$A$ car is moving with a speed of $150 \ km/h$ and after applying the brakes,it travels $27 \ m$ before it stops. If the same car is moving with a speed of one-third of the initial speed,then it will stop after traveling how many meters?

$A$ 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 $x=0$.

$A$ 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:

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

$A$ car moving with a speed of $40 \, km/h$ can be stopped by applying brakes after at least $2 \, m$. If the same car is moving with a speed of $80 \, km/h$,what is the minimum stopping distance in meters?

$A$ body starts from rest and moves with constant acceleration for $t$ seconds. It travels a distance $x_{1}$ in the first half of the time and $x_{2}$ in the next half of the time. Then: