An automobile travellingat `40 km//h`, can be stopped at distance of 40 m by applying brakes. If the same autombile is travelling at`80 km//h`, the minimum stopping distance, in metres is (assume no skidding) :

Options

A. 150m
B. 100m
C. 75m
D. 160m

Answer

D

Explanation

If a is the retardation then
`v^(2)-u^(2)=-2as`
To stop after travelling a distance S we can write the following (v=0)
`2aS=u^(2)`
`rArr (S_(1))/(S_(2))=(u_(1)^(2))/(u_(2)^(2))rArr S_(2)=(u_(2)^((2)))/(u_(1)^(2))(40)=((80)/(40))^(2)(40m)`
160m