Question

Derive $v^2-u^2=2as$?

Answer 1

We will use both of the equations of motion to reach the third equation of motion. This will require a bit of algebra.
$S=ut+\frac{1}{2}a{t}^{2}andv=u+at$, include the time variant $t$
There will be some situations when we do not have any information about time and so it would be a good idea to derive an equation that does not have a $t$ term.
To do this, we rearrange our first equation to get
$t=\frac{v-u}{a}$
and use this to replace $t$ wherever it appears in the second equation. So
$S=ut+\frac{1}{2}a{t}^2$ becomes,
$S=u\left(\frac{v-u}{a}\right)+\frac{1}{2}a{\left(\frac{v-u}{a}\right)}^2$
$\Rightarrow 2aS=2u(v-u)+{(v-u)}^2$
$\Rightarrow 2aS=2uv-2u^2-v^2-2uv-u^2$
$\Rightarrow 2aS=v^2-u^2$
$\Rightarrow v^2=u^2+2aS$