The displacement `x` of particle moving in one dimension, under the action of a constant force is related to the time `t` by the equation ` t = sqrt(x) +3`
where `x is in meters and t in seconds` . Find
(i) The displacement of the particle when its velocity is zero , and
(ii) The work done by the force in the first ` 6 seconds`.

Options

A. `9 J`
B. `6 J`
C. `0J`
D. `3 J`

Answer

C

Explanation

`x = (t - 3)^(2) implies v = (dx)/(dt) = 2 (t - 3)`
or `t = 0: v_(1) = - 6 m//s and at t = 6 sec: v_(2) = 6 m//s`
so, change in kinetic energy `= W = (1)/(2) mv_(2)^(2) - (1)/(2) mv_(1)^(2) = 0`