A man 45 m behind the bus starts accelerating from the rest with acceleration 2.5. With what minimum velocity should the man starts running to catch the bus:
A) 12
B) 14
C) 15
D) 16

Hint:Acceleration is defined as the rate of change of velocity with respect to time. A body which is accelerated can be easily solved by using Newton's law of motion relations, there are three relations that can be used to solve the problem for an accelerating body.

Formula used:The formula of the second relation of the Newton’s law of motion is given by,
$s = ut + \dfrac{1}{2}a{t^2}$
Where s is the distance a is the acceleration u is initial velocity and t is the time taken.

Complete step-by-step answer:
It is given in the problem that a man is m behind the bus and the bus starts accelerating with an acceleration of $2 \cdot 5\dfrac{m}{{{s^2}}}$ and we need to find the minimum velocity with which the man should start so that he can catch the bus.
For the motion of the bus. Applying Newton’s law of motion second relation.
$s = ut + \dfrac{1}{2}a{t^2}$
Where s is the distance a is the acceleration u is initial velocity and t is the time taken.
As the bus started from rest therefore,
$ \Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
$ \Rightarrow s = 0 + \dfrac{1}{2}\left( {2 \cdot 5} \right){t^2}$
$ \Rightarrow s = \dfrac{1}{2}\left( {2 \cdot 5} \right){t^2}$………eq. (1)
Now the man has to cover a distance of$\left( {s + 45} \right)m$.
Let minimum velocity of the man is$u$.The distance can be expressed in terms of the distance and time as,
$ \Rightarrow s + 45 = \left( u \right)t$………eq. (2)
Replacing the value of s from equation (1) to above equation (2)
$ \Rightarrow s + 45 = u \cdot t$
$ \Rightarrow \dfrac{1}{2}\left( {2 \cdot 5} \right){t^2} + 45 = u \cdot t$
\[ \Rightarrow u = \dfrac{1}{2}\left( {2 \cdot 5} \right)t + \dfrac{{45}}{t}\]………eq. (3)
For minimum velocity,
$ \Rightarrow \dfrac{{du}}{{dt}} = 0$
$ \Rightarrow \dfrac{{d\left[ {\dfrac{1}{2}\left( {2 \cdot 5} \right)t + \dfrac{{45}}{t}} \right]}}{{dt}} = 0$
$ \Rightarrow - \dfrac{{45}}{{{t^2}}} + 1 \cdot 25 = 0$
$ \Rightarrow 1 \cdot 25{t^2} = 45$
$ \Rightarrow {t^2} = \dfrac{{45}}{{1 \cdot 25}}$
$ \Rightarrow {t^2} = 36$
$ \Rightarrow t = \sqrt {36} $
$ \Rightarrow t = 6s$
Replace the value of time $t = 6s$ in equation (3),
\[ \Rightarrow u = \dfrac{1}{2}\left( {2 \cdot 5} \right)t + \dfrac{{45}}{t}\]
\[ \Rightarrow {u_{\min .}} = \dfrac{1}{2}\left( {2 \cdot 5} \right) \times \left( 6 \right) + \dfrac{{45}}{6}\]
\[ \Rightarrow {u_{\min .}} = 15\dfrac{m}{s}\].
The minimum velocity of the man is \[{u_{\min .}} = 15\dfrac{m}{s}\].

The correct option for this problem I option C.

Note:The Newton’s law of motion relations can only be used if the body accelerating has a uniform acceleration if the accelerating body does not has a uniform acceleration then we cannot apply Newton’s law of motion relations as the relations in the Newton’s law of motion are formulated assuming that the value of the acceleration is constant.