Average Speed of a CaterpillarDate: 09/08/99 at 06:24:33 From: Tabrez Subject: Average Speed A caterpillar crawls at 21 in/hr to a vegetable patch. After being overindulgent, he returns over exactly the same distance at 7 in/hr. What is the caterpillar's average speed over the entire journey? If you get an answer, please let me know the formula for it and how you got it. Is it an unsolvable question because it does not give the distance? What can we do with this question - any ideas or suggestions? Thanks in advance. Date: 09/08/99 at 13:05:32 From: Doctor Rick Subject: Re: Average Speed Hi, Tabrez. We do have enough information to solve this problem, even without knowing the distance. All we need to know is that the two distances (out and back) are the same. One trap in this sort of problem is to think that you can just average the two speeds (21 in/hr and 7 in/hr) to get 14 in/hr, because the distances are the same. But when we take average speeds, we need to take the average with respect to time, not distance. In other words, if the two legs of the trip took the same _time_, then we could average the speeds directly; but not if they cover the same _distance_. Here is what to do. Suppose we knew the distance of each leg - I'll call this distance D inches. How long does each leg take? On the way out, at 21 in/hr, we have distance = speed x time time = distance / speed T1 = (D/21) hours On the way back, at 7 in/hr, the time comes out to T2 = (D/7) hours The total time for the out-and-back trip is T = T1 + T2 = D/21 + D/7 The total distance covered in this time is 2 x D. The average speed for the entire trip is thus speed = distance / time = 2D / (D/21 + D/7) You can simplify this expression, and D will vanish - the average speed does not depend on the distance. Here is another way to think about it, if you aren't comfortable with the algebra. On the trip back, the caterpillar goes 1/3 as fast, so it takes 3 times as long to go the same distance. Remember I said before that we could average the speeds over _equal_ times? We can divide the time for the trip back into 3 equal parts; now the whole trip is divided into 4 equal time periods. We can thus average the speeds: 21 + 7 + 7 + 7 -------------- = ? 4 You'll get the same answer either way. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 09/09/99 at 07:26:44 From: Intl Agency Subject: Re: Average Speed Thanks a lot. |
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