In previous lessons, students learned to solve problems in which time is expressed as distance/rate. Such problems involved round trips, as well as the speed of objects with and against river currents or wind speed—and vice versa. The central idea in these problems was that of “time = time”. So far, these problems have involved one variable. Now we come to problems in which both wind/current, and the speed of objects are to be found.
A typical problem is: “A motorboat went 18 miles downstream in 1 hour. The return trip took 3 hours. Find the rate of the boat in still water and the speed of the current.”
This is typically taught as a “distance = distance” problem. If we let r equal the speed of the boat in still water and c the speed of the current, then the downstream speed of the boat is r + c and the upstream speed is r – c. Since distance = rate × time, the problem can be stated as:
I teach it this way, but I also like to assign homework problems that include a mixture of round-trip, and one and two- variable current/wind problems so that students don’t just learn the latest problem and forget previous techniques.
There is a mathematical connection between the one and two-variable problems which I present here, but I believe it may be more of a distraction for most students. In this section, therefore I focus only on the “distance = distance” approach. I did have a student who had made such connection herself. For those teachers who may have such students and/or want to present the connection as an introduction, I include the discussion here.
Using the same problem as above, we are given neither the speed of the current, nor the speed of the boat in still water. Let r = speed in still water and c= speed of the current.
We write two equations: time downstream = 1 hr, and time upstream = 3 hrs, where time is expressed as distance/rate.
For both equations, we multiply both sides by the respective denominators to obtain the equation we started with:
This is now in “distance = distance” form .
Problem 1 requires students to know the squares from 11 through 20; those who do not will find this problem difficult. Problem 2 is a “time = time” problem that students have solved before and serves as a segue to the lesson that follows. Problem 3 requires students to remember how to work with negative exponents as well as the rules for the power of a product. Problem 4 is an absolute value inequality with which students should be familiar. Problem 5 requires plugging in the values to see which one does not make the inequality true.
Two Variable Wind/Current Problems. Problem 2 of the Warm-Ups gave us the speed of the airplane, and asked to find the speed of the wind. I remind students of that problem. “Today, we’ll look at similar problems, where we do not know either one; neither the speed of the wind or current, or the speed of the plane, boat, or whatever. For such problems we use ‘distance = distance’. ” I project a problem on the board.
An airplane takes 3 hours flying 1200 miles into the wind. With the same wind, the return trip takes 2 hours. What is the speed of the wind and the speed of the plane in still air?
“It may be helpful to use the rate × time = distance table we’ve used before. Help me fill it in. We don’t know the speed of the plane in still air. Let’s call that r. And we don’t know the speed of the wind, so we’ll call that w. How would we represent the downwind speed of the plane? How about upwind?”
They may be hesitant, in which case a prompt may help: “If the speed of the plane were 300 mph, and we didn’t know the wind speed, how would I represent the speed of the plane?” A few hands will typically go up, and I’m told the answer is 300 + w. “And if I knew neither?” I hear r + w and getting that, they figure the upwind is r – w. The other data in the problem are stated, and we fill in the table:
“Since distance equals rate × time, then I can write two equations.”
“I wrote it with time first, multiplied by the rate on the left hand side, and we’re given the distance of 1200 which I write on the right hand side. Now in most of these problems, but not all, we can simplify so we’re not dealing with such big numbers. Both the 2 and the 3 go into 1200, so what can I do to simplify?”
Students can see they can divide both sides in the two equations by 2 and 3 respectively, obtaining:
I have them solve it; it is obvious that the elimination method is the best choice and they get the answer: r = 500 mph, and w = 100 mph.
Examples. I work through the next one with them. It becomes obvious after that how to set them up and solve them. These examples are straightforward to give students the general feel and confidence in solving them.
I give more examples as necessary to ensure students are comfortable with solving the problems.
More Complex Problems. “Now let’s deal with more complicated problems, where we have time expressed as a fraction of an hour.”
“Anyone remember how we get rid of the fractions? Remember how we solved problems like 3/5(x) = 9? What do we multiply both sides by?”
They recall that both sides are multiplied by the reciprocal and I have them do that:
“Now solve the equations.” The answers are r = 315 mph; w = 45 mph
Homework. The problems should be a mix of two variable and one variable wind/current type problems. For the two variable problems some should have the students solving for distance as in Problem 3 above.