Consider a particle moving along the positive direction of X-axis. The velocity of the particle is given by `v = alphasqrt(x)` (`alpha` is a positive constant). At time t = 0, if the particle is located at x = 0, the time dependence of the velocity and the acceleration of the particle are respectively 

Options

A. `(alpha^2)/(2) t and (alpha^2)/(2)`
B. `alpha^2 t and alpha^2`
C. `(alpha)/(2) t and (alpha)/(2)`
D. `(alpha^2)/(4) t and (alpha^2)/(4)`

Answer

A

Explanation

Given that ,
`v= alpha sqrt(x)`8
`(dx)/(dt) = alpha sqrx (therefore v = (dx)/(dt))`
`(dx)/(sqrtx) = alpha dt`
`int (dx)/(dt) = alpha int dt`
`rArr 2 sqrtx = alpha t + c`
At ` t = 0, x =0`
`rArr c=0`
`therefore 2 sqrtx = alpha t`
On squaring, `rArr 4x = alpha^2 t^2`
`rArr x = (alpha^2 t^2)/(4)`
On differentiating,
`(dx)/(dt) = v= (alpha^2)/(4) ( 2t^2 - 1)`
`= (alpha^2 t)/(2)`
`rArr alpha = (dv)/(dt)`
`= (dv)/(dt) ((dx)/(dt))`
`= v ((dv)/(dx))`
`= alpha sqrtx ((d)/(dx) (alpha sqrtx))`
`= (alpha sqrtx(alpha) ( x-^(1//2)))/(2)`
`therefore a = (alpha^2)/(2)`