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

Answer

B

Explanation

Data: 72km/h`=(72xx1000)/(3600)=20m//s=u`
and v=0 and `mu=0.5`
The forward motion of the car is opposed by the force of friction (F)`=muR=mumg`
`:.` The retarding force=m(-a)
`:." "-ma=mumg`
`:." "a=-mug=0.5xx10=-5m//s^2`
Using `v^2=u^2+2`as, we get
`0=20xx20+2(-5)s`
`:.` 10s=400 `:.` s=40m