Question

In a car race, car A takes time $t$ less than car B at the finish and passes the finishing point with speed $v$ more than that of the car B. Assuming both the cars starts from rest and travel with the constant acceleration ${a_1}$ and ${a_2}$ respectively. Show that $v = \sqrt {{a_1}{a_2}} t$ .

Answer 1

We need to use Newton’s laws of kinematics, ${v^2} - {u^2} = 2as$ and $s = ut + \dfrac{1}{2}a{t^2}$ . Since these are two bodies in motion, these laws of kinematics can be used for each car separately to find a relation between the two.

Formulas used We will be using the formula ${v^2} - {u^2} = 2as$ where $v$ is the final velocity of the body, $u$ is the initial velocity of the body, $a$ is the acceleration the body acquired and $s$ the displacement travelled by the body. Also we would be using $s = ut + \dfrac{1}{2}a{t^2}$ where $t$ will be the time taken to travel the distance $s$ .

Complete Answer:
We all know that there are few laws of kinematics that will be used to solve most of the problems that are related to kinematics. The third law of kinematics states ${v^2} - {u^2} = 2as$ . But if you investigate in accordance with this problem you can see that the bodies are starting from rest and hence $u = 0$
If $u = 0$ then the law would be, ${v^2} - 0 = 2as$ . Simplifying the equation further and applying square root on both sides we get,
$v = \sqrt {2as}$ .
Similarly, you can find that the second law of kinematics states, $s = ut + \dfrac{1}{2}a{t^2}$ . Again, here $u = 0$ because the cars begin at rest. Hence the equation becomes, $s = (0)t + \dfrac{1}{2}a{t^2}$ . Simplifying it further,
$\Rightarrow {t^2} = \dfrac{{2s}}{a}$ or $t = \sqrt {\dfrac{{2s}}{a}}$ .
Now, in the problem, since Car A takes time $t$ less than car B, let us assume that the time taken by car B is $T$ and so time taken by car A becomes $T - t$ . It also passes the finish line with a speed $v$ more than B. So let the speed of car B be $v'$ and the speed of car A be $v + v'$ .
Now let us apply the laws of kinematics on each car, starting with car A,
$v + v' = \sqrt {2{a_1}s}$ and $T - t = \sqrt {\dfrac{{2s}}{{{a_1}}}}$
For car B,
$v' = \sqrt {2{a_2}s}$ and $T = \sqrt {\dfrac{{2s}}{{{a_2}}}}$
Now substitute the values of $v' = \sqrt {2{a_2}s}$ and $T = \sqrt {\dfrac{{2s}}{{{a_2}}}}$ in the equations for car A.
$v + (\sqrt {2{a_2}s} ) = \sqrt {2{a_1}s}$ and $\sqrt {\dfrac{{2s}}{{{a_2}}}} - t = \sqrt {\dfrac{{2s}}{{{a_1}}}}$
$\Rightarrow v = \sqrt {2{a_1}s} - \sqrt {2{a_2}s}$ and $\sqrt {\dfrac{{2s}}{{{a_2}}}} - \sqrt {\dfrac{{2s}}{{{a_1}}}} = t$
Divide the final equations to give you the value of $\dfrac{v}{t}$
$\dfrac{{\sqrt {2{a_1}s} - \sqrt {2{a_2}s} }}{{\sqrt {\dfrac{{2s}}{{{a_2}}}} - \sqrt {\dfrac{{2s}}{{{a_1}}}} }} = \dfrac{v}{t}$
$\Rightarrow \dfrac{v}{t} = \dfrac{{\sqrt {{a_1}} - \sqrt {{a_2}} }}{{\sqrt {\dfrac{1}{{{a_2}}}} - \sqrt {\dfrac{1}{{{a_1}}}} }} \times \dfrac{{\sqrt {2s} }}{{\sqrt {2s} }}$
Taking L.C.M of the denominator and solving further we get,
$\dfrac{v}{t} = \dfrac{{\sqrt {{a_1}} - \sqrt {{a_2}} }}{{\sqrt {{a_1}} - \sqrt {{a_2}} }} \times \sqrt {{a_1}{a_2}}$
Eliminating the common terms, we get,
$\dfrac{v}{t} = \sqrt {{a_1}{a_2}}$
$\Rightarrow v = \sqrt {{a_1}{a_2}} \times t$ .

Note:
Although we changed most of the terms for each car A and B, we can see that the distance was never changed, this is because both the cars travelled the same distance from rest to complete the rest.