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.

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)The acceleration of the particle at t = 2 sec. is ...... Here x is in meter.
Answer» `12m/s ^(2)` `8m//s^(2)` `7m//s^(2)` None of these Solution :Here, position of PARTICLE, `x =-2 5t +t ^(3)` is given. `thereforex = 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)`