The position of a particle moving along a straight line is given by `x =2 - 5t + t ^(3).` Find the acceleration of the particle at `t =2 s.` (x is metere).

Explanation

`x (t) = 2 - 5t + t ^(3)` is given,
`therefore v = (dx)/(dt) = (d)/(dt) [2- 5t +t ^(3)]`
`therefore v =-5 + 3t ^(2)` and acceleration
`a = (dv)/(dt) = (d)/(dt) [-5 + 3t ^(2) ] = 6t` acceleration at 2s, `a = 6 xx 2 =12 ms ^(-2)`