The velocity of a particle moving in the positive direction of the x axis varies as `v=alphasqrtx`, where `alpha` is a positive constant. Assuming that at the moment `t=0` the particle was located at the point `x=0`, find:
(a) the time dependence of the velocity and the acceleration of the particle,
(b) the mean velocity of the particle averaged over the time that the particle takes to cover the first s metres of the path.

Answer

(a) `v=alpha^2t//2`, `w=alpha^2//2`; (b) `<< v >>=alphasqrt(s//2)`.

Explanation

As particle is in unidirectional motion it is directed along the x-axis all the time. As at `t=0`, `x=0`
So, `Deltax=x=s`, and `(dv)/(dt)=w`
Therefore, `v=alphasqrtx=alphasqrts`
or, `w=(dv)/(dt)=(alpha)/(2sqrts)(ds)/(dt)=(alpha)/(2sqrts)`
`=(alphav)/(2sqrts)(alphaalphasqrt5)/(2sqrts)=(alpha^2)/(2)` (1)
As, `w=(dv)/(dt)=alpha^2/2`
On integrating, `underset0oversetvintdv=underset0oversettintalpha^2/2dt` or, `v=alpha^2/2t` (2)
(b) Let s be the time to cover first s m of the path. From the Eq.
`s=intvdt`
`S=underset0oversettintalpha^2/2dt=alpha^2/2t^2/2` (using 2)
or `t=2/alphasqrt5`
The mean velocity of particle
`lt v gt =(intv(t)dt)/(intdt)=(underset0overset(2sqrtsalpha)intalpha^2/2tdt)/(2sqrts//alpha)=(alphasqrts)/(2)`