Consider a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of static friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is `[g=10 ms^(-1)]`

Options

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

Answer

B

Explanation

Friction =`muMg`
Hence retardation `=mu g=0.5xx10=5ms^(-2)`
Using ` v^(2)-u^(2)=2ax`
Taking ` v=0, u=20ms^(-1)`
we get x=40 ,