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/
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