Q. A car is moving along a straight horizontal road with a speed $v_{0}$. If the coefficient of friction between the tyres and the road is $\mu$, the shortest distance in which the car can be stopped is
A) $\frac{v_{0}^{2}}{2\mu g}$
B) $\frac{v_{0}}{\mu g}$
C) $\left( \frac{v_{0}}{\mu g} \right)^{2}\ $
D) $\frac{v_{0}}{\mu}$
Explanation
(a) Relating force $F = ma = \mu\ R = \mu\ mg$
$\therefore a = \mu g$
Now from equation of motion $v^{2} = u^{2} - 2as$
$\Rightarrow 0 = u^{2} - 2as \Rightarrow s = \frac{u^{2}}{2a} = \frac{u^{2}}{2\mu\ g}$
$\therefore s = \frac{v_{0}^{2}}{2\mu g}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.