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/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.