In a car race on straight road, car A takes a time t less than car...

Given,
Initially both car is at rest, so ${u}_{1} = {u}_{2} = 0$
Acceleration of car $A \text{ and } B$ is ${a}_{1} \text{ and } {a}_{2}$

Let us assume,
Time of reach to destination of car $A$ is, ${t}_{1} = {t}_{0}$
Time of reach to destination of car $B$ is, ${t}_{2} = {t}_{0} + t$

Using second equation of motion, we have
${u}_{1}t + \dfrac{1}{2}{a}_{1}{t}_{1}^{2} = {u}_{2}t + \dfrac{1}{2}{a}_{2}{{t}_{2}}^{2}$
$⇒ 0 \times t + \dfrac{1}{2}{a}_{1}{t}_{0}^{2} = 0 \times t + \dfrac{1}{2}{a}_{2}{\left({t}_{0} + t\right)}^{2}$
$⇒ \sqrt{\dfrac{{a}_{1}}{{a}_{2}}}{t}_{0} = {t}_{0} + t$
$⇒ {t}_{0} = \dfrac{t}{\sqrt{\dfrac{{a}_{1}}{{a}_{2}}} - 1}$

From first equation of motion, we have
$\begin{cases}{v}_{1} = {a}_{1}{t}_{0} \text{ and } & {v}_{2} = {a}_{2}\left({t}_{0} + t\right)\end{cases}$
$⇒ v = {v}_{1} - {v}_{2} = \left({a}_{1} - {a}_{2}\right){t}_{0} - {a}_{2}t$
$⇒ v = \left({a}_{1} - {a}_{2}\right)\dfrac{t}{\sqrt{\dfrac{{a}_{1}}{{a}_{2}}} - 1} - {a}_{2}t$
$⇒ v = t\left(\dfrac{{a}_{1}\sqrt{{a}_{2}}}{{\sqrt{a}}_{1} - {\sqrt{a}}_{2}} - \dfrac{{a}_{2}{\sqrt{a}}_{2}}{{\sqrt{a}}_{1} - {\sqrt{a}}_{2}} - {a}_{2}\right)$
$⇒ v = t\left(\dfrac{{a}_{1}\sqrt{{a}_{2}} - {a}_{2}{\sqrt{a}}_{1}}{{\sqrt{a}}_{1} - {\sqrt{a}}_{2}}\right)$
$⇒ v = \sqrt{{a}_{1} . {a}_{2}}t$