$A$ 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 of the particle in $m/s$ when its acceleration is zero.

$A$ lift of mass $M = 500 \, kg$ is descending with a speed of $2 \, ms^{-1}$. Its supporting cable begins to slip,thus allowing it to fall with a constant acceleration of $2 \, ms^{-2}$. The kinetic energy of the lift at the end of a fall through a distance of $6 \, m$ will be $........... \, kJ$.

$A$ particle of mass $m$ is constrained to move on the $x$-axis. $A$ force $F$ acts on the particle. $F$ always points toward the position labeled $E$. For example,when the particle is to the left of $E$,$F$ points to the right. The magnitude of $F$ is constant except at point $E$ where it is zero. The system is horizontal. $F$ is the net force acting on the particle. The particle is displaced a distance $A$ towards the left from the equilibrium position $E$ and released from rest at $t=0$. Find the minimum time it will take to reach from $x=-A/2$ to $x=0$.

$A$ particle starts its motion from rest under the action of a constant force. If the distance covered in the first $10 \ s$ is $S_1$ and that covered in the first $20 \ s$ is $S_2$,then:

The displacement of a particle starting from rest at $t=0$ is given by $s=9 t^2-2 t^3$. The time in seconds at which the particle will attain zero velocity is (in $s$)

$A$ 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.