1. Let the accelerations be modeled mathematically

   Since acceleration of car $P$ increases linearly with time, $a_P(t)=\alpha t+\beta$ where $\alpha>0$.

   Car $Q$ has constant acceleration, $a_Q(t)=\gamma$ where $\gamma$ is constant.

2. Find velocities

   Integrating acceleration with respect to time,

   For car $P$: $v_P(t)=\int a_P(t)\,dt=\frac{\alpha}{2}t^2+\beta t+C_1$

   For car $Q$: $v_Q(t)=\int a_Q(t)\,dt=\gamma t+C_2$

3. Find positions

   Integrating again,

   For car $P$: $x_P(t)=\int v_P(t)\,dt=\frac{\alpha}{6}t^3+\frac{\beta}{2}t^2+C_1 t+C_3$

   For car $Q$: $x_Q(t)=\int v_Q(t)\,dt=\frac{\gamma}{2}t^2+C_2 t+C_4$

4. Use the condition that they cross at $t=0$

   At $t=0$, both cars are at the same position for the first time, so $x_P(0)=x_Q(0)$ which gives $C_3=C_4$

5. Form the relative position

   Let $\Delta x(t)=x_P(t)-x_Q(t)$

   Then, $\Delta x(t)=\frac{\alpha}{6}t^3+\left(\frac{\beta-\gamma}{2}\right)t^2+(C_1-C_2)t$

   Since $C_3=C_4$, the constant term is zero.

   So, $\Delta x(t)=t\left[\frac{\alpha}{6}t^2+\frac{\beta-\gamma}{2}t+(C_1-C_2)\right]$

6. Interpret crossings

   Crossings occur when $x_P=x_Q$, i.e. when $\Delta x(t)=0$

   This equation is a cubic polynomial in $t$, but because one root is already $t=0$, the remaining roots come from a quadratic equation.

   Therefore, the total number of possible real crossing times is at most:

   • one root at $t=0$,
   • plus at most two more real roots from the quadratic.

   Hence maximum possible number of crossings is $1+2=3$

7. Can 3 actually occur?

   Yes. A quadratic can have two distinct positive real roots for suitable choices of parameters, so along with the root $t=0$, total crossings can be $3$.

Therefore, the maximum possible number of crossings is: $\boxed{3}$