Two cars P and Q start from a point at the same time in a straight line and their positions are represented by

Question

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?

Accepted Answer

Velocity is defined as the rate of change of position with respect to time ($v = \frac{dx}{dt}$) .

1.  **Find the velocity of car P:**
    $$ x_P(t) = at + bt^2 $$
    $$ v_P = \frac{dx_P}{dt} = \frac{d}{dt}(at + bt^2) = a + 2bt $$

2.  **Find the velocity of car Q:**
    $$ x_Q(t) = ft - t^2 $$
    $$ v_Q = \frac{dx_Q}{dt} = \frac{d}{dt}(ft - t^2) = f - 2t $$

3.  **Equate velocities:**
    Given that the cars have the same velocity:
    $$ v_P = v_Q $$
    $$ a + 2bt = f - 2t $$

4.  **Solve for time (t):**
    Rearrange terms containing $t$ to one side:
    $$ 2bt + 2t = f - a $$
    $$ 2t(b + 1) = f - a $$
    $$ t = \frac{f - a}{2(1 + b)} $$

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