Power `k = F. v = (ma) v = m(dv)/(dt) v = (dv)/(dt)`
Rearranging we get `vdv = (k)/(m) dt`
Integrating both side.
`underset(0)overset(v)intvdv=(k)/(m)underset(0)overset(t)intdt rArr (v^(2))/(2) =(k)/(m)t`
`implies v = sqrt((2 kt)/(m))` ...(i)
Force acting on partical `F = m(dv)/(dt)` ...(ii)
Differentiating equation (i) w.r.t. time
`(dv)/(dt) = (sqrt((2k)/(m))) (dt^(1//2))/(dt) = (sqrt((2k)/(m))) (1)/(2) t^(-1//2)`
`= (sqrt((k)/(2m))) t^(-1//2)`
Subsituting the value of `(dv)/(dt)` in equation (ii) we get
`F = m (dv)/(dt) = m(sqrt((k)/(2m))) t^(-1//2) = (sqrt((mk)/(2))) t^(-1//2)`