The stopping distance of a vehicle moving with a given acceleration is proportional to the ....... of its speed.

The motion of a particle is described by the equation $v = at$. The distance travelled by the particle in the first $4 \ s$ is: (in $a$)

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

$A$ particle initially at rest moves along the $x$-axis. Its acceleration varies with time as $a = 4t$. If it starts from the origin,the distance covered by it in $3 \ s$ is $........... \ m$.

$A$ car moving along a straight highway with a speed of $126 \; km/h$ is brought to a stop within a distance of $200 \; m$. What is the retardation of the car (assumed uniform in $m/s^2$)?

$A$ particle starts from the origin at time $t=0$ and moves in the positive $x$-direction. Its velocity $v$ varies with time as $v=10t \text{ cm/s}$. The distance covered by the particle in $8 \text{ s}$ will be: (in $\text{ cm}$)