Two cars start in a race with velocities u₁ and u₂ and travel in a straight line with acceleration ‘a’ and b. If both reach the finish line at the same time, the range of the race is

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$\frac{2 (u_{1} - u_{2})}{(β - α)^{2}} (u_{1} β - u_{2} α)$

$\frac{2 {(u_{1} - u_{2})}^{2}}{(β - α)^{2}}$

$\frac{2 (u_{1} - u_{2})}{β + α} (u_{1} α - u_{2} β)$

$\frac{2 u_{1} u_{2}}{βα}$

answer is A.

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Detailed Solution

$s = u_{1} t + \frac{1}{2} {at}^{2}$ (for first car) ...... (1)
$s = u_{2} t + \frac{1}{2} {bt}^{2}$ (for second car) ....... (2)
(1) = (2)
$\begin{array}{l} 0 = (u_{1} - u_{2}) t + \frac{1}{2} (a - b) t^{2} \\ t = \frac{2 (u_{2} - u_{1})}{(b - a)} \end{array}$
substitute above equation in (1)
$s = \frac{2 u_{1} (u_{2} - u_{1})}{(b - a)} + \frac{1}{2} a {[\frac{(u_{2} - u_{1})}{(b - a)}]}^{2}$
simplifying we get $s = \frac{2 (u_{1} - u_{2}) (u_{1} b - u_{2} a)}{(b - a)^{2}}$
$a = α; b = β$