Two cars P and Q are moving on a road in the same direction. Accleration of car P increases linearly with
time whereas car Q moves with a constant accleration. Both cars cross each other at time t=0, for the first
time. The maximum possible number of crossing(s) (including the crossing at t=0) is ________.

Two cars P and Q move on a road in the same direction. Car P has an acceleration that increases linearly with time, and car Q has constant acceleration. They cross each other at $$t = 0$$ for the first time, and we need to find the maximum possible number of crossings.

We begin by setting up the position equations. Let both cars be at the same position at $$t = 0$$ and define the relative position $$\Delta x = x_P - x_Q$$.

For car P the acceleration is $$a_P = \alpha t$$ (linearly increasing, where $$\alpha > 0$$ is a constant). Integrating gives the velocity $$v_P = v_{P0} + \frac{\alpha t^2}{2}$$ and the position $$x_P = x_0 + v_{P0}t + \frac{\alpha t^3}{6}$$.

Similarly, for car Q the constant acceleration is $$a_Q = \beta$$. Integrating gives the velocity $$v_Q = v_{Q0} + \beta t$$ and the position $$x_Q = x_0 + v_{Q0}t + \frac{\beta t^2}{2}$$.

Subtracting these positions, the relative position becomes

$$\Delta x = x_P - x_Q = (v_{P0} - v_{Q0})t + \frac{\alpha t^3}{6} - \frac{\beta t^2}{2}$$

Letting $$c = v_{P0} - v_{Q0}$$ (the initial relative velocity) simplifies this to

$$\Delta x = ct + \frac{\alpha t^3}{6} - \frac{\beta t^2}{2}$$.

Crossings occur when $$\Delta x = 0$$, so we set

$$t\left(c + \frac{\alpha t^2}{6} - \frac{\beta t}{2}\right) = 0$$

This equation yields the trivial root $$t = 0$$ (the first crossing) and the quadratic

$$\frac{\alpha}{6}t^2 - \frac{\beta}{2}t + c = 0$$.

This quadratic can have at most two real roots. In order for both of these roots to be positive (since $$t > 0$$), three conditions must be satisfied:

- Discriminant $$\left(\frac{\beta}{2}\right)^2 - 4 \cdot \frac{\alpha}{6} \cdot c \geq 0$$, i.e., $$\frac{\beta^2}{4} \geq \frac{2\alpha c}{3}$$

- Sum of roots $$\frac{\beta/2}{\alpha/6} = \frac{3\beta}{\alpha} > 0$$ (true since $$\alpha, \beta > 0$$)

- Product of roots $$\frac{c}{\alpha/6} = \frac{6c}{\alpha} > 0$$, which requires $$c > 0$$

When these conditions are met for appropriate values of $$\alpha$$, $$\beta$$, and $$c$$, the quadratic has two positive real roots, corresponding to two additional crossings beyond the one at $$t = 0$$.

To see that no more crossings are possible, note that $$\Delta x / t$$ is a quadratic function, so after factoring out the root at $$t = 0$$ the remaining quadratic factor can have at most two positive zeros. Therefore, the total number of crossings cannot exceed three.

In conclusion, the maximum possible number of crossings (including the one at $$t = 0$$) is $$\boxed{3}$$.