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Two cars P and Q start from a point at the same time in a straight line and their positions are represented by $x_{P} (t) = (at + {bt}^{2})$ and $x_{Q} (t) = (ft - t^{2})$ . At what time do the car A has double velocity of B ?

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$\frac{a + f}{2 (b - 1)}$

$\frac{2 f - a}{2 b + 4}$

$\frac{f - a}{2 (1 + b)}$

$\frac{a - f}{1 + b}$

answer is C.

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Detailed Solution

$\begin{array}{l} V_{P} = 2 V_{Q} \\ a + 2 bt = 2 [f - 2 t] \\ 2 bt + 4 t = 2 f - a \\ t = \frac{2 f - a}{2 b + 4} \end{array}$

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