Consider an analog clock whose minute and hour hand move smoothly and constantly
ID: 2865629 • Letter: C
Question
Consider an analog clock whose minute and hour hand move smoothly and constantly. The minute then moves one revolution every hour and the hour hand moves one revolution every 12 hours. Assume that "now" the minute hand is pointing at 12 and the hour hand is pointing at one. When the minute hand reaches 1, the hour hand has moved to 1 + 1/12. When the minute hand reaches 1 + 1/12, the hour hand has moved to 1 + 1/12 + 1/12^2, etc... Use the sum of a geometric sequence to determine the exact time at which the minute hand and the hour hand are in the same location. Do not round off - provide the precise time they are together.Explanation / Answer
by the time the minute hand gets to five past one the hour hand will have moved slightly past 1 o'clock. The hour hand after all does not wait at 1 o'clock for an hour but rather it moves to 2 o'clock over the course of the next hour. The first overlap will be a little after five past one.
The first approach is probably the most obvious; it's the most mechanically intuitive. We'll use a variable 't' the time in hours. We know the speeds of the hands by definition the Minute Hand will move one full rotation or 360° per hour. The Hour Hand makes a full rotation in 12 hours and will therefore move at 30° per hour. At our first overlap just after five past one the Minute Hand will have done one full rotation plus the bit we are interested in. The Hour Hand will have done just a part rotation of 't' times it's speed.