Question

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. $\left(\frac{v_0^2}{\mu g}\right)$
• D. $\frac{v_0}{\mu }$

Answer 1

Answer: (A)

$\frac{v_0^2}{2\mu g}$

Initial kinetic energy of the car =$\frac{m{v}_o^2}{2}$

Work done against friction = $\mu mgs$

From conservation of energy

$\mu mgs$ =$\frac{m{v}_o^2}{2}$

On solving we get

s = $\frac{v_o^2}{2\mu g}$

Stopping distance, s = $\frac{v_o^2}{2\mu g}$