List of top Physics Questions on Kinematics asked in Tripura JEE

The distance covered by a particle moving in a straight line path at time $ t $ (in seconds) is given by $ s = (t^3 - 6t^2 + 3t + 4) \, \text{m}$. The velocity of the particle when its acceleration is zero, will be
The position of a particle at time $ t $ is given by $ \vec{r} = \vec{r_0}(1 + at) $, where $ \vec{r_0} $ and $ a $ are two constants. When will the particle come back to its starting position?