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(m)/(s^2)`
B. `22(m)/(s^2)`
C. `12(m)/(s^2)`
D. `10(m)/(s^2)`

Answer

B

Explanation

`v=4t^3-2t` (given) `a=(dv)/(dt)=12t^2-2`
and `x=int_0^tvdt=int_0^t(4t^3-2t)dtt^4-t^2`
When particle is at 2 m from the origin `t^4-t^2=2impliest^4-t^2-2=0(t^2-2)(t^2+1)=0impliest=sqrt2`sec
acceleration at `t=sqrt2` sec given by,
`a=12t^2-2=12xx2-2=22(m)/(s^2)`