At exactly six o'clock, a spider started to walk at a constant speed from the hour hand anticlockwise round the edge of the clock face. When it reached the minute hand, it turned round (assume the turn was instantaneous) and walked in the opposite direction at the same constant speed. It reached the minute hand again after 20 minutes. At what time was this second meeting?

On the second leg of the spider's trip, its speed relative to the minute hand is 1/20 rpm and the minute hand itself, in "absolute" terms is moving at 1/60 rpm. We want the absolute speed (i.e., relative to the clock face) of the spider, s. So

s - 1/60 = 1/20 rpm

s = 1/20 + 1/60 = 4/60 = 1/15 rpm

On the first leg of the spider's trip, its speed in rpm was added to that of the minute hand (whose speed is 1/60 rpm).

1/15 + 1/60 = 5/60 = 1/12 rpm

It has to make 1/2 revolution relative to the minute hand, so, using time = distance / rate:

t = (1/2) rev / (1/12) rpm = 6 min

So it took 6 minutes to reach the minute hand the first time, and then 20 minutes later it met the second time, for a total of 26 minutes.  It was 6:26.

Posted by Charlie on 2006-05-05 09:27:30