The displacement x of a particle moving in one dimension, under the action of a constant force is related to timet by the equation `t = sqrt(x) + 3`
where x is in metre and tin second. Calculate : (a) the displacement of the particle when the velocity is zero, (b) the work done by the force in the first 6's.

Explanation

As `t = sqrt(x) + 3 ` i.e., `x = (t - 3)^2 " " ....(i)`
So, `v = ((dx)/(dt)) = 2(t - 3) " " ....(ii)`
(a) v will be zero when `2(t-3)=0 i.e., t=3`
Substituting this value of t in Eq. (i), `x = (3-3)^2 = 0`
i.e., when velocity is zero, displacement is also zero.
(b) From Eq. (ii), `(v)_(t = 0) = 2 (0 - 3) = - 6 m//s and (v)_(t = 6) = 2(6 - 3) = 6 m//s`
So, from work-energy theorem
`W = Delta KE = 1/2 m [v_f^2 - v_i^2] = 1/2 m [6^2 - (-6)^2] = 0`
i.e., work done by the force in the first 6 is zero.