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. (a) `18m`
B. (b) Zero
C. (c) `9m//2`
D. (d) `36m`

Answer

B

Explanation

Here `x=(t-3)^2=t^2-6t+9`
`v=(dx)/(dt)=2t-6`
At `t=0`, `v=2xx0-6=-6`
At `t=6s`, `v=2xx6-6=+6`
Initial and final KE are same, hence no work is done.