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

Here, `t=sqrt(x)+3`
`:. T-3=sqrt(x)`
or `x=(t-3)^(2)` …(i)
Velocity, `upsilon=(dx)/(dt)=2(t-3)`
When `upsilon=0, 2(t-3)=0, t=3sec. `
From (i), `x=(t-3)^(2)=(3-3)^(2)=Zero .`
Now, at `t=0, upsilon=2(6-3)=-6m//s,` and
at `t=6s, upsilon=2(6-3)=-6m//s`
Work done `=` final KE `-` initial KE
`=(1)/(2)m(-6)^(2)-(1)/(2)m(-6)^(2)`
Work done `=` Zero