$A$ body is moving with uniform acceleration,describing $40 \, m$ in the first $5 \, s$ and $65 \, m$ in the next $5 \, s$. Its initial velocity will be . . . . . . $m/s$.

$A$ particle of mass $m$ is constrained to move on the $x$-axis. $A$ force $F$ acts on the particle. $F$ always points toward the position labeled $E$. For example,when the particle is to the left of $E$,$F$ points to the right. The magnitude of $F$ is constant except at point $E$ where it is zero. The system is horizontal. $F$ is the net force acting on the particle. The particle is displaced a distance $A$ towards the left from the equilibrium position $E$ and released from rest at $t=0$. Find the minimum time it will take to reach from $x=-A/2$ to $x=0$.

The distance covered by a body of mass $5\,g$ having linear momentum $0.3\,kg\,m/s$ in $5\,s$ is $..........\,m$.

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

The distance travelled by a particle is proportional to the square of time. Then,the particle travels with:

$A$ driver applies the brakes on seeing the red traffic signal $400 \ m$ ahead. At the time of applying brakes,the vehicle was moving with $15 \ m/s$ and retarding at $0.3 \ m/s^2$. The distance of the vehicle from the traffic light one minute after the application of brakes is: (in $m$)