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. `18 m`
B. zero
C. `9 m//2`
D. `36 m`

Answer

B

Explanation

Here `t = sqrtx+ 3 or x = (t - 3)^(2) = t^(2) - 6t + 9`
`v = (dx)/(dt) = 2t + 6`
at `t = 0, v = 2 xx 0 - 6 = - 6`
at `t = 6 s, v = 2 xx 6 - 6 = +6`
Initial and final `KE` ar esame hence no work is done.