Question

A person is at a distance $x$ from the bus when the bus begins to move with a constant acceleration $a$ . What is the minimum velocity with which the person should run towards the bus so as to catch it?
A)$2ax$
B)$\sqrt {2ax} $
C)$ax$
D)$\sqrt {ax} $

Answer 1

Hint This is a pretty simple question which only requires us to use the equations of motion. These are:
$s = ut + \dfrac{1}{2}a{t^2}$
${v^2} - {u^2} = 2as$
$v = u + at$
Where $s$ is the displacement.
$u$ is the initial velocity.
$v$ is the final velocity.
$t$ is the time.
$a$ is the acceleration.

 Complete step-by-step solution:Let the distance moved by the bus until the man catches the bus be $y$ . The only thing common in the motion of the bus and man will be the time taken. Time taken by the man to cover distance $\left( {x + y} \right)$ will be equal to the time taken by the bus to cover the distance $y$ . You can also equate the velocity of the man and bus when the man gets on the bus.
Complete step by step answer:
A pictorial representation to understand this problem well is as follows:

Here, $A$ is the initial position of man, $B$ is the initial position of the bus and $C$ is the position at which man catches the bus.
Try to visualize the situation to proceed. Both the man and the bus start from the rest. Man runs with constant velocity while the bus moves with constant acceleration. Man should have a velocity fast enough such that his velocity matches the velocity of the bus after a certain displacement. Let this distance be $y$ . We are using the terms distance and displacement interchangeably because the motion is in one direction.
We know that initially the distance between the man and the bus is $x$ .
Now, let us focus upon the situation at point $C$ . At point $C$, velocity of bus as well as the man attains the same value due to which the man is able to catch the bus. Moreover, time taken by man to reach this point from $A$ is equal to the time taken by the bus to reach this point from $B$ . We will make use of these results along with the three equations of motion which are:
$s = ut + \dfrac{1}{2}a{t^2}$
${v^2} - {u^2} = 2as$
$v = u + at$
Where $s$ is the displacement.
$u$ is the initial velocity.
$v$ is the final velocity.
$t$ is the time.
$a$ is the acceleration.
Velocity of bus at point $C$
$
  {V_{bus}} = u + at \\
  {V_{bus}} = at \\
 $
(Since, the initial velocity of the bus is zero.)
$ \Rightarrow t = \dfrac{{{V_{bus}}}}{a}{\text{ or }}\dfrac{{{V_{man}}}}{a}$
Because at point $C$ , the velocity of the bus is equal to the velocity of the man.
So, $t = \dfrac{{{V_{man}}}}{a}$ (1)
Using the relation ${v^2} - {u^2} = 2as$ for bus at point $C$
${V_{man}}^2 - 0 = 2ay$
$y = \dfrac{{{V_{man}}^2}}{{2a}}$ (2)
Using relation $s = ut + \dfrac{1}{2}a{t^2}$ to describe the motion of man at point $C$ , we get
$x + y = {V_{man}}t$
(Since man moves with constant velocity.)
Substituting the values of $t$ and $y$ from equation (1) and (2) in the above relation, we get
$
  x + \dfrac{{{V_{man}}^2}}{{2a}} = \left( {{V_{man}}} \right)\left( {\dfrac{{{V_{man}}}}{a}} \right) \\
  x = \dfrac{{{V_{man}}^2}}{a} - \dfrac{{{V_{man}}^2}}{{2a}} \\
  x = \dfrac{{{V_{man}}^2}}{{2a}} \\
  2ax = {V_{man}}^2 \\
   \Rightarrow {V_{man}} = \sqrt {2ax} \\
 $
So, option B is correct.

Note: You need to visualize the situation to understand and solve it. This solution is nothing but repeated use of equations of motions keeping in mind the conditions such as equal velocities and time taken at point $C$ as in this question.