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

Options

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

Answer

B

Explanation

Initial kinetic energy of the car = `1/2 mv^2`
Work done against friction = `mu mg s`
From work - energy theorem
`mu m g s = 1/2 mv^2 " or " s = ((v^2)/(2 mu g))`
Stopping distance , `s = ((v^2)/(2 mu g))`
Given, `v = 72 km//h = 72 xx 5/18 = 20 m//s`
`mu = 0.5 and g = 10 ms^(-2)`
`:. s = (20 xx 20)/(2 xx 0.5 xx 10) = 40 m` .