If a particle moves according to the equation $v = at$,what distance will it cover in the first $4 \, s$ (in $a$)?

If the velocity of a particle is given by $v = (180 - 16x)^{1/2} \text{ m/s}$,then its acceleration will be ....... $\text{m/s}^2$.

$A$ particle moves along a straight line such that its displacement at any time $t$ is given by $s = t^3 - 6t^2 + 3t + 4$ meters. The velocity when the acceleration is zero is ........ $m s^{-1}$.

$A$ train moves from rest with a uniform acceleration $a$. After attaining a maximum speed $v$,it starts moving with uniform retardation $a$. Assuming $s$ is the total distance covered in the unidirectional motion of the train,find its total time of journey and maximum speed.

$A$ lift is coming down from the $8^{th}$ floor and is just about to reach the $4^{th}$ floor. Taking the ground floor as the origin and the upward direction as positive for all quantities,which one of the following is correct?

$A$ particle starts from the origin at time $t=0$ and moves in the positive $x$-direction. Its velocity $v$ varies with time as $v=10t \text{ cm/s}$. The distance covered by the particle in $8 \text{ s}$ will be: (in $\text{ cm}$)