Two buses $$P$$ and $$Q$$ start from a point at the same time and move in a straight line and their positions are represented by $$x_P(t) = \alpha t + \beta t^2$$ and $$x_Q(t) = ft - t^2$$. At what time, both the buses have same velocity?

We are given the positions of two buses:

$$x_P(t) = \alpha t + \beta t^2$$

$$x_Q(t) = ft - t^2$$

The velocity of each bus is the derivative of its position with respect to time.

Velocity of bus P:

$$v_P = \frac{dx_P}{dt} = \alpha + 2\beta t$$

Velocity of bus Q:

$$v_Q = \frac{dx_Q}{dt} = f - 2t$$

For same velocity, we set $$v_P = v_Q$$:

$$\alpha + 2\beta t = f - 2t$$

$$2\beta t + 2t = f - \alpha$$

$$t(2\beta + 2) = f - \alpha$$

$$t = \frac{f - \alpha}{2(\beta + 1)} = \frac{f - \alpha}{2(1 + \beta)}$$

Hence, the correct answer is Option D.