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

Hint: 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.