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

Options

A. `12m/s ^(2)`
B. `8m//s^(2)`
C. `7m//s^(2)`
D. None of these

Answer

A

Explanation

Here, position of particle, `x =-2 5t +t ^(3)` is given.
`therefore x = t ^(3) - 5t +2`
So, velcity` v = (dx)/(dt) = (d)/(dt) (t ^(3) - 5t +2)`
Thus, acceleration ` = (dv)/(dt) = (d)/(dt) (3t ^(2) -5)`
`therefore a = 6t`
Taking `t =2s` in above formula, we get
`a =12 m//s^(2)`