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).

The position of a particle moving along a straight line is given by x

1.	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).
Answer» Solution :`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)`