$A$ particle is moving along the $Y$-axis. The position of the particle from the origin as a function of time $(t)$ is given as $y(t) = 10 t e^{-2 t}$. How far is the particle from the origin when it stops momentarily? ($y$ is given in units of metre and $t$ is in units of second)

$A$ 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 $t=0$. The velocity-time graph of the particle is:

Suggest a suitable physical situation for the following $x-t$ graph.

Figures $(i)$ and $(ii)$ below show the displacement-time graphs of two particles moving along the x-axis. We can say that

$y = (P t^2 - Q t^3) \ m$ is the vertical displacement of a ball which is moving in a vertical plane. Then the maximum height that the ball can reach is,

$A$ particle starts from rest and moves in a straight line. It travels a distance $2L$ with uniform acceleration and then moves with a constant velocity for a further distance of $L$. Finally,it comes to rest after moving a distance of $3L$ under uniform retardation. Then the ratio of average speed to the maximum speed $\left(\frac{\bar{V}}{V_m}\right)$ of the particle is