The velocity of a particle moving in the `x` direction varies as `V = alpha sqrt(x)` where `alpha` is a constant. Assuming that at the moment `t = 0` the particle was located at the point `x = 0`. Find the acceleration.

Explanation

`v=alphasqrt(x)=alphax^(1//2) [Here v=fn(x)]`
`(dv)/(dx)=alpha.(1)/2x^(-1//2)`
Acceleration `a=v(dv)/(dx)=alphax^(1//2) alpha/2x^(-1//2)=alpha^(2)/2`
Acceleration is constant. Using equation of motion,
`v=u+at=0+alpha^(2)/2t`
`=(alpha^(2)t)/2`