The displacement of a particle as a function of time is shown in the figure. The figure shows that

If $t = \sqrt{x} + 4$,then $\left(\frac{dx}{dt}\right)_{t=4}$ is:

The position $x$ of a particle moving in one dimension under the influence of a constant force is given by $t = \sqrt{x} + 3$,where $x$ is in meters and $t$ is in seconds. Find the displacement of the particle in $m$ when its velocity becomes zero.

For the displacement-time graph shown in the figure,the ratio of the magnitudes of the (constant) speeds during the first two seconds and the next four seconds is

$A$ 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 a constant $F_0$ 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: