It is noon. All three hands of your watch are in the 12 o\'clock position. After

Question

It is noon. All three hands of your watch are in the 12 o'clock position. After how many seconds will the second hand for the first time be again in the same position as the minute hand?  
(Note that unlike with the previous "catching up" problems for linear motion, the total "distance" moved in radians will never be the same. What is the requirement for the difference in angular displacements, such that they end up in the same position on the dial of the watch?)

This is my work so far:

2pi/60 * (t-60) = 2pi/3600 + t

I know I need to solve for t but I cannot figure out the algebraic steps involved to get t=61 from this. Please help and please show your work in detail so I can better understand. Thank you

Explanation / Answer

For the second hand, ws = (?/30) rad/sec
For the minute hand, wm = (?/1800) rad/sec

The angle for the second hand to catch the minute hand:

(?/1800)*t = (?/30)*t - 2?

Solving for t,

t = 61.017 sec

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