The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Rate of Travel

Date: 05/15/2003 at 09:04:48
From: Wendy
Subject: Rate of travel 

A passenger on a train traveling at 135 km/h walks toward the back of 
the train at a rate of 7 km/h. What is the passenger's rate of travel 
with respect to the ground?

What does "with respect to the ground" mean? I think it is a trick 
question because the person is walking 7 km/h no matter what the 
speed of the train.

Date: 05/15/2003 at 14:38:20
From: Doctor Ian
Subject: Re: Rate of travel 

Hi Wendy,

Suppose there is a flatcar being pulled along at 5 mph, with two 
guys riding on it:

      \A/           \B/
       |             |
      / \           / \
    ---------------------   -- 5 mph -->
     o                 o

There is another guy standing by the tracks watching:

      \A/           \B/
       |             |
      / \           / \
    ---------------------   -- 5 mph -->
     o                 o

                    / \

Now, 5 miles per hour is the same as 88 feet per second, which is the
same as 22 feet every 15 seconds. So if we wait 15 seconds, the
situation looks like this:

               \A/           \B/
                |             |
               / \           / \
             ---------------------   -- 5 mph -->
              o                 o

                    / \

That is, both guys on the train have moved 22 feet as seen by the 
guy on the ground. However, suppose that as the train moves, guy 
B starts walking toward guy A at a speed of 5 mph. Now, if we wait 
15 seconds, the situation looks like this:

               \A/  \B/
                |    | 
               / \  / \
             ---------------------   -- 5 mph -->
              o                 o

                    / \

That is, from the point of view of guy A, guy B has been moving 
at 5 miles per hour. But from the point of view of guy C, guy B 
hasn't moved at all. He's still in the same place he was before, even 
though the _train_ has moved. 

Does this make sense?  

The language we use to describe this situation is that guy B has 
moved at 5 miles per hour 'with respect to' guy A (i.e., as seen 
from guy A's point of view), and that guy B has moved at 0 miles 
per hour 'with respect to' guy C.  

Since guy C is stationary on the ground, we also say that guy B 
has moved 0 mph with respect to the ground. And since guy A is 
stationary on the train, we say that guy A has moved 0 mph with 
respect to the train, but 5 mph with respect to the ground.  

In many (perhaps most) cases, we leave the observer unspecified. If a
policeman gives you a ticket for driving 80 miles per hour, he means
'with respect to the ground', even though he won't specify that. (If
he's right behind you at 80 mph when he turns his siren on, you're not
moving at all with respect to him!)

Now, suppose we go back to our train, but this time, instead of
walking at 5 mph, guy B walks toward guy A at 3 mph with respect 
to guy A. What does guy C see?  

Well, imagine that there is a fence blocking his view of the train, so
all he can see is the heads of guys A and B. What guy C will see 
is guy A moving to the right at 5 mph, and guy B moving to the 
right at 2 mph (i.e., 5 mph minus 3 mph). 
The point of all this is that whenever you say that something is
moving at a certain speed, it implies that the speed is a speed that
would be measured BY SOME OBSERVER. This is necessary, because as we
see, different observers will end up making different measurements. 

Note that by thinking in terms of relative speeds, certain kinds of
problems can be made much easier. For example, suppose two cars start 
a race at the same time, around a track that is 1 1/2 miles long, and 
one moves at 151 mph while the other moves at 154 mph. How long will 
it take the faster car to lap the slower one?  

If you try to solve this using their speeds relative to the ground,
it's a mess. But you can reason this way: the faster car is moving at 
3 mph WITH RESPECT TO the slower car. This is true whether their 
speeds with respect to the ground are 151 and 154 mph, or 18 and 15
mph, or 46.3 and 49.3 mph.  

And to lap the slower car, the faster car needs to get ahead by the
length of the track. In other words, we want to know how long it takes 
to go 1 1/2 miles at 3 mph. And that's an easy problem to solve: it 
takes half an hour, or 30 minutes. 

Does this help? 

- Doctor Ian, The Math Forum 
Associated Topics:
Middle School Word Problems

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.