$A$ truck moving with a constant velocity $12 \,m/s$ crosses a car moving from rest with uniform acceleration $2 \,m/s^2$. The distance the car has to travel from the starting point to cross the truck again is (in $\,m$)

The velocity of a car accelerating uniformly increases from $20 \ m/s$ to $80 \ m/s$ while covering a distance of $200 \ m$. The time taken is $.... \ s$.

Consider a particle moving along the positive direction of the $X$-axis. The velocity of the particle is given by $v = \alpha \sqrt{x}$ (where $\alpha$ is a positive constant). At time $t = 0$, the particle is located at $x = 0$. Find the time dependence of the velocity and the acceleration of the particle, respectively.

The velocity of a particle is $v = v_0 + gt + ft^2$. If its position is $x = 0$ at $t = 0$,then its displacement after unit time $(t = 1)$ is

$A$ particle initially at rest is moving along a straight line with an acceleration of $2 \,m/s^2$. At a time of $3 \,s$ after the beginning of motion, the direction of acceleration is reversed. The time from the beginning of the motion in which the particle returns to its initial position is

For a body moving with uniform acceleration along a straight line, the variation of its velocity $(v)$ with position $(x)$ is best represented by: