD$$ $$=$$ $$t$$ and $$BD$$ $$=$$ $$at$$, we get

$\begin{array}{l}s\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}u×t+\frac{1}{2}\phantom{\rule{0.147em}{0ex}}\left(t×\mathit{at}\right)\\ \\ s\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\mathit{ut}+\frac{1}{2}\phantom{\rule{0.147em}{0ex}}a{t}^{2}\end{array}$
Position-velocity relation:
The area enclosed within the trapezium $$OABC$$ gives the distance ($$s$$) travelled by the object in time ($$t$$), moving under uniform acceleration ($$a$$).

$\begin{array}{l}s\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\mathit{area} \mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{trapezium}\phantom{\rule{0.147em}{0ex}}\mathit{OABC}\\ \\ s\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{\left(\mathit{OA}\phantom{\rule{0.147em}{0ex}}+\phantom{\rule{0.147em}{0ex}}\mathit{BC}\right)\phantom{\rule{0.147em}{0ex}}×\mathit{OC}}{2}\end{array}$

Substituting $$OA$$ $$=$$ $$u$$, $$BC$$ $$=$$ $$v$$ and $$OC$$ $$=$$ $$t$$, we get

$s=\frac{\left(u+v\right) t}{2}\to \left(a\right)$

From the velocity-time relation (from the second equation of motion),

$t=\frac{v-u}{a}\to \left(b\right)$

Substituting equation (b) on (a) we get,

$s=\frac{\left(v+u\right)\left(v-u\right)}{2a}$

(or)

$\begin{array}{l}2 a\phantom{\rule{0.147em}{0ex}}s\phantom{\rule{0.147em}{0ex}}={v}^{2}-{u}^{2}\\ \\ {v}^{2}={u}^{2}+2\phantom{\rule{0.147em}{0ex}}a\phantom{\rule{0.147em}{0ex}}s\end{array}$