$A$ moving body is covering distances which are proportional to the square of the time. Then the acceleration of the body is

The velocity of a particle moving in the positive direction of the $x$-axis varies as $v = 5 \sqrt{x}$. Assuming that at $t = 0$,the particle was at $x = 0$,what is the acceleration of the particle in $m/s^2$?

$A$ bullet fired into a fixed target loses half of its velocity after penetrating $3\,cm$. How much further will it penetrate before coming to rest,assuming that it faces constant resistance to motion? (in $cm$)

$A$ 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

Match the relations in Column-$I$ with the equations in Column-$II$ for uniformly accelerated motion.

Column-$I$	Column-$II$
$(1)$ Velocity-time relation	$(a)$ $v = v_0 + at$
$(2)$ Velocity-displacement relation	$(b)$ $S = v_0t + \frac{1}{2}at^2$
	$(c)$ $v^2 = v_0^2 + 2as$

$A$ body moving with uniform acceleration travels a distance of $25 \ m$ in the fourth second and $37 \ m$ in the sixth second. The distance covered by the body in the next two seconds is: (in $m$)