Question

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`.

Answer 1

As `t=sqrtx+3`
i.e. `x=(t-3)^2` …(i)
So, `v=(dx//dt)=2(t-3)` …(i)
(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)=-6m//s`
and `(v)_(t=6)=2(6-3)=6m//s`
So, from work-energy theorem
`w=DeltaKE=1/2m[v_f^2-v_i^2]=1/2m[6^2-(-6)^2]=0`
i.e. work done by the force in the first 6 s is zero.