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

Options

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

Answer

D

Explanation

`s = t^2 - 6t + 3t + 4`
Velocity `v = (ds)/(dt) = 3t^2 - 6 xx 2t + 3 xx 1 + 0`
=`3t^2 - 12t + 3` ….(i)
Acceleration `a = (dv)/(dt) = 3 xx 2t - 12 xx 1 + 0 = 6t - 12`
Acceleration is zero at time t given by `6t-12=0`
`rArr t = 2 seconds`
So, velocity v at `t = 2` seconds is
`v = (3t^2 - 12t + 3)_(t = 2s)`
=`3 xx (2)^2 - 12 xx 2 + 3 = -9 m//s`.