A car running with a velocity 72 kmph on a level road, is stopped after travelling a distance of 30 m after disengaging its engine $\left( g={ 10ms }^{ -2 } \right)$. The coefficient of friction between ;& ;the road and the tyres is.

Velocity of car=$u= 72kmph= 20m/s$

Stopping distance of car $= s=30m$

here accceleration $(a)$ is negative because frictional force will be the stopping force and equal to $\mu g$

Final veocity$v=0$

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

$\implies a=\dfrac{u^2}{2s}$

$\implies \mu g= \dfrac{u^2}{2s}$

$\implies \mu =\dfrac{u^2}{ 2sg}=\dfrac{400}{600}=0.67$