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# A body $A$ moves with a uniform acceleration $a$ and zero initial velocity. Another body $B$ , starts from the same point moves in the same direction with a constant velocity $v$ . The two bodies meet after a time $t$ . The value of $t$ is(A) $2v/a$(B) $v/a$(C) $v/2a$(D) $\sqrt {v/2a}$

Last updated date: 20th Jun 2024
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Hint: We will use the equations of motion. Out of all the equations, we will select the appropriate equation connecting all the parameters. Finally, we will find the appropriate relation.

Formulae Used: $s = ut + {\text{ }}\raise.5ex\hbox{\scriptstyle 1}\kern-.1em/ \kern-.15em\lower.25ex\hbox{\scriptstyle 2} {\text{ }}a{t^2}$
Where, $s$ is the displacement of the body, $u$ is the initial velocity of the body, $t$ is the time taken and $a$ is the acceleration of the body.

Step By Step Solution
For the body $A$ ,
$u = 0$
$a = a$
Thus, the formula turns out to be,
$s = {\text{ }}\raise.5ex\hbox{\scriptstyle 1}\kern-.1em/ \kern-.15em\lower.25ex\hbox{\scriptstyle 2} {\text{ }}a{t^2} \cdot \cdot \cdot \cdot (1)$
Now,
For the body $B$ ,
$u = v$
$a = 0$
Thus, the formula turns out to be,
$s = vt \cdot \cdot \cdot \cdot (2)$
Then,
Equating $(1)$ and$(2)$, we get
$\raise.5ex\hbox{\scriptstyle 1}\kern-.1em/ \kern-.15em\lower.25ex\hbox{\scriptstyle 2} {\text{ }}a{t^2} = vt$
$\Rightarrow t = 2v/a$

$v = u + at$
$s = ut + {\text{ }}\raise.5ex\hbox{\scriptstyle 1}\kern-.1em/ \kern-.15em\lower.25ex\hbox{\scriptstyle 2} {\text{ }}a{t^2}$
${v^2} - {u^2} = 2as$