Question

Consider a car moving along a straight horizontal road with a speed of $72km{h}^{-1}$. If the coefficient of static friction between road and tyres is $0.5$, the shortest distance in which the car can be stopped is

• A. 30 m
• B. 40 m
• C. 72 m
• D. 20 m

Answer 1

Answer: (B)

Initial kinetic energy of the car $=\frac{1}{2}m{v}^2$
Work done against friction = $\mu mgs$
From conservation of energy
$\mu mgs=\frac{1}{2}m{v}^2$ or $s=\left(\frac{v^2}{2\mu g}\right)$
$\therefore$ Stopping distance, $s=\left(\frac{v^2}{2\mu g}\right)$
Given, $v=72km{h}^{-1}=72\times \frac{5}{18}=20m{s}^{-1}$
$\mu =0.5$ and $g=10m{s}^{-2}$
$\therefore s=\frac{20\times 20}{2\times 0.5\times 10}=40m$