Swimming Upstream and Down

Date: 09/19/97 at 02:05:37
From: Sarah C. Brown
Subject: Algebra

A man swims 200 meters against the flow of a stream in 3 minutes 
(180 seconds). Then he returns downstream 150 meters with the flow in 
.7 minutes. How fast is the man swimming and how fast is the water 
traveling?

I believe the man is not swimming downstream at all, since the problem 
does not state it. I think he is just "going with the flow." Therefore 
I think the speed of the water could be found with:

 Velocity x Time = Distance or V= D/T. 

Is that right? Can this same formula be used for how fast he is 
swimming?

Date: 09/19/97 at 05:25:39
From: Doctor Mitteldorf
Subject: Re: Algebra

Dear Sarah,

If, as you say, he is "going with the flow" downstream then you can
use the formula you quoted to get the speed. .7 minutes is 42 seconds, 
so he goes 150 meters in 42 seconds, or 3.57 meters/sec.

One thing that's not explained in the problem is how the swimmer's 
speed and the water speed are related. When you swim, you push against
the water. You never touch the ground at all. 

That means that your swimming speed depends on the water speed in an 
easy way: if you swim with the water, your speed is just swim speed + 
speed of water current.

If you swim against the flow, then your speed is just swim speed - 
speed of current.

If he goes upstream 200 meters in 180 seconds, his speed is 
1.11 meters/sec. But the current is still flowing at 3.57 meters/sec.  
So he's going 4.68 meters/sec.

                  -----------------------------------

Now, as you suggest, the problem isn't really clear about whether he's
swimming or just coasting on the way back. If he's swimming, then it's
a different story. Then we don't know either his speed or the water 
speed. But we do know the sum of his speed and the water speed is 
3.57, and we know the difference between his speed and the water speed 
is 1.11.

I don't know if you've studied much algebra yet, but what you have 
here is a famous algebra problem. Suppose you know 1 fact (equation) 
about 1 unknown (number). You can often solve your equation and find 
the unknown number.  In this case, you have two facts, two equations, 
but thereare two things you don't know: the swimmer's speed and the 
water speed.

With a little cleverness, you can often solve the two equations 
together and find both quantities.

   x + y = 3.57
   x - y = 1.11

A famous trick used to solve this kind of equation is to add the two
equations together. After all, if two things are equal, and another 
two things are equal, you should be able to add the things on the left 
and add  the things on the right, and you'll have another pair of 
equal things.

If you add x+y and x-y, you have two x's, but the two y's cancel each 
other out, so you make the equation:

   2*x = 4.68 meters/sec

So the swimmer's speed is half of this, or 2.34.

Now that you know the swimmer's speed, you can use it to calculate 
the speed of the water. If x+y=3.57 and x=2.34, then y must = 1.23. 
This is how fast the water is going. Suppose you used the other 
equation to get y: x-y=1.11. Do you get the same answer?

I think you should ask whoever wrote this problem whether he meant 
that the swimmer was swimming on the way back or just coasting.

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