Starting from rest, the acceleration of a particle is `a=2(t-1)`. The velocity of the particle at t=5s is :-

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.