Question

In a car race on straight road, car $\text{A}$ takes time $t$ less than car $\text{B}$ at the finish and passes finishing point with a speed $v$ more than that of car $\text{B}$. Both the cars start from rest and travel with constant acceleration $\alpha$ and $\beta$ respectively. Then $v$ is equal

• A. $\left(\sqrt{2\alpha \beta }\right)t$
  
• B. $\left(\sqrt{\alpha \beta }\right)t$
  
• C. $\left(\frac{\alpha +\beta }{2}\right)t$
• D. $\left(\frac{2\alpha +\beta }{\alpha +\beta }\right)t$

Answer 1

The correct option is B $\left(\sqrt{\alpha \beta }\right)t$


Let, the velocity of car $\text{A}$ and $\text{B}$ are $v_A$ and $v_B$ respectively at the finish point.
So,
$v_A=v_B+v$

Let, time taken by car $\text{A}$ and $\text{B}$ are $t_A$ and $t_B$ respectively to finish the race. So,
$t_B=t_A+t$

For car $\text{A}:$
Initial speed, $u=0$
Acceleration, $a_A=\alpha$
Distance travelled,
$S=u{t}_A+\frac{1}{2}{a}_A{t_A}^2$

$=0+\frac{1}{2}\alpha (t_B-t{)}^2$

$(t_B-t)=\sqrt{\frac{2S}{\alpha }}....\left(1\right)$

For car $\text{B}:$

Initial speed, $u=0$
Acceleration, $a_B=\beta$
Distance travelled,
$S=u{t}_B+\frac{1}{2}{a}_B{t_B}^2$

$S=0+\frac{1}{2}\beta t{{}_B}^2$

$t_B=\sqrt{\frac{2S}{\beta }}....\left(2\right)$

Subtracting equation $\left(2\right)$ from $\left(1\right)$ gives,
$t=\sqrt{\frac{2S}{\beta }}-\sqrt{\frac{2S}{\alpha }}$

$\Rightarrow t=\sqrt{2S}[\frac{1}{\sqrt{\beta }}-\frac{1}{\sqrt{\alpha }}]....\left(3\right)$

Now using,

$v{{}_A}^2-u^2=2\alpha S$
$\Rightarrow v_A=\sqrt{2\alpha S}....\left(4\right)$


$v{{}_B}^2-u^2=2S\beta$
$\Rightarrow v_B=\sqrt{2\beta S}....\left(5\right)$

From $\left(4\right)$ and $\left(5\right)$,

$v_A-v_B=\sqrt{2\alpha S}-\sqrt{2\beta S}$
$\Rightarrow v_B+v-v_B=\sqrt{2S}[\sqrt{\alpha }-\sqrt{\beta }]$
$\Rightarrow v=\sqrt{2S}[\sqrt{\alpha }-\sqrt{\beta }]....\left(6\right)$

From equation $\left(3\right)$ and $\left(6\right)$,
$\frac{v}{t}=\frac{\sqrt{2S}[\sqrt{\alpha }-\sqrt{\beta }]}{\sqrt{2S}[\frac{1}{\sqrt{\beta }}-\frac{1}{\sqrt{\alpha }}]}$

$\Rightarrow v=\frac{(\sqrt{\alpha }-\sqrt{\beta })\sqrt{\alpha \beta }}{(\sqrt{\alpha }-\sqrt{\beta })}t$

$\Rightarrow v=\left(\sqrt{\alpha \beta }\right)t$

Hence, option (b) is the correct answer.