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 moves with constant acceleration. Let $v_1, v_2, v_3$ be the average velocities in successive time intervals $t_1, t_2$ and $t_3$. Then:

Two cars leave one after the other and travel with an acceleration of $0.4\, m/s^2$. Two minutes after the departure of the first,the distance between the cars becomes $1.9\, km$. The time interval between the departures of the cars is ........$s$.

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 is moving along a straight line with initial velocity $v_0$. Its acceleration $a$ is constant. After $t$ seconds,its velocity becomes $v$. The average velocity of the body over the given time interval is

$A$ particle moves along a straight line in such a way that its acceleration is increasing at the rate of $2 m/s^3$. Its initial acceleration and velocity were $0$. The distance covered by it in $t = 3 s$ is ........ $m$.