Consider a car moving along a straight horizontal road with a speed of 72 km / h . If the coefficient of kinetic friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is `[g=10ms^(-2)]`

Options

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

Answer

B

Explanation

`a = mu g = 0.5 xx 10 = 5 m//s^(2)`
Stopping distance = S = `(u^2)/(2a) = ((72 xx (5)/(18))^(2))/(2 xx 5) = ((20)^(2))/(10) = 40` m