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

Options

A. at `t=2` sec, the position of the particle is `x=9 m`
B. particle acceleration at `t=2` sec is `1 m//s^(2)`
C. particle acceleration is `½ ms^(2)` through out the motion
D. particle will never go in negative direction from its starting position.

Answer

A, C, D

Explanation

`v=sqrt(x)`
`(dx)/(dt)=sqrt(x)impliesint(dx)/x^(1//2)=int dt implies 2sqrt(x)=t+c`
But at `t=0,x=4,impliesc=4`
`impliesx=((t+4)^(2))/(4)impliesx=((6)^(2))/(4)=(36)/(4)=9`
`a=v(dv)/(dx)=sqrt(x)xx(1)/(2sqrt(x))= ½ m//s^(2)`.