A body of mass 'm' is travelling with a velocity 'u' . When a constant retarding force 'F' is applied , it comes to rest after travelling a distance `s_1` . If the initial velocity is 2u , with the same force F , the distance travelled before it comes to rest is `s_(2)` . then

Options

A. `s_(2) = 2 s_(1)`
B. `s _(2) = (s_(1))/(2)`
C. `s_(2) = s_(1)`
D. `s_2 = 4 s_1`

Answer

D

Explanation

By applying constant retarding force F the body is brought to rest so v = 0 .
Retardation , `a = - (F)/(m)`
If s be distance travelled by the body before it comes to rest (called stopping distance) . Using `3^(rd)` equation of motion
`v^(2) - u^(2) = 2as`
`(0)^(2) - u^(2) = 2 (- (F)/(m) ""s)`
`s = (u^(2) m)/(2 F)`
`s prop u^(2)` For the same of m.F.
`therefore (s_(1))/(s_(2)) = ((u)/(2u))^(2) = (1)/(4)`
or `s_(2) = 4 s_(1)`