Upstream and Downstream Speed

Date: 08/16/98 at 22:12:55
From: Maggy 
Subject: Motion Problems

It takes a ferryboat 2 hours to travel 30 miles going downstream and 3 
hours for the return trip upstream. What is the ferryboat's speed in 
still water?

I know that rate * time = distance. How do you figure the current 
speed or still speed?

In my book it gives another example in the form of a chart, but once 
you have the chart and equations it doesn't show you how to solve your 
equations. It just shows the answer. I've tried to figure out the 
example and I don't see how they get their answer.

--------------------------------
Downstream     m+r     2     30
Upstream       m-r     3     30


2(m + r) = 30
3(m - r) = 30

So do you multiply 2*m and 2*r and 2*30 and do the same for 3? Or do 
you take 2m = 60 and then divide and come up with m = 30 as your 
answer?

I'm probably wrong altogether so any help you can give me will be 
great.

Maggy

Date: 08/17/98 at 16:54:50
From: Doctor Peterson
Subject: Re: Motion Problems

Hi, Maggy! Your equations look just fine. How to solve them depends 
partly on what you have learned. They are a pair of "simultaneous 
equations," for which there are standard methods you can use. I'll 
start you out with the method of substitution, which is probably 
easiest to follow in case you haven't met simultaneous equations 
before.

You have:

    2(m + r) = 30
    3(m - r) = 30

where m is the rate of the boat relative to the water, and r is the 
rate of the river. (What does m stand for, by the way? It doesn't 
matter, but I'm curious.)

The first thing to do is to simplify both equations, by multiplying 
out using the distributive rule:

    2m + 2r = 30
    3m - 3r = 30

Then we can simplify further by dividing both sides of the first 
equation by 2 and of the second equation by 3. (We could have done this 
all in one step, but I didn't happen to see that at first.)

    m + r = 15
    m - r = 10

Now you're supposed to solve for m, so we might as well eliminate r. 
We can do that by solving for r in terms of m in either equation. 
I'll pick the first:

    r = 15 - m

Then we can substitute this into the other equation:

    m - (15 - m) = 10

This gives you one equation in one unknown. See if you can solve this, 
then check to make sure it works out right. You might also want to see 
how fast the river is flowing, as an extra check. And how long would 
it take the boat to go 30 miles in still water?

If this method of substitution is new to you, it may feel a little 
strange at first. First you think of m as if it were some known 
quantity, so you can solve the first equation for the unknown r. Then, 
once r becomes "known" (in terms of m), you plug it into the other 
equation, and suddenly "remember" that you really don't know m yet 
after all, but now you have an equation that you can solve for m. It 
stretches your brain a little, doesn't it?

By the way, if you're not quite comfortable with how you got the 
original equations, there's a nice way to visualize the concept of 
relative speed. Have you ever been on one of those moving walkways in 
an airport that work like a horizontal escalator, carrying you down a 
long hallway but allowing you to walk forward as well? A boat on a 
flowing river, or an airplane flying through a wind, behaves the same 
way. If I "stand still" on the walkway, I find myself moving forward. 
If I walk forward along the walkway, my actual speed will be the sum 
of my normal walking speed and the speed of the walkway. If I were to 
walk on a walkway going in the wrong direction, at more than the 
walkway's speed, I would eventually get where I was going, but at a 
speed that much slower than my walking speed. If I walked slower than 
the walkway, I would find myself moving with a negative velocity. And 
if I walked at exactly the speed of the walkway, I would never go 
anywhere. I find it easier to picture this on a solid walkway than in 
a liquid or a gas.

If you have learned about simultaneous equations but are having 
trouble with them, search the Dr. Math archives for that subject and 
you'll find plenty of examples and explanations. If you haven't 
learned about them yet, I wouldn't worry too much about this problem 
for now. You'll get there.

- Doctor Peterson, The Math Forum
Check out our web site! http://mathforum.org/dr.math/