A particle moves along a staight line such that its displacement at any time t is given by `s=t^3-6t^2+3t+4m`. Find the velocity when the acceleration is 0.

Options

A. `3ms^(-1)`
B. `-12ms^(-1)`
C. `42ms^(-1)`
D. `-9ms^(-1)`

Answer

D

Explanation

Velocity, `v= (ds)/(dt) = 3t^(2) - 12t + 3`
Acceleration, `a= (dv)/(dt) = 6t-12`
For a=0, we have 0=6
`t- 12 or t= 2s`. Hence, at t= 2s the velocity will be `v= 3 xx 2^(2) -12 xx 2 + 3 = -9 ms^(-1)`