Question

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 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 ________.

Answer 1

3

Answer 2

Solution Explanation:
Let the displacements of cars P and Q (starting from the meeting point at t = 0) be:

$$s_P = v_{P0}\,t + \frac{A}{6}t^3,\quad s_Q = v_{Q0}\,t + \frac{1}{2}a\,t^2,$$

where car P’s acceleration is $a_P = At$ and car Q’s acceleration is constant $a$. Setting the displacement difference to zero for crossing,

$$s_P - s_Q = (v_{P0} - v_{Q0})t - \frac{1}{2}at^2 + \frac{A}{6}t^3 = 0.$$

Factor $t$ out:

$$t\left[(v_{P0} - v_{Q0}) - \frac{1}{2}at + \frac{A}{6}t^2\right] = 0.$$

This indicates one root at $t = 0$ (the first crossing) and a quadratic in $t$:

$$\frac{A}{6}t^2 - \frac{1}{2}at + (v_{P0} - v_{Q0}) = 0.$$

By suitably choosing the initial velocity difference $(v_{P0} - v_{Q0})$ and the constants $A$ and $a$, it is possible for the quadratic to have two distinct positive real roots. Thus, including $t=0$, the maximum number of distinct crossings is 3.