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	$(b)$ $S = v_0t + \frac{1}{2}at^2$
	$(c)$ $v^2 = v_0^2 + 2as$

$A$ vehicle moving with $15 \,km/hr$ comes to rest by covering $5 \,m$ distance by applying brakes. If the same vehicle moves at $45 \,km/hr$, then by applying brakes, it will come to rest by covering a distance of: (in $\,m$)

$A$ particle executes the motion described by $x(t) = x_0 (1 - e^{-\gamma t})$ for $t \geqslant 0$ and $x_0 > 0$.
$(a)$ Where does the particle start and with what velocity?
$(b)$ Find the maximum and minimum values of $x(t)$,$v(t)$,and $a(t)$. Show that $x(t)$ increases with time,$v(t)$ decreases with time,and $a(t)$ increases with time.

$A$ particle initially at rest starts moving from reference point $x=0$ along the $x$-axis,with velocity $v$ that varies as $v=4 \sqrt{x} \ m/s$. The acceleration of the particle is . . . . . . $m/s^2$.

An object moving along the $x$-axis with a uniform acceleration has a velocity $\vec{v} = (12 \ cm \ s^{-1}) \hat{i}$ at $x = 3 \ cm$. After $2 \ s$,if it is at $x = -5 \ cm$,then its acceleration is:

$A$ car accelerates from rest at a constant rate of $2 \ m/s^2$ for some time. Then it retards at a constant rate of $4 \ m/s^2$ and comes to rest. If it remains in motion for $3 \ s$,then the total displacement covered by it is $.... \ m$.