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Exchanging Hands |
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Suppose a guest arrives for a visit sometime between 3pm and 4pm on one day, and departs several days later sometime between 7am and 8am, and suppose that at the instant of departure the positions of the hour hand and the minute hand on the clock are exactly transposed from their positions at the instant of arrival. What was the exact time of arrival? |
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Let H(t) and M(t) denote the angular positions of the hour and minute hand respectively at the time t, so we have |
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Letting t1 and t2 denote the arrival and departure times respectively (measured from some arbitrary reference time), we have H(t2) = M(t1) and M(t2) = H(t1). Since the pattern of the hand positions repeats every 12 hours, we can equivalently consider the case of arrival and departure within the same 12 hour period, with t initialized at the beginning of that period. For the case when t1 is between n and n+1, and t2 is between m and m+1, all measured in hours, the above conditions can we written as |
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Thus the basic angular speed ω cancels out, and we immediately have t2 – t1 = (12/13)(m−n). To determine the individual arrival and departure times, we solve the system of equations |
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This gives the result |
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In the stated problem we have n=3 and m=7, as depicted in the figure below. |
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Therefore, the arrival time is 3 and 87/143 hours, and the departure time is 7 and 43/143 hours. In terms of hours:min:sec notation these times are 03:36:30.21 and 07:18:02.52. The clocks at these two times appear as shown below. |
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