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

$s=t^3-6t^2+3t+4$

Velocity $v=\frac{ds}{dt}$= $3t^2-12t+3$

Acceleration $a=\frac{dv}{dt}=6t-12$

Acceleration is zero at time t given by 6t – 12 = 0

$\Rightarrow t=2$seconds

So, velocity v at t = 2seconds is

$v={\left(3t^2-12t+3\right)}_{t=2s}=-9m/s$