A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by `(4t^3-2t)`, where t is in sec and velocity in m/s. what is the acceleration of the particle when it is 2 m from the origin?

Options

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

Answer

2

Explanation

`v_(t)=4t^(3)-2t, " " (dx)/(dt)=4t^(3)-2t`,
`int_(0)^(x) dx=int_(0)^(t) (4t^(3)-2t) dt`
`x=t^(4)-t^(2)" " "When" x=2,t=?`
`2=(t^(4)-t^(2))`
`t^(4)-t^(2)-2=0, " " t^(2)=y` ltbRgt `y^(2)-y-2=0`
`y=(1+pmsqrt(1-4xx1xx-2))/2=(1+pmsqrt(9))/2=(1+3)/2`
`(1-3)/2`
`y=2,-1" " t^(2)=2, -1`
`t^(2)=-1` (impossible)
`t=pmsqrt(2)` (possible), `a=(dv)/(dt)=12t^(2)-2`
`a=24-2=22 m//s^(2)`