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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 cars have the same velocity?
A
$\frac{a + f}{2(1 + b)}$
B
$\frac{f - a}{2(1 + b)}$
C
$\frac{a - f}{1 + b}$
D
$\frac{a + f}{2(b - 1)}$
Explanation
Differentiate both positions and equate.
Detailed Solution
$v_P = a + 2bt$ and $v_Q = f - 2t$
As $v_P = v_Q$: $a + 2bt = f - 2t$
$\Rightarrow t = \frac{f - a}{2(1 + b)}$
As $v_P = v_Q$: $a + 2bt = f - 2t$
$\Rightarrow t = \frac{f - a}{2(1 + b)}$