Solve Dis Riddles

What time do ends of wall clock hands approach each other the fastest?

Assume the hour hand is shorter than the minute hand.  In the rotating reference frame in which the minute hand is stationary, the hour hand’s tip is moving at a constant speed, so the distance between the tips decreases most quickly when the hour hand’s tip is moving directly towards the minute hand’s tip—that is, when the minute and hour hands are the hypotenuse and adjacent leg of a right triangle.

That happens when the angle θ$\theta$ between the hands satisfies cosθ=yx$\cos \theta =\frac{y}{x}$.  θ$\theta$ starts at 0$0$ and increases at a constant rate vx−vy$v_x-v_y$, so this angle is achieved at the times t=θvx−vy=1vx−vy(2πn−cos−1yx)$t=\frac{\theta }{v_x-v_y}=\frac{1}{v_x-v_y}(2\pi n-{\cos }^{-1}\frac{y}{x})$.

(At t=1vx−vy(2πn+cos−1yx)$t=\frac{1}{v_x-v_y}(2\pi n+{\cos }^{-1}\frac{y}{x})$, the right triangle is flipped and the distance is increasing most quickle151

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