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(A) For body $A$,the distance covered in time $t$ with initial velocity $u = 0$ and acceleration $a$ is given by the equation of motion: $s_A = ut + \

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$2v/a$

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(A) For body $A$,the distance covered in time $t$ with initial velocity $u = 0$ and acceleration $a$ is given by the equation of motion: $s_A = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}at^2 = \frac{1}{2}at^2$.For body $B$,the distance covered in time $t$ with constant velocity $v$ is: $s_B = vt$.Since both bodies start from the same point and meet after time $t$,their displacements must be equal: $s_A = s_B$.Equating the two expressions: $\frac{1}{2}at^2 = vt$.Dividing both sides by $t$ (assuming $t 
eq 0$): $\frac{1}{2}at = v$.Solving for $t$,we get: $t = \frac{2v}{a}$.

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 $A$ moves with a uniform acceleration $a$ and zero initial velocity. Another body $B$ starts from the same point and moves in the same direction with a constant velocity $v$. The two bodies meet after a time $t$. The value of $t$ is:

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

$(c)$ $v^2 = v_0^2 + 2as$

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