JEE PYQ: Motion in a Plane - Question ID 00437b0d3a73 (JEE Main 2019)
Select Option
Step-by-step Explanation
In problems involving relative motion in a plane, we treat the positions and velocities of the two moving objects as vectors. The key idea is:
- Express the position of Ship B relative to Ship A as a vector that changes with time.
- Express the velocity of Ship B relative to Ship A as a constant vector .
- The distance between the two ships at any time is the magnitude of . To find the time at which this distance is minimum, we minimize (which avoids dealing with square roots).
The fundamental formula is: where is the initial relative position vector and is the relative velocity vector.
Step-by-Step Derivation:1. Define the coordinate system and initial positions
- Let point east and point north.
- At , Ship A is at the origin: .
- Ship B is initially at km relative to Ship A.
2. Write the velocity vectors
- Ship A’s velocity: km/hr.
- Ship B’s velocity: km/hr (since it moves west).
- Relative velocity of B with respect to A:
3. Express the relative position vector at time
4. Compute the squared distance
5. Expand and simplify
6. Minimize by calculus
To find the minimum, set the derivative with respect to to zero: Solving gives7. Match with the given options
The closest option is hours, which corresponds to choice C. Common Traps & Exam Tip:1. Sign errors in relative velocity: Many students forget to subtract from or mix up the order. Always compute . 2. Minimizing instead of : Differentiating the square root is messier; minimizing the squared distance is algebraically cleaner. 3. Unit consistency: Ensure all distances are in km and velocities in km/hr so the time comes out in hours. 4. Approximation errors: Carry exact fractions (e.g., ) until the final step to avoid rounding mistakes.
Exam Tip: When the question asks for the time of minimum distance, always set up the relative position vector and minimize its squared magnitude. This method is robust and avoids trigonometric complications.
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