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 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:

Hint:

In such problems, consider the velocity-time relationship for both cars and solve for when their positions are equal to find the crossing points.

Answer 1

Correct Answer: 3

Let the acceleration of car P be \( a_P = k t \), which increases linearly with time. Let the acceleration of car Q be constant \( a_Q = a \). The velocity of car P at time \( t \) is given by: \[ v_P = \int a_P \, dt = \int k t \, dt = \frac{k t^2}{2} \] The velocity of car Q is given by: \[ v_Q = \int a_Q \, dt = a t \] At \( t = 0 \), both cars are at the same position. As time progresses, the velocities of both cars change, and they may cross each other again. Case I:
- \( v_P \) increases quadratically, while \( v_Q \) increases linearly.
- The cars will cross twice, once when \( t = 0 \) and again when the positions of the cars match at a later time.
Case II:
- For a constant acceleration \( a_Q = a \), car P crosses car Q when the difference in their velocities leads to the second crossing.
Thus, the total number of crossings is 3, including the crossing at \( t = 0 \).