Question

Two cars A and B are travelling in the same direction with velocities ${v_{\text{A}}}$ and ${v_{\text{B}}}$ where $\left( {{v_{\text{A}}} > {v_{\text{B}}}} \right)$ . When the car ${\text{A}}$ is at a distance $d$ behind the car ${\text{B}}$ the driver of the car ${\text{A}}$ applies brakes producing a uniform retardation $a$ . There will be no collision when:
(A) $d < \dfrac{{{{\left( {{v_{\text{A}}} - {v_{\text{B}}}} \right)}^2}}}{{2a}}$
(B) $d < \dfrac{{v_{\text{A}}^2 - v_{\text{B}}^2}}{{2a}}$
(C) $d > \dfrac{{{{\left( {{v_{\text{A}}} - {v_{\text{B}}}} \right)}^2}}}{{2a}}$
(D) $d > \dfrac{{v_{\text{A}}^2 - v_{\text{B}}^2}}{{2a}}$

Answer 1

First of all, we will find the initial relative velocities of the two cars. After that for the final case, the velocities of the two cars will be the same, so we will also find the final relative velocity. We will also find the stopping distance and then compare.

Complete step by step solution:
In the given question, we are supplied the following data:
There are two cars ${\text{A}}$ and ${\text{B}}$ which are travelling in the same direction.
The velocities of the two cars are ${v_{\text{A}}}$ and ${v_{\text{B}}}$ where $\left( {{v_{\text{A}}} > {v_{\text{B}}}} \right)$ .The driver of the car ${\text{A}}$ applies brakes producing a uniform retardation $a$ when the car ${\text{A}}$ is at a distance $d$ behind the car ${\text{B}}$ .We are asked to find out the condition for which there will be no collision.
To begin with, we need to first find the initial relative speed of the two cars and the final relative speed of both the cars again. We will find out the distance through which the first car moves before it comes to halt. This is the main technique that we will apply to solve this problem.Let us proceed to solve this problem.Let us first find the initial relative velocity which is given by the difference of the velocities of the two cars.
$u = {v_{\text{A}}} - {v_{\text{B}}}$
Where,
$u$ indicates the initial relative velocity of the two cars.
${v_{\text{A}}}$ indicates the velocity of the first car.
${v_{\text{B}}}$ indicates the velocity of the second car.
Again, for the final relative velocity:
In the final case the velocity of both the cars will be equal.
Then we can write,
$v = {v_{\text{A}}} - {v_{\text{B}}} \\\ \Rightarrow v = {v_{\text{A}}} - {v_{\text{A}}} \\\ \Rightarrow v = 0$
Now, we use one of the equations from the laws of motion:
${v^2} = {u^2} - 2as$ …… (1)
Where,
$v$ indicates the final relative velocity.
$u$ indicates the initial relative velocity.
$a$ indicates the retardation which is negative.
$s$ indicates the distance moved by the first car before coming to complete halt.
Now, we substitute the required values in the equation (1) and we get:
${v^2} = {u^2} - 2as \\\ \Rightarrow {0^2} = {\left( {{v_{\text{A}}} - {v_{\text{B}}}} \right)^2} - 2as \\\ \Rightarrow 2as = {\left( {{v_{\text{A}}} - {v_{\text{B}}}} \right)^2} \\\ \Rightarrow s = \dfrac{{{{\left( {{v_{\text{A}}} - {v_{\text{B}}}} \right)}^2}}}{{2a}}$
Therefore, the distance moved by the first car before coming to complete halt is $\dfrac{{{{\left( {{v_{\text{A}}} - {v_{\text{B}}}} \right)}^2}}}{{2a}}$ .
So, to avoid collision, the stopping distance $s$ must be less than the actual initial distance $d$ between the two cars.
Hence, we can write:
$d > s$
$\therefore d > \dfrac{{{{\left( {{v_{\text{A}}} - {v_{\text{B}}}} \right)}^2}}}{{2a}}$

The correct option is (C).

Note: While solving this problem, keep in mind that here relative velocities come into play as this is the case of the two cars combined. In the final case, the velocities of both the cars become equal after the brakes are applied. Hence in the final case, the final relative velocity is equal which means both are moving at the same velocity. If the stopping distance is greater than the actual initial distance between the two cars then the collision was bound to take place.