Question

In a car race on straight road, car A takes a times t less than car B at the finish and passes finishing point with speed ‘v’ more than that of car B. Both the car start from rest and travel with constant acceleration $ {{a}_{1}} $ and $ {{a}_{2}} $ respectively. Then ‘v’ is equal to?
\[A.\dfrac{{{a}_{1}}+{{a}_{2}}}{2}t\]
\[B.\sqrt{2{{a}_{1}}{{a}_{2}}t}\]
 $ C.\dfrac{2{{a}_{1}}{{a}_{2}}}{{{a}_{1}}+{{a}_{2}}}t $
\[D.\sqrt{{{a}_{1}}{{a}_{2}}t}\]

Answer 1

Hint: Two Newton’s laws of motion, will be necessary to solve this question, which are $ s=ut+\dfrac{1}{2}a{{t}^{2}} $ and $ {{v}^{2}}-{{u}^{2}}=2as $ . The relation given in the question between velocities of both the cars and the relation between the times of completing the race, will solve the problem, when we find the velocity of the velocity and time of completion of race of car A.

Complete step-by-step answer:
Let’s consider the time taken by car B to finish the race be ‘T’. Given that car A takes ‘t’ amount of time less than that of car B, Hence, time taken by car A to finish the race is ‘T-t’. Further, by considering the velocity of car B as ‘u’ upon finishing the race, we will get the velocity of car A upon finishing the race as ‘u+v’.
As per the question, the constant acceleration of car A is $ {{a}_{1}} $ and that of car B is $ {{a}_{2}} $ .
We will now use 2 Newton’s laws of motion.
 $ s=ut+\dfrac{1}{2}a{{t}^{2}} $ and $ {{v}^{2}}-{{u}^{2}}=2as $ , where s is the distance, u is the initial velocity and v is the final velocity, a refers to the acceleration of the body and t is the time.
In the given information, initial velocity = 0. Hence, $ s=\dfrac{1}{2}a{{t}^{2}}\Rightarrow {{t}^{2}}=\dfrac{2s}{a}\Rightarrow t=\sqrt{\dfrac{2s}{a}} $ and $ {{v}^{2}}=2as\Rightarrow v=\sqrt{2as}. $
Hence, the information that we have for car B becomes: $ u=\sqrt{2{{a}_{2}}s} $ and $ T=\sqrt{\dfrac{2s}{{{a}_{2}}}} $
Similarly, the information for car B becomes: $ u+v=\sqrt{2{{a}_{1}}s} $ and $ T-t=\sqrt{\dfrac{2s}{{{a}_{1}}}} $
 $ \therefore (v+u)-u=\sqrt{2{{a}_{1}}s}-\sqrt{2{{a}_{2}}s}\Rightarrow v=\sqrt{2{{a}_{1}}s}-\sqrt{2{{a}_{2}}s} $ and\[T-(T-t)=\sqrt{\dfrac{2s}{{{a}_{2}}}}-\sqrt{\dfrac{2s}{{{a}_{1}}}}\Rightarrow t=\sqrt{\dfrac{2s}{{{a}_{2}}}}-\sqrt{\dfrac{2s}{{{a}_{1}}}}\]
We can further simplify them to, \[v=\sqrt{2s}(\sqrt{{{a}_{1}}}-\sqrt{{{a}_{2}}})\] and $ t=\sqrt{2s}(\dfrac{1}{\sqrt{{{a}_{2}}}}-\dfrac{1}{\sqrt{{{a}_{1}}}})\Rightarrow t=\sqrt{2s}(\dfrac{\sqrt{{{a}_{1}}}-\sqrt{{{a}_{2}}}}{\sqrt{{{a}_{1}}{{a}_{2}}}}) $
 $ \therefore t=\dfrac{v}{\sqrt{{{a}_{1}}{{a}_{2}}}}\Rightarrow v=\sqrt{{{a}_{1}}{{a}_{2}}}t. $
Hence, the required additional velocity with which the car A crosses the finish line more than that of B is $ v=\sqrt{{{a}_{1}}{{a}_{2}}}t. $

Note: The question can be very confusing, if you don’t mark each of the variables correctly.
In this question, I have considered the final velocity of car B as $ 'u', $ however, earlier to explain the Newton’s laws of motion, I have also used $ 'u' $ . For Newton’s laws of motion, it refers to the initial velocity of the body.
Don’t get confused with the notations that are used.