Q. A car is moving along a straight horizontal road with a speed$v_{0}$. If the coefficient of friction between 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) Retarding force $F = ma = \mu R = \mu\ mg$
$a = \mu g$
Now, from equation of motion, $v^{2} = u^{2} - 2as$
$\therefore\ 0 = u^{2} - 2as$
$\therefore\ \ s = \frac{u^{2}}{2a} = \frac{u^{2}}{2\mu g} = \frac{v_{0}^{2}}{2\mu g}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.