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Question: A body \(A\) moves with a uniform acceleration \(a\) and zero initial velocity. Another body \(B\) ,...

A body AA moves with a uniform acceleration aa and zero initial velocity. Another body BB , starts from the same point moves in the same direction with a constant velocity vv . The two bodies meet after a time tt . The value of tt is
(A) 2v/a2v/a
(B) v/av/a
(C) v/2av/2a
(D) v/2a\sqrt {v/2a}

Explanation

Solution

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+ \raise.5ex1/\lower.25ex2 at2s = ut + {\text{ }}\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} {\text{ }}a{t^2}
Where, ss is the displacement of the body, uu is the initial velocity of the body, tt is the time taken and aa is the acceleration of the body.

Step By Step Solution
For the body AA ,
u=0u = 0
a=aa = a
Thus, the formula turns out to be,
s= \raise.5ex1/\lower.25ex2 at2⋅⋅⋅⋅(1)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 BB ,
u=vu = v
a=0a = 0
Thus, the formula turns out to be,
s=vt⋅⋅⋅⋅(2)s = vt \cdot \cdot \cdot \cdot (2)
Then,
Equating (1)(1) and(2)(2), we get
\raise.5ex1/\lower.25ex2 at2=vt\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} {\text{ }}a{t^2} = vt
⇒t=2v/a\Rightarrow t = 2v/a

Hence, the answer is (a).

Additional Information: The equations of motions helps us to connect all the parameters related to motion such as displacement, distance, speed, velocity time and acceleration. The usage of this equation depends on the given situation and the required parameter to evaluate.
The equations are:
v=u+atv = u + at
Here, the final parameter to find the final velocity (/speed) when the initial velocity (speed), acceleration and time are given.
s=ut+ \raise.5ex1/\lower.25ex2 at2s = ut + {\text{ }}\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} {\text{ }}a{t^2}
Here, the final parameter to find is the displacement (in some cases distance), when the initial velocity (/speed), acceleration and time are known.
v2−u2=2as{v^2} - {u^2} = 2as
Here, the motive is to relate all the parameters.
Thus, every equation has its own purpose at the same time restrictions to be used. These equations come in very handy for solving any type of situation in motion.

Note: We took the displacement same for both the bodies. This is because they started from the same position.