$A$ graph between the square of the velocity of a particle and the distance $s$ moved by the particle is shown in the figure. The acceleration of the particle is $...........m/s^2$.

The relation between time $t$ and distance $x$ of a particle is $t = ax^2 + bx$, where $a$ and $b$ are constants. If $v$ is the velocity of the particle, then its acceleration is

$A$ particle is moving with constant acceleration $a$. The following graph shows the $v^{2}$ versus $x$ (displacement) plot. The acceleration of the particle is $...... \text{m/s}^{2}$.

$A$ particle is moving along a straight line with constant acceleration. At the end of the $10^{th}$ second,its velocity becomes $20 \, m/s$,and in the $10^{th}$ second,it travels a distance of $10 \, m$. Then the acceleration of the particle will be ........ $m/s^2$.

If the velocity of a body related to displacement $x$ is given by $v = \sqrt{5000 + 24x} \; \text{m/s}$,then the acceleration of the body is $\dots \dots \; \text{m/s}^2$.

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