Two Planes and Not Enough InformationDate: 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 apart? 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 apart? 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 http://mathforum.org/dr.math/ |
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