Question

Consider a car moving along a straight horizontal road with a speed of ${ 72\, km\, h^{-1}}$. 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)

40 m

Answer 2

Initial kinetic energy of the car $= \frac{1}{2} mv^2$
Work done against friction = $?mgs$
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, $v = {72 \, km \, h^{-1} = 72 \times \frac{5}{18} = 20 \, m \, s^{-1}}$
$\mu = 0.5$ and $g = {10 \, m \, s^{-2}}$
$\therefore \:\:\: {s = \frac{20 \times 20}{2 \times 0.5 \times 10} = 40 \, m}$