Average Speed of a Caterpillar

Date: 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.