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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 + bt² and x_q(t) = ft – t². At what time do the cars have the same velocity?

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

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

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

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

answer is D.

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Detailed Solution

$\begin{array}{l} x_{p} = at + {bt}^{2}; x_{q} = ft - t^{2} \\ \frac{{dx}_{p}}{dt} = a + 2 bt; \frac{{dx}_{q}}{dt} = f - 2 t \end{array}$

when, $v_{p} = v_{q}$

$a + 2 bt = f - 2 t \Rightarrow t = \frac{f - a}{2 (b + 1)}$

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