Question

Two cars A and B are moving in the same direction. Car A is moving with speed 10 \[m{{s}^{-1}}\] and at a distance 80 m behind car B which starts from rest and with uniform acceleration a. The maximum value of ‘a’ for which the car A reach car B is –
\[\begin{align}
  & \text{A) }\dfrac{5}{8}m{{s}^{-2}} \\
 & \text{B) }\dfrac{7}{8}m{{s}^{-2}} \\
 & \text{C) }\dfrac{5}{16}m{{s}^{-2}} \\
 & \text{D) }\dfrac{3}{8}m{{s}^{-2}} \\
\end{align}\]

Answer 1

Hint: We need to understand how the motion of the two bodies are related to each other with respect to change in time in order to find the acceleration of the second body so that the first one can overtake in spite of the acceleration of the car A.

Complete Step-by-Step Solution:
We are given the situation in which two cars A and B are moving with respect to each other. The car A is moving with a constant speed, whereas the car B starts from rest and is moving with a constant acceleration which is ahead of car A by 80 m at the time of starting. We are told that the car A overtakes the car B at some point of time. We know that for the car A to overtake B, the acceleration of car B should not be greater than a particular value.

We know that the distance travelled by the car A when it reaches the car B can be given as –
\[{{x}_{A}}={{x}_{B}}+80\]
Where, \[{{x}_{A}}\] is the distance travelled by car A and \[{{x}_{B}}\] is the distance travelled by car B.

Now, we can find the distance travelled by A when it reaches the car B after ‘t’ time as –
\[{{x}_{A}}=10t\]

Now, applying the equations of motion for the car B which started from rest and moving with acceleration ‘a’, we get –
\[\begin{align}
  & S=ut+\dfrac{1}{2}a{{t}^{2}} \\
 & \Rightarrow {{x}_{B}}=(0)t+\dfrac{1}{2}a{{t}^{2}} \\
 & \Rightarrow {{x}_{A}}-80=\dfrac{1}{2}a{{t}^{2}} \\
\end{align}\]
\[\begin{align}
  & \Rightarrow 10t=\dfrac{1}{2}a{{t}^{2}}+80 \\
 & \Rightarrow a{{t}^{2}}-20t+160=0 \\
 & \therefore t=\dfrac{-(-20)\pm \sqrt{{{(-20)}^{2}}-4a\times 160}}{2\times a} \\
\end{align}\]

For the acceleration to be such that car A overtakes B, the discriminant should be positive, i.e.,
\[\begin{align}
  & {{(-20)}^{2}}-640a\ge 0 \\
 & \Rightarrow -640a\ge 400 \\
 & \therefore a\le \dfrac{5}{8}m{{s}^{-2}} \\
\end{align}\]
This is the maximum acceleration that the car B can have.

The correct answer is option A.

Note:
We should keep in mind that the car from behind is travelling at speed which is constant. Therefore, the acceleration of the car moving in the front should not result in a velocity greater than this by the time the car A overtakes the car B as the motions are relative.