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(B) The position of car $P$ at any time $t$ is given by $x_P(t) = at + bt^2$.The velocity of car $P$ is $v_P(t) = \frac{dx_P(t)}{dt} = a + 2bt$ ... $(

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

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(B) The position of car $P$ at any time $t$ is given by $x_P(t) = at + bt^2$.The velocity of car $P$ is $v_P(t) = \frac{dx_P(t)}{dt} = a + 2bt$ ... $(i)$.Similarly,for car $Q$,the position is $x_Q(t) = ft - t^2$.The velocity of car $Q$ is $v_Q(t) = \frac{dx_Q(t)}{dt} = f - 2t$ ... $(ii)$.Given that the cars have the same velocity,$v_P(t) = v_Q(t)$.Therefore,$a + 2bt = f - 2t$.Rearranging the terms to solve for $t$: $2bt + 2t = f - a$.$2t(b + 1) = f - a$.Thus,$t = \frac{f - a}{2(1 + b)}$.

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?

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