# Question d67f9

Feb 20, 2017

Let time taken to finish the race by car $A \mathmr{and} B$ be $= {t}_{A} \mathmr{and} {t}_{B}$ respectively.

We need to assume that cars start from rest and have respective acceleration $= {a}_{1} \mathmr{and} {a}_{2}$.

Using the kinematic equation
$s = u t + \frac{1}{2} a {t}^{2}$ we get
Race track $s = \frac{1}{2} {a}_{1} {t}_{A}^{2}$
Also $s = \frac{1}{2} {a}_{2} {t}_{B}^{2}$

Equating the length of race track in case of both cars
${a}_{1} {t}_{A}^{2} = {a}_{2} {t}_{B}^{2}$ ......… (1)

Also given is
${t}_{B} = {t}_{A} + t$ ........... (2)

Using the kinematic equation
$v = u + a t$ we get velocity at the end of race for
car $A = {a}_{1} {t}_{A} \mathmr{and}$ for car $B = {a}_{2} {t}_{B}$

Given is
${a}_{1} {t}_{A} = {a}_{2} {t}_{B} + v$ .....(3)

From this equation we need to eliminate ${t}_{A} \mathmr{and} {t}_{B}$ to get the desired expression. (3) can be rewritten as
$v = {a}_{1} {t}_{A} - {a}_{2} {t}_{B}$

Substituting value of ${a}_{1}$ from (1)
$v = \frac{{a}_{2} {t}_{B}^{2}}{t} _ A - {a}_{2} {t}_{B}$
=>v= a_2(t_B/t_A)(t_B –t_A)#

Using (2) we get
$v = {a}_{2} t \left({t}_{B} / {t}_{A}\right)$

Using (1) we get
$v = {a}_{2} t \sqrt{{a}_{1} / {a}_{2}}$

Squaring both sides we get
${v}^{2} = {t}^{2} {a}_{1} {a}_{2}$