Students started solving word problems algebraically in seventh grade. In eighth grade algebra, so far, I have introduced fairly straightforward word problems with specific techniques for translating into algebra. We now progress into problems that involve uniform motion which many students new to algebra find difficult. Some of the difficulty comes from the wordiness of the problems—students need to identify what the problem is asking. Another source of difficulty is identifying what is being equated.
My aim in teaching uniform motion problems, as well as other word problem types, is to show the basic structures of these problem types and how they translate into equations. In this unit, all uniform motion problems take the form of distance = distance. (Later when they have worked with algebraic fractions (rational expressions), such problems take the form of time = time.)
This section addresses motion in opposite directions—i.e., two objects heading toward each other, or away from each other. The next section addresses objects moving in the same direction. A third category, round trips, is addressed later in the section on algebraic fractions. The main reason is that round trip problems at this stage are difficult for students who at this stage in an algebra 1 course are in the process of getting their footing in expressing equations in one variable.
Solving round trip problems in terms of “distance = distance” is best for problems that ask for distance rather than time taken for a round trip. When students reach the point of learning how to work with algebraic fractions, they have had more experience with equations and are better prepared for solving both types of problems—even if it is slightly better.
Warm-Ups.
1. There are 900 students at Wayne High School. There are 20 more girls than boys. How many girls are there? Answer: Let x = number of boys; then x + 20 = number of girls. x + (x+20) = 900; 2x = 880; x = 440, x + 20 = 460
2. A car traveled on a freeway for 2 hours at 80 mi/hr. He then traveled for another 3 hours at 70 mi/hr. How far did he travel in all? (Hint: Distance = rate × time) Answer: 80 × 2 + 70 × 3 = 160 + 210 = 370 miles.
3. A motorcyclist took 7 hours to travel between Town X and Town Y at an average speed of 35 mi/hr. A car took 5 hours for the same trip. Find the speed of the car. (Hint: Speed = Distance/time) Answer: Motorcycle travels 7 × 35 = 245 miles. Speed of car: 245/5 = 49 mi/hr.
4. Solve the inequality; indicate whether it is “or” or “and”. |x – 45| > 5 Answer: x-45 > 5; x > 50, OR x-45 < -5; x < 40
5. Solve and write as one inequality rather than two. |x +3| < 4 Answer: x < 1; x +3> -4; x > -7; -7 < x < 1
Students will likely need guidance for Problems 2 and 3. The distance = rate x time formula applies. Prompt for Problem 2: “How do you find his distance for the first part? How far does he travel in 1 hour going 80 mi/hr?” For Problem 3: Distance divided by total time to travel the distance = average speed. Prompt: “How far does the motorcycle travel in 7 hours?”
Distance = Rate × Time. Problems 2 and 3 relied on the D = r ∙ t formula. I make a point of saying the obvious which I’ve realized is not always obvious to all, that 60 mi/hr means 60 miles/1 hour. It is a unit rate, which students have had in seventh grade.
“We will be working on uniform rate problems—that is, problems in which the objects are going at a steady speed. In all such problems, the solution relies on the formula distance = rate times time. I’m going to give you a problem now that relies on this formula.”
Two cars head towards each other on the same road. They start at the same time. One starts from the north, and travels south on this road at 70 mph. The other starts from the south and travels north at 80 mph. They meet at a certain point on that road. How far apart were they one hour before they met on that road?
Classes vary in their response. Some students get it right away, and others can’t see how it is solved. I often hear “There’s not enough information” to which I reply, “Yes, there is.”
I draw a vertical line and put a dot somewhere near the middle of the line. “The dot is where the two cars meet. Can you tell me how far away from this dot the car going south is one hour prior to getting there?”
Next prompt if I hear nothing: “How fast is the southbound car going?” They answer 70 mi/hr. “So how far does the car travel in one hour?”
This generally provides the hint they need and I usually hear a collective “OHHH”
One hour previous to the cars meeting, the southbound car is 70 mi north of the meeting point. The northbound car is 80 mi south of that point; so they are 150 miles apart.
Some Problems. “Today we will be looking at problems in which, like the problem I just gave you, there are two objects—cars, ships, bikes, people—traveling in opposite directions. They start at the same time, travel at a uniform speed, and are either heading towards each other, or away from each other.”
Don rides at 15 mph and Nancy rides at 12 mph. They go in opposite directions but start from the same point. How long until they are 135 miles apart?
“This is an opposite-direction problem but unlike the one I just gave you, they are starting from the same place and heading away from each other.”
I draw a picture to illustrate:
“The total length of the line is 135 miles—that’s how far apart they are after some amount of time. What are we trying to find?”
We are trying to find the amount of time, which students easily see. “The diagram represents the total distance Don and Nancy travel. If I find the sum of the distances that Don and Nancy travel in this diagram, what is that total?”
If there is any hesitation, I tell them to look at the problem again. They will see that the total distance is 135 miles.
“So Don’s distance + Nancy’s distance = Total distance (135 miles)”
“For purposes of illustration and only for such purposes, let’s use a guess and check approach.”
I draw a chart on the board and fill it in with them. “Distance = rate × time, so in 1 hour how far does Don go, riding at 15 mi/hr? And how far does Nancy go?”
Easy to see they go 15 and 12 miles respectively for a sum of 27 miles. Does this equal 135 miles? No, it doesn’t so 1 hour is not the answer to the problem. We do the same for 2 and 3.
“Well this is rather tedious and inefficient. There is a better way. Let’s use algebra.” I have them copy this chart:
“Instead of plugging in numbers for the time and trying them out, what do we use when we don’t know a number in an equation?”
I usually hear “x” which is fine, although I like to use t just as a reminder that the variable represents time. “Let’s fill out the chart. How fast is Don going? And since we don’t know for how long, what do we put in the ‘time’ column? And how do I write 15 times x?” Continuing in this fashion the completed chart looks like this:
“So Don’s distance after riding for x hours is what?” 15x
“And Nancy’s?” 12x
Plugging in these terms into Don’s distance + Nancy’s distance = 135 we obtain the following equation:
More Examples. I work with them on these, with diminishing support as they gain confidence.
1. Two hikers are 14 miles apart. The hikers start walking towards each other at 7 AM. One hiker is walking at 4 mph, and the other at 3 mph. How long will it take to meet?
“In this problem, they are walking towards each other. They start at the same time, and each walks at a constant rate of speed. “Would you say that the time it takes for them to meet is the same for each person?”
There will be general agreement.
“So what we have is this situation:”
“Each arrow represents a hiker. This is another distance = distance problem, and it’s set up the same way as the one we just solved:”
First hiker’s distance + Second hiker’s distance = Total distance (14 miles).
“Now you can use a table like we did before, but some of you may not need one. Let’s see what you come up with.”
They work in their notebooks and I go around offering help and guidance. The equation I want to see is: 3t + 4t = 14 Solving it, x = 2 hours.
“Suppose I had asked you what time they will meet? How can we do that? Look at the problem.” Seeing that they start at 7 AM, students figure out that the time they meet has to be 9 AM.
2. Two cars traveling in opposite directions meet on a highway. One averages 80 miles per hour and the other 70 miles per hour. In how many hours will they be 450 miles apart?
The wording is slightly different in this problem. One prompt may be “Is the problem asking for the time it takes when they are going toward each other or away from each other?” It is a “traveling away from each other” problem. The equation is 80t + 70t = 450; t = 3 hours.
3. Two ships are sailing toward each other and are 120 nautical miles apart. If the rate of one ship is 4 knots greater than the rate of the other, and if they meet in 3 hours, find the rate of each ship.
In this problem a table is useful so I have them construct it. We are given the time—3 hours—but not the rates of the ship. This confuses students who look at this as if it’s a betrayal. I explain that some problems ask for time, others for rate, and still others for distance. “You will get used to it,” I tell them, but generally they don’t look convinced.
“If one ship is going x knots, then how do we represent the speed of the other ship?” It is x +4. The equation is 3x + 3(x+4) = 120; 3x + 3x + 12 = 120; 6x = 108; x = 13 knots, x + 4 = 17 knots.
Homework. I typically assign 5 or 6 problems and I’ll work on the first two with the class.