Question

Derive the third equation of motion.

Answer 1

Step 1: Deriving the third equation of motion

Let the initial velocity be $u$ , final velocity be $v$, time taken be $t$, distance travelled be $s$ and the acceleration of the body be $a$.

From the first equation of motion, we have $v=u+at$….(I)

From the second equation of motion, we have $s=ut+\frac{1}{2}a{t}^2$….(ii)

Squaring equation (I) then

$v^2={(u+at)}^2$

or $v^2=u^2+a^2{t}^2+2uat\left[\mathrm{As}{(a+b)}^2=a^2+b^2+2ab\right]$

or $v^2=u^2+2a\left(\frac{1}{2}a{t}^2+ut\right)$

or $v^2=u^2+2as(\mathrm{from}(\mathrm{ii}\left)\right)$

Hence, $v^2=u^2+2as$ is the third equation of motion.