Two cars P and Q are moving on a road in the same direction. Acceleration of car P increases linearly with time whereas car Q moves with a constant acceleration. 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 ________ .

We analyze the motion of two cars, P and Q, and determine the number of times they can cross each other.

Step 1: Define Motion Equations

Car P: Acceleration increases linearly with time, i.e.,

$$a_P = k t$$

where $k$ is a constant.

Since acceleration is the derivative of velocity, integrating:

$$v_P = \int k t \, dt = \frac{k t^2}{2} + v_{P0}$$

Again integrating to get position:

$$x_P = \int \left( \frac{k t^2}{2} + v_{P0} \right) dt$$ $$x_P = \frac{k t^3}{6} + v_{P0} t + x_{P0}$$

Car Q: Moves with constant acceleration $a_Q$.

The velocity equation is:

$$v_Q = a_Q t + v_{Q0}$$

The position equation is:

$$x_Q = \int (a_Q t + v_{Q0}) dt$$ $$x_Q = \frac{a_Q t^2}{2} + v_{Q0} t + x_{Q0}$$

Step 2: Find the Intersection Points (Crossings)

To find when the two cars cross each other, we set their positions equal:

$$\frac{k t^3}{6} + v_{P0} t + x_{P0} = \frac{a_Q t^2}{2} + v_{Q0} t + x_{Q0}$$

Since both cars cross at $t = 0$, assume $x_{P0} = x_{Q0} = 0$, giving:

$$\frac{k t^3}{6} + v_{P0} t = \frac{a_Q t^2}{2} + v_{Q0} t$$

Rearrange:

$$\frac{k t^3}{6} - \frac{a_Q t^2}{2} + (v_{P0} - v_{Q0}) t = 0$$

Factorizing:

$$t \left( \frac{k t^2}{6} - \frac{a_Q t}{2} + (v_{P0} - v_{Q0}) \right) = 0$$

One root is $t = 0$ (given). The remaining equation:

$$\frac{k t^2}{6} - \frac{a_Q t}{2} + (v_{P0} - v_{Q0}) = 0$$

is a quadratic equation in $t$:

$$\frac{k}{6} t^2 - \frac{a_Q}{2} t + (v_{P0} - v_{Q0}) = 0$$

A quadratic equation can have two real roots at most (including $t = 0$, three total crossings are possible).

Step 3: Conclusion

The maximum possible number of crossings is $$\mathbf{3}$$