Question

A motor cycle 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. $\frac{v^2}{\mu }$
• C. ${\left[\frac{v_0}{\mu g}\right]}^2$
• D. $\frac{v_0}{\mu g}$

Answer 1

The correct option is A $\frac{v_0^2}{2\mu g}$

From $v^2-u^2=2as$
$0-v_0^2=2\left(\mu g\right)s\therefore s=\frac{v_0^2}{2\mu g}$