Question

In a car race, car A takes a time of t s, less than car B at the finish and passes the finishing point with a velocity v more than car B. Assuming that the cars start from rest and travel with constant accelerations $a_1$ and $a_2$, respectively, show that v = $\sqrt{a_1{a}_{2}t}$

Answer 1

In the car race, the cars start from zero velocity and accelerate with constant accelerations. Here well discuss an important concept of uniformly accelerated motion. If a body starts from rest, i.e., with zero initial velocity and accelerates with an acceleration a after travelling distance 5, its velocity can be given by speed equation $v^2=u^2+2as$.

As we have u = 0 $\Rightarrow$ v = $\sqrt{2as}$

For the time taken to travel this distance 5, we use equation as

$s=ut+\frac{1}{2}a{t}^2$

Here again we have u = 0,s=$\frac{1}{2}a{t}^2$ or t =$\sqrt{\frac{2s}{a}}$

In the questions given, car A starts with acceleration $a_1$ and car B with acceleration $a_2$. If car B reaches the finishing point at time T and with speed u, car A will reach at time T- t and with speed u + v as given in the question.

As both the cars start from rest and cover the same distance, say s, we have

$ForcarA$ v + u =$\sqrt{2a_{1}s}$ and T - t = $\sqrt{\frac{2s}{a_1}}$

$ForcarB$ u =$\sqrt{2a_{2}s}$ and T = $\sqrt{\frac{2s}{a_2}}$

From above equations, eliminating the terms of u and T, we get

$v=\sqrt{2a_{1}s}-\sqrt{2a_{2}s}$

$t=\sqrt{\frac{2s}{a_2}}-\sqrt{\frac{2s}{a_1}}$

Dividing the above equations, we get $v=\sqrt{a_1{a}_{2}t}$