A Body start from origin and move along x-axis such that velocity at any instant is given by `4t^(3)-2t` is in second and velocity in m/s Find acceleration of the particle when it is at a distance of 2 m from the origin:

Options

A. `28 ms^(-2)`
B. `22ms^(-2)`
C. `12 ms^(-2)`
D. `10ms^(-2)`

Answer

b

Explanation

`v=4t^(3)-2timplies(dx)/(dt)=4t^(3)-2t`
`implies int dx =int (4t^(3)-2t)dt`
`x=(4t^(4))/(4)-(2t^(2))/(2)implies x=t^(2)(t^(2)-1)` when x=2 then 2 =`t^(4)-t^(2) implies t^(4)-t^(2)-2 =0`
`implies t^(4)-2t^(2)+t^(2)-2=0`
`t^(2)(t^(2)-2)+1(t^(2)-2)=0`
`implies (t^(2)-2)(t^(2)+1)=0 implies t^(2)=2 implies t=sqrt(2)`
Now `a=(dv)/(dt)=12t^(2)-2 implies a=12xx2-2=22m//s^(2)`