The positions of two cars A and B are given b | vedclass

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(B) Given the position functions for cars $A$ and $B$:$X_A(t) = at + bt^2$$X_B(t) = ft - t^2$The velocity of a particle is the time derivative of its

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

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(B) Given the position functions for cars $A$ and $B$:$X_A(t) = at + bt^2$$X_B(t) = ft - t^2$The velocity of a particle is the time derivative of its position function,$v = \frac{dX}{dt}$.For car $A$:$v_A = \frac{d}{dt}(at + bt^2) = a + 2bt$For car $B$:$v_B = \frac{d}{dt}(ft - t^2) = f - 2t$To find the time when both cars have the same velocity,we set $v_A = v_B$:$a + 2bt = f - 2t$Rearranging the terms to solve for $t$:$2bt + 2t = f - a$$t(2b + 2) = f - a$$t(2(b + 1)) = f - a$$t = \frac{f - a}{2(1 + b)}$Thus,the cars will have the same velocity at time $t = \frac{f - a}{2(1 + b)}$.

The positions of two cars $A$ and $B$ are given by $X_A = at + bt^2$ and $X_B = ft - t^2$. At what time will both cars have the same velocity?

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