Clock Hands Diametrically OppositeDate: 10/10/2001 at 20:05:42 From: Jonathan Thomas Subject: A Problem of the week At what time between two and three o'clock are the hands of a clock diametrically opposite? I have tried drawing a clock out to actually see when this will occur, but it is very messy and hard to follow. I was wondering what would be an easier way to do this? Date: 10/12/2001 at 13:37:06 From: Doctor Ian Subject: Re: A Problem of the week Hi Jonathan, The minute hand moves 360 degrees in one hour, right? So that's 360 degrees in 3600 seconds, or 1/10 of a degree per second. The hour hand moves 360 degrees in 12 hours, or 30 degrees in an hour. That's 30 degrees in 3600 seconds, or 1/120 of a degree per second. At 2 o'clock, the minute hand is straight up (zero degrees), and the hour hand is at the 2 (one 6th of the way around, or 60 degrees). Suppose that some number of seconds - let's call it T - passes. Where is the minute hand? It's at T/10 degrees. Where is the hour hand? It's at (60 + T/120) degrees. (Do you see why?) So after T seconds, the angle between them is angle = T/10 - (60 + T/120) Let's do some simple checks. After half an hour (1800 seconds), angle = 1800/10 - (60 + 1800/120) = 180 - (60 + 15) = 115 degrees Does that make sense? Sure, the hour hand should have moved halfway from the 2 to the 3, a distance of 15 degrees. After an hour (3600 seconds), angle = 3600/10 - (60 + 3600/120) = 360 - (60 + 30) = 270 degrees which also makes sense. The minute hand is straight up again, and the hour hand is 270 degrees behind it (or 90 degrees in front of it, depending on how you look at it). So the formula seems to be okay. Now, what we want is to know the value of T that gives us a difference of 180 degrees: 180 = T/10 - (60 + T/120) If you solve the equation for T, you'll know how many seconds after the hour it takes to move the two hands to opposite directions. Can you take it from here? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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