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 ________.

Solution

Physics sir · 4 steps

Write the gap between the cars as a function of time. It is a cubic, and a cubic has at most three zeros.

1. Relative acceleration. Car P has a_P = a₀ + kt and car Q has a constant a_Q. Their relative acceleration a_P - a_Q = (a₀ - a_Q) + kt is linear in t.
2. Integrate twice. Integrating once gives a quadratic relative velocity, and integrating again gives a cubic relative displacement. The cars are side by side at t = 0, so the constant term is zero.

   x_P - x_Q = k/6t³ + (a₀ - a_Q)/2t² + u t = t(k/6t² + (a₀ - a_Q)/2t + u)

3. Count the roots. A crossing happens whenever x_P - x_Q = 0. One root is t = 0, and the quadratic in the bracket can give two more. With suitable values of u, a₀ - a_Q and k, both of those roots are real and positive, so both are real later crossings.
4. So. A cubic cannot have more than 3 zeros, so the cars can cross at most 3 times, counting the crossing at t = 0.

The common slip: 2. Leaving out the crossing at t = 0 gives 2. The question asks you to include it, so the answer is 3.

The idea to keep. To count crossings, write the separation as a polynomial in time. Its degree caps the number of meetings: a linear relative acceleration gives a cubic, so at most 3 meetings.

Answer from the NTA final answer key. Solution by PrepWiser.