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Two Planes and Not Enough Information

Date: 08/10/2004 at 17:36:30
From: Crystal
Subject: Two Planes Leave Chicago

Two planes leave Chicago, one traveling due east and the other due
west.  The first travels 100 miles per hour faster than the second. 
How long will it be until they are 2000 miles apart? 

I figured out that if plane two is going at X mph, then plane one is 
traveling at X + 100 mph.  But I can't figure out the value of x. 
Please help.  Thank you!

Date: 08/11/2004 at 10:20:45
From: Doctor Ian
Subject: Re: Two Planes Leave Chicago

Hi Crystal,

Let's think about a simpler problem for a moment:  

  Suppose you and I start at some point, standing back to 
  back, and I walk away from you at a speed of 10 feet 
  per second.  How long will it take until we're 40 feet

Now consider this twist on the problem:

  Suppose you and I start at some point, standing back to 
  back, and walk away from each other at a speed of 5 feet 
  per second.  How long will it take until we're 40 feet

Do you see why these are really asking the same thing?  Since we're
concerned about the distance between us, we don't really care how fast
either of us is moving relative to the _ground_.  We just care how
fast we're moving relative to each other. 

So I could make lots of variations on the problem:  I move at 8 feet
per second, and you move at 2 feet per second; I move at 4 feet per
second, and you move at 6 feet per second; and so on.  In fact, I
could make a version like this,

  Suppose you and I start at some point.  I walk at a speed 
  of 12 feet per second, and you move at a speed of 2 feet 
  per second IN THE SAME DIRECTION.  How long will it take 
  until we're 40 feet apart? 

and it's _still_ the same, because I'm still moving away from _you_ at
10 feet per second.  Does this make sense? 

These problems are a classic application of 

  distance = rate * time

That is, it's the same as

  At a speed of 10 feet per second, how long does it take to 
  move 40 feet?

We can write an equation

  40 feet = 10 feet/sec * x seconds
  40 = 10 * x
   4 = x

Clearly after 4 seconds we will be 40 feet apart.

Having said all this, let's look at the original problem.  As you 
said, if the first plane is traveling at speed x, then the second
travels at speed (x + 100).  This is the same as if one of the planes
stays on the ground, and the other moves away at a speed of 
(x + x + 100).  So after h hours, the distance between the planes will be 

  distance = rate * time
  2000 = (x + x + 100) * h

But there's a problem here!  It's not possible to find a unique
solution.  For example, I could say that x is 200 miles per hour. 
Then the time is

  2000 = (200 + 200 + 100) * h
  2000 = 500 * h
     4 = h  so it will take 4 hours.           
On the other hand, I could say that v is 150 miles per hour:

  2000 = (150 + 150 + 100) * h
  2000 = 400 * h
     5 = h  so now it will take 5 hours.

And there are infinitely many other solutions that will work.  What
the equation is telling you is that there is a relationship between
the value of x (the speed of the slower plane) and the value of h (the
amount of time required); but it's not enough information to specify a
_unique_ value for either quantity.  

There are (at least) a couple of possibilities here.  One is that 
there's more to the problem than you've said, and the extra
information would allow you to nail down a particular time.  Another 
is that you're not really expected to come up with a unique value for
the time, i.e., you're supposed to end up with an equation that you
can use to determine the time that corresponds to any chosen value of
x.  To do that, you'd want to rearrange the equation

  2000 = (x + x + 100) * h

to make it look like 

  h = [something involving only x's and numbers]

I hope this helps!  Write back if you'd like to talk more about this,
or anything else. 

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

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