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

Car P's acceleration is \( a_P = k t \), increasing linearly with time. Car Q has constant acceleration \( a_Q = a \).Car P's velocity at time \( t \) is calculated as:\[v_P = \int a_P \, dt = \int k t \, dt = \frac{k t^2}{2}\]Car Q's velocity is given by:\[v_Q = \int a_Q \, dt = a t\]Both cars start at the same position at \( t = 0 \). As time passes, their velocities change, potentially leading to further crossings.Scenario I:
- \( v_P \) exhibits quadratic growth, whereas \( v_Q \) grows linearly.
- The cars will cross twice: once at \( t = 0 \), and again when their positions are equal at a later time.
Scenario II:
- With \( a_Q = a \) being constant, car P crosses car Q when the difference in their velocities results in a second crossing.
Therefore, including the initial crossing at \( t = 0 \), there are a total of 3 crossings.