If the speed of a vehicle become 3 times for a given deceleration its stopping distance become .....

Options

A. 2 times
B. 9 times
C. 1/9` times
D. 3 times

Answer

B

Explanation

Stopping distance `(d _(s)) = (v_(0) ^(2))/(2a)`
`therefore dS prop v _(0) ^(2) [because 2a` constant]
`therefore (ds_(2))/( ds_(1)) = [ ((v _(0))_(2))/( (v_(0)) _(1))]`
`therefore (ds _(2))/( ds _(1)) = (3) ^(2)`
`therefore (ds _(2))/( ds _(1)) = 9`