This lesson is a follow-up to the previous one which addressed uniform motion with motion in opposite directions. In this lesson, we look at problems with motion in same direction, which are typically “catch-up” problems—e.g, how long will it take a faster car to catch up with a slower one that had a head start of so many miles, or so many hours.
I use an approach which I saw in Foerster’s Algebra 1 textbook, and which I find students can follow easier than other methods.
Today’s Warm-Ups include a problem with motion in opposite directions, but also a problem in which there is motion in one direction at two different rates. The equation for this problem is similar to that used in the opposite direction problems, and takes the form of Distance 1 + Distance 2 = Total distance. This problem is included to broaden the concept of “distance = distance” beyond that of opposite motion or today’s “catch-up” type problems.
Homework problems and some of the examples in today’s lesson include opposite motion as well as same-direction, so that problem solving ability is not limited to the latest thing students have learned.
Note: I have omitted discussion in these draft sections of other word problems types such as consecutive numbers and some types of mixture problems. I am focusing on key areas of algebra in this book so some will have to be omitted. That said, the Warm-Up problems may include problems that will have been addressed in an algebra course, even though they have not been discussed in these sections.
Warm-Ups.
1. Two planes start towards each other at the same times at airports that are 1,380 miles apart. One plane travels at 320 mph and the other at 600 mph. In how many hours will they meet? Answer: 320t + 600 t = 1380; 920t=1380; t = 3/2 or 1 ½ hrs.
2. In Problem 1, how far will the slow plane have traveled before meeting the other plane? Answer: d = r ∙ t: d = 320(3/2) = 480 miles.
3. Two trains travel toward each other and are 388 miles apart. One train averages 47 mph. It takes 4 hours for the trains to meet. What is the speed of the other train? Answer: 47(4) + 4x = 388; 188 + 4t = 388; 4t = 200; t = 50 mph.
4. Mr. Smith travels 665 miles, half of the time by car at 45 mph and half the time by train at 50 mph. How long did the trip take? Answer: Let t = time for each half of the trip; 45t + 50t = 665; 95t = 665; t = 7 hours; so time for total trip = 14 hrs
5. Find four consecutive even numbers whose sum is 100. Answer, Let x = first number, then the next three numbers are x + 2, x+4, and x + 6. x + x +2 + x + 4 + x + 6 = 100; 4x + 12 = 100; 4x = 88; x = 22, x + 2 = 24, x + 4 = 26, x + 6 = 28.
Problem 2 is an extension of Problem 1 and requires students to use the d = rt formula. Problem 3 gives the time it takes for the trains to meet so that students are solving for speed. Possible prompt: “We have distance of one train + distance of the other train = 388. How do we represent the distance traveled by the train going 47 mph?” “How can we use distance = r × t to represent the distance traveled by the train whose speed we don’t know?” Problem 4 defines a trip that is split into two halves by time, so that t is the same for both halves. Since the problem asks for total time, it would be 2t or 14 hour. For Problem 5, as mentioned earlier, we will have covered consecutive problems. Consecutive even numbers advance by 2; e.g., 2, 4, 6…etc.
Going Over Homework. There are always questions regarding the previous homework assignment, so I pick the more difficult problems and go over them.
Relative Speeds. “Yesterday we talked about motion in opposite directions. Today we’ll be talking about motion in the same direction.”
After this announcement which usually elicits growns, I then present a situation with which all students are familiar. “You’ve been in a car that is passing another, and when you look at the car being passed, it appears that you’re traveling at a slow speed relative to that car. So if you pass the car going 65 mph and the car you’re passing is going 60, what speed does it look like you’re going?”
Students immediately respond “Five miles per hour.”
“This is what we call ‘relative speed’. The speed of the faster car to the slower car in that example is 5 mph.” I give several other examples so they are comfortable with the concept.
“Now suppose that there is a car 10 miles ahead of you. You are going 70 mph and the car ahead of you is going 60 mph. What is your speed relative to the car ahead of you?” Ten mph.
“Suppose now that you have telescopic vision and you watch the 10 mile space in front of you get shorter and shorter. It appears you are traveling the 10 miles at 10 mph. What is our formula for distance that we’ve been using?”
They should know it is distance = rate × time.
“I want to know how long it will take to go 10 miles at the speed of 10 mph—that’s the speed it looks like we’re going relative to the slower car. Can you figure that out?”
Some students will figure it out and I’ll ask them to explain what they did. Generally the explanation comes back as “If you’re going 10 mph, it takes 1 hour to go 10 miles.”
“This is correct,” I’ll say, “but I would like it stated as an equation. Distance between the cars is 10 mph. But instead of 70 for the rate, I’m going to use relative rate. Using d = rt, let’s plug in the values that we know.” I’ll ask for someone to do this.
The equation is 10 = 10t.
Solving Problems with Motion in Same Direction. “This principle is used to solve problems where we have motion in the same direction. Let’s say we have a car that has gone 200 miles, and averages 60 miles per hour. How long would it take you to catch up to that car if you go after it at 100 miles per hour? Let’s write out the steps.”
I then write down the steps which they copy in their notebooks:
Step 1: Find the relative speeds.
I ask them what is the speed of your car at 100 mph relative to the 60 mph car? Answer: 100 – 60 = 40 mph
Step 2: What is the gap in distance?
In this problem, there is a 200 mile gap—the faster car must close the 200 mile distance between the two cars in order to catch up.
Step 3: Relative speed ∙ time = gap in distance
This is the form the equation will take, and it translates to distance = distance. The faster car closes the gap of 200 miles at a relative speed of 40 mph. We are solving for the time it takes at 40 mph to go 200 miles, or 40t = 200.
Step 4: Solve the equation: t = 5 hrs for this problem.
Examples. Problems are projected on the board. I have students work in their notebooks.
1. Mr. Smith starts driving from San Luis at 60 mph and has gone 120 miles when his wife starts after him at 80 mph. How long will it take Mrs. Smith to catch up to the Mr. Smith? Answer: Relative speed of faster car compared to slower: 80-60 = 20 mph; Gap in distance: 120 miles; 20t = 120; t = 6 hrs
(Note: I once had a student set up the equation in the way I used to teach this type of problem. The distance that Mrs. Smith drives in catching up to Mr. Smith is equal to the 120 mile gap plus the distance Mr. Smith drives in the time it takes for her to catch up. Translated to an equation. Thus, 80t is the total distance Mrs. Smith drives. It is equal to 60t + 120. The equation is then: 80t = 60t + 120, which simplifies to the same equation used in the method above: 20t = 120. I found that this method was confusing to many students, so I used the relative speed method, which they found easier to understand and work with.)
2. Two airplanes start from the same airport at the same time and travel in opposite directions at 350 mph and 325 mph respectively. In how many hours will they be 2,025 miles apart? (I ask the students “Is this a catch-up problem or opposite direction problem? They see it is opposite direction and solve accordingly. I tell them it is important to decide what type of problem it is before rushing in to solve.) Answer: 350t + 325t = 2,025; 675t = 2025; t = 3 hrs.
3. A freight train left Beeville at 5 AM traveling 30 mph. At 7 AM an express train traveling 90 mph left the same station. When did the express overtake the freight? Students say they don’t know what the gap in distance is. “Is there information in the problem that you can use to figure that out? How much time elapses between the time the freight train leaves and the express train starts out?” They generally can piece it together that the gap is 30 mph times the 2 hours it is traveling prior to the express train starting—a 60 mile gap. Equation is then 60t = 60; t = 1 hr.
Homework. Homework is a mix of same and opposite direction problems. In some problems, students need to find the gap distance; in others, students need to find the rate of the one of the vehicles. Since I leave time for students to start working on homework, I spend time providing hints and guidance on the various problems.