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A particle P is located at A. Another particle Q is located at B 200 m south of A. P starts moving at 2 m/s towards east at some moment and simultaneously Q starts moving at 4 m/s. (i) If Q moved first towards A and on reaching A quickly changed direction and moved towards east, find the time it takes to catch up with P. (ii) If instead Q aimed at a point C east of A at which it arrives at the same time as P, find the elapsed time. (iii) What should be the minimum speed of Q to catch up with P before it moves 50 m from its initial position? (Assume Q catches up by aiming at the position P would reach at the same time as it does)
Correct Answer: 0
PLACEHOLDER ANSWER - the question has three answers. (i) Q takes 50 s to reach A, when P is 100 m ahead; closing at $$4 - 2 = 2\text{ m/s}$$ takes another 50 s, so 100 s in all. (ii) If they meet after $$t$$ seconds, $$\sqrt{200^2 + (2t)^2} = 4t$$, so $$12t^2 = 40000$$ and $$t = \frac{100}{\sqrt{3}} \approx 57.7\text{ s}$$. (iii) P covers 50 m in 25 s, so Q must cover $$\sqrt{200^2 + 50^2} = 50\sqrt{17} \approx 206.2\text{ m}$$ in 25 s, a speed of at least $$2\sqrt{17} \approx 8.25\text{ m/s}$$.
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