If the velocity of the particle is given by `v=sqrt(x)` and initially particle was at `x=4m` then which of the following are correct.

Options

A. At `t=2 s`, the position of the particle is at `x=9m`
B. Particle acceleration at `t=2 s` is `1 m//s^(2)`
C. Particle acceleration is `1//2 m//s^(2)` through out the motion
D. Particle will never go in negative direction from it's starting position

Answer

A, C, D

Explanation

`v=sqrt(x), " " .x_(4)^(x)int(dx)/sqrt(x)=underset(t=0)overset(t)(int)dtrArr [2sqrt(x)]_(4)^(x)=t`
`rArr x=((t+4)/2)^(2) at t=2 rArr x=9m`
`a=v(dv)/(dx)=sqrt(x)xx1/(2sqrt(x))=1/2 m//s^(2)`
at `x=4rArrv 2m//s` & it increases as x increases so it never becomes negative.