Rate of River Current
Date: 10/31/2001 at 09:44:32 From: Dave Normansell Subject: Current of a river Ryan is canoeing upstream while his sister Katie sleeps in the back of the canoe. As they pass a big tree, one of its branches knocks Katie's hat into the water. Katie wakes up 5 minutes later and discovers that her hat is missing. Ryan instantly turns around and paddles downstream and they find the hat 1 mile downstream from the big tree. What is the rate of the current?
Date: 10/31/2001 at 12:34:03 From: Doctor Peterson Subject: Re: Current of a river Hi, Dave. This is an interesting problem, because it seems to have insufficient data (one equation for two unknowns), until you are just about to solve it, at which point everything simplifies out and you are left with just the information you are asked for and nothing else. In fact, you CAN'T tell how fast he paddles relative to the water, only how fast the water flows! Here's one way to set it up: Define variables v = speed of boat relative to water, in miles per minute w = speed of water relative to shore, in miles per minute (I use miles per minute to match the units given in the problem.) Now write expressions for these quantities: How fast does the canoe go relative to shore, going upstream? How fast does it go, downstream? How far does it go in the five minutes before it turns around? How long does it take the hat to drift one mile? How far does the canoe have to go to catch the hat? How long does it take to catch the hat? It helped me to draw a graph of the situation, with distance upstream on the vertical axis and time on the horizontal axis: D turned around ^ + | lost / \ | hat / \canoe | + \ | / \ \ | / hat\ \ | / canoe + | / caught hat +--------------------------------------->T Now you can write an expression for the speed of the boat downstream, based on the last two answers. But you already know what that is in terms of v and w, so you can write an equation. As I said, this doesn't seem like enough, but it's all you have, so you go ahead and simplify it, and Voila! - you'll find the answer. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 10/31/2001 at 17:44:39 From: Doctor Anthony Subject: Re: Current of a river Let u = speed of river current (miles/min) v = speed of canoe in still water Speed of canoe going upstream = v-u Distance travelled upstream = 5(v-u) (in 5 minutes) Distance travelled by canoe downstream = 5(v-u) + 1 If t = time travelling downstream, then with speed (v+u) we get (v+u)t = 5(v-u) + 1 ........(1) also u(t+5) = 1 (distance travelled by hat in t+5 minutes) t+5 = 1/u t = 1/u - 5 substitute for t in (1) (v+u)(1/u - 5) = 5(v-u) + 1 (v+u)(1-5u)/u = 5v-5u+1 v+u - 5uv - 5u^2 = 5uv - 5u^2 + u v = 10uv 1 = 10u and so u = 1/10 miles/min = 6 mph So the current is flowing at 6 mph. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
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