Derive the third equation of motion:$v^2-u^2=2as$.

Step 1: To prove:

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

Where u is initial velocity, v is final velocity, a is the acceleration and s is the displacement, t is the time.

Step 2: Derivation:

For getting the third equation of motion, we use both of the equations of motion

$$\begin{gathered} S=ut+\frac{1}{2}a{t}^2\_\_\_\_\_\_\_\_\_\_\left(1\right)(2ndequationofmotion) \\ v=u+at\_\_\_\_\_\_\_\_\_\_\_\_\_\left(2\right)(1stequationofmotion) \\ t=(v–u)/a \end{gathered}$$

Substitute the value of t in equation (1),$S=ut+1/2a{t}^2$,

We get,

$S=u\left(\frac{v-u}{a}\right)+\frac{1}{2}a{\left(\frac{v-u}{a}\right)}^2$

$$\begin{gathered} \Rightarrow 2aS=2u(v-u)+{(v-u)}^2 \\ \Rightarrow 2aS=2uv-2u^2+v^2-2uv+u^2 \\ \Rightarrow 2aS=v^2-u^2 \\ \Rightarrow v^2-u^2=2as \end{gathered}$$

Hence, $v^2-u^2=2as$