The relation $3t = \sqrt{3x} + 6$ describes the displacement of a particle in one direction,where $x$ is in $m$ and $t$ is in $s$. The displacement,when velocity is zero,is ......... $m$.

$A$ 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:

$A$ car starts from rest and moves with uniform acceleration $a$ on a straight road from time $t = 0$ to $t = T$. After that,a constant deceleration $a$ brings it to rest. In this process,the average speed of the car is

The accompanying graph of position $x$ versus time $t$ represents the motion of a particle. If $p$ and $q$ are both positive constants,the expression that best describes the acceleration $a$ of the particle is

Consider a car initially at rest, starts to move along a straight road first with acceleration $5 \ m \ s^{-2}$, then with uniform velocity and finally, decelerating at $5 \ m \ s^{-2}$, before coming to a stop. Total time taken from start to end is $t = 25 \ s$. If the average velocity during that time is $72 \ km \ hr^{-1}$, the car moved with uniform velocity for a time of: (in $s$)

The $x-t$ graph of a particle moving along a straight line is shown in the figure. The distance-time graph of the particle is correctly shown by: