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`.

Explanation

From `t=sqrt(x)+3 , sqrt(x)=t-3`
Squaring both sides, `x=t^(2)-6t+9` …(i)
Now, `v=(dx)/(dt)=2t-6 and a =(dv)/(dt)=2.`
When `v=0, 2t-6=0, 2t=6,t=3s`
From (i), `x=3^(2)-6(3)+9=0, i.e.,`
displacement is zero.
(ii) From (i), at `t=0,x=0-0+9=9m`
and at `t=6,x=6^(2)-6xx6+9=9m`
`:.` In the first six seconds, net displacement of the particle is zero.
As work done `=` force `xx` displacement.
`:.` Work done in first six seconds `=` zero