1. Classic Math Problems with Distance, Rate, and Time

2. Today’s Learning Goal
We will apply the steps to reading an algebra word problem to solving problems involving distance, rate, and time.
We will learn how to use a picture and a rate-time-distance table to help us solve these problems.

3. Rate, Time, and Distance
In previous lessons, we worked a lot with rate, time, and distance. Consider the following two examples:
A plane flies at 550 mph for 6 hours.
Jerome walks 2.5 meters per second for 9 seconds.
D = 6*550
D = 9*2.5
For each of these, how would you find the distance covered?
Nice…multiply the rate times the time!
We have seen that distance equals rate times time. In short, D = r*t.

4. Rate, Time, and Distance Problems
For rate, time, and distance problems, it is often convenient to draw a diagram because there are usually two moving objects.
Consider the following example:
One train leaves Chicago for Boston and at the same time another train leaves Boston for Chicago on the same track but traveling at different speeds.
The sketch for this would look something like this:
Train 1
Train 2
Chicago
Boston
Notice that because the trains left at the same time, we drew the trains coming at each other.

5. Rate, Time, and Distance Problems
Different problems will have different sketches that correspond to the direction and distance of each object that is moving.
Consider this example:
Mr. Nemuth leaves New York for a drive to Long Island. Later he returns to New York.
What would the sketch of this look like?
New York
Long Island
New York
Long Island
Because Mr. Nemuth traveled at different times, we did not draw his travel lines coming at each other.

6. Rate, Time, and Distance Problems
Consider this example:
Two planes leave New York at 10am. One plane is heading for Europe and the other plane is heading in the opposite direction.
What would the sketch of this look like?
New York
Because the planes left at the same time, we drew the travel lines on the same line.

7. Rate, Time, and Distance Problems
Consider this last example:
One train leaves Los Angeles for Chicago at 40 mph. Two hours later, a second train leave Los Angeles for Chicago at 75 mph.
What would the sketch of this look like?
Los Angeles
Chicago
Los Angeles
Chicago
(2 hours later)
Again, because the trains left at different times, we did not draw the travel lines on the same line.

8. Rate, Time, and Distance Problems
Now that we have seen how to draw diagrams for each type of problem, let’s look at a real problem:
A car leaves San Francisco for Los Angeles traveling an average of 60 mph. At the same time, another car leaves Los Angeles for San Francisco traveling 50 mph. If it is 440 miles between San Francisco and Los Angeles, how long before the two cars meet?
First, let’s draw a sketch of the movement in the problem. What will the sketch look like?
San Francisco
Los Angeles
Car 1
Car 2
440 miles

9. What known information can we fill in from the problem? 60
San Francisco
Car 1
Car 2
Los Angeles
A car leaves San Francisco for Los Angeles traveling an average of 60 mph. At the same time, another car leaves Los Angeles for San Francisco traveling 50 mph. If it is 440 miles between San Francisco and Los Angeles, how long before the two cars meet?
440 miles
Another important tool to use when solving rate, time, and distance problems is a table like the following:
Rate
Time
Distance
Car 1
Car 2
What known information can we fill in from the problem? 60 50
In rate, time, and distance problems, both the times or both the rates are given. So, start by filling in the known values first.

10. What is the unknown for this problem?
San Francisco
Car 1
Car 2
Los Angeles
A car leaves San Francisco for Los Angeles traveling an average of 60 mph. At the same time, another car leaves Los Angeles for San Francisco traveling 50 mph. If it is 440 miles between San Francisco and Los Angeles, how long before the two cars meet?
440 miles
What is the unknown for this problem?
Great…the time it takes for the cars to meet!
Rate
Time
Distance
Car 1
Car 2
What is true about the travel time for each car before they meet? 60 x
60x 50 x
50x
Nice…the times are the same.
Let x be the time that each car travels. If D = r*t, then how can we express the distance for each car using what we have in the table?

11. Fantastic…the total distance traveled will be the 440 miles! 60 x 60x
San Francisco
Car 1
Car 2
Los Angeles
A car leaves San Francisco for Los Angeles traveling an average of 60 mph. At the same time, another car leaves Los Angeles for San Francisco traveling 50 mph. If it is 440 miles between San Francisco and Los Angeles, how long before the two cars meet?
440 miles
So, 60x is the distance car 1 travels and 50x is the distance that car 2 travels. What is true about the total distance that they travel when they collide?
Rate
Time
Distance
Car 1
Car 2
Fantastic…the total distance traveled will be the 440 miles! 60 x
60x 50 x
50x
What equation can we write using that 50x and 60x have to add up to be a total of 440 miles?
Awesome…50x + 60x = 440!

12. Yes…combine the x-terms together! 50x + 60x = 440
San Francisco
Car 1
Car 2
Los Angeles
A car leaves San Francisco for Los Angeles traveling an average of 60 mph. At the same time, another car leaves Los Angeles for San Francisco traveling 50 mph. If it is 440 miles between San Francisco and Los Angeles, how long before the two cars meet?
440 miles
Now that we have our equation, we can solve the problem. What would you do first?
Yes…combine the x-terms together!
50x + 60x = 440
110x =
440
When we divide both sides by 110, what is the x-value?
110
x = 4
What does the solution of x = 4 mean in terms of this situation?
Super…after four hours, the two cars would have met!

13. Rate, Time, and Distance Problems
For rate, time, and distance problems, if it says that the two object leave at the same time or meet at the same time, then the times are both the same.
For these problems, do not put anything in the distance box until all other information is filled in. The distance box can be filled in using the D = r*t relationship.
Every rate, time, and distance problem has some kind of relationship between the distances. For the last problem, both distances added up to be the total of 440 miles. You have to watch for this relationship!

14. Rate, Time, and Distance Problems
Let’s try another problem:
Mr. Derbyshire makes a business trip from his house to Loganville in 2 hours. One hour later, he returns home in traffic at a rate 20 mph less than his rate going. If Mr. Derbyshire is gone a total of 6 hours, how fast did he travel on each leg of the trip?
First, let’s draw a sketch of the movement in the problem. What will the sketch look like?
House
Loganville
House
Loganville
(1 hour later)

15. Mr. Derbyshire makes a business trip from his house to Loganville in 2 hours. One hour later, he returns home in traffic at a rate 20 mph less than his rate going. If Mr. Derbyshire is gone a total of 6 hours, how fast did he travel on each leg of the trip?
House
Loganville
House
Loganville
(1 hour later)
Again, we will use the rate, time, and distance table to help us. What known information can we fill in?
Rate
Time
Distance
First Leg 2
Second Leg 3
In this problem, both times were given. So, we put the known information into the table first.

16. What is the unknown for this problem? 2x
Mr. Derbyshire makes a business trip from his house to Loganville in 2 hours. One hour later, he returns home in traffic at a rate 20 mph less than his rate going. If Mr. Derbyshire is gone a total of 6 hours, how fast did he travel on each leg of the trip?
House
Loganville
3(x – 20)
House
Loganville
(1 hour later)
What is the unknown for this problem?
Correct…the rate that the car goes on each leg!
Rate
Time
Distance
First Leg x 2 2x
Second Leg
x – 20 3
3(x – 20)
If we let x be the rate that he went on the first leg, how can we express his rate on the second leg?
Using D = r*t, how can we express the distance for each leg using what we have in the table?

17. Do we know what the distance is from his house to Loganville? 2x
Mr. Derbyshire makes a business trip from his house to Loganville in 2 hours. One hour later, he returns home in traffic at a rate 20 mph less than his rate going. If Mr. Derbyshire is gone a total of 6 hours, how fast did he travel on each leg of the trip?
House
Loganville
3(x – 20)
House
Loganville
(1 hour later)
Do we know what the distance is from his house to Loganville?
No…the distance is not given!
Rate
Time
Distance
First Leg x 2 2x
Second Leg
x – 20 3
3(x – 20)
But, we know something about the relationship between the distance in the first leg and the distance in the second leg. What is the relationship?
Excellent…the distances from both legs are equal!

18. What is the last step to solve the equation?2x
Mr. Derbyshire makes a business trip from his house to Loganville in 2 hours. One hour later, he returns home in traffic at a rate 20 mph less than his rate going. If Mr. Derbyshire is gone a total of 6 hours, how fast did he travel on each leg of the trip?
House
Loganville
3(x – 20)
House
Loganville
(1 hour later)
Because the distances are equal on both legs of the trip, what is the equation we can write?
2x = 3(x – 20)
Now, we just need to solve the equation. If we used the distributive property first, what would be the resulting equation? 2x = 3x
– 60
-2x = x
– 60
+60
If we subtracted 2x from both sides, what will be the resulting equation?
60 = x
What is the last step to solve the equation?
Perfect…add 60 to both sides.

19. Did we answer the question in the problem yet? 2x
Mr. Derbyshire makes a business trip from his house to Loganville in 2 hours. One hour later, he returns home in traffic at a rate 20 mph less than his rate going. If Mr. Derbyshire is gone a total of 6 hours, how fast did he travel on each leg of the trip?
House
Loganville
3(x – 20)
House
Loganville
(1 hour later)
Did we answer the question in the problem yet?
No…the question asked about his rate on both legs of the trip!
60 = x
Looking back at the table we filled in before, we see that x = 60 mph is his rate for the first leg. What is his rate for the second leg of the trip?
Fantastic…40 mph because x – 20 gives us his rate on the second leg!
Rate
Time
Distance
First Leg
Second Leg 2 3 x
x – 20 2x
3(x – 20) x
x – 20

20. Partner Work
You have 20 minutes to work on the following questions with your partner.

21. For those that finish early
Solve the following problems:
1. Two planes leave New York at 10am, one heading for Europe at 600 mph and one heading in the opposite direction at 150 mph (so it isn’t a jet). At what time will they be 900 miles apart? How far has each traveled?
2. A freight train starts from Los Angeles and heads for Chicago at 40 mph. Two hours later, a passenger train leaves the same station for Chicago traveling 60 mph. How long before the passenger train overtakes the freight train?

22. Big Ideas from Today’s Lesson
With rate, time, and distance problems, usually both rates or both times are given in the problem.
Use D = r*t to express the distances of each object as algebraic expressions.
The relationship between the distances has to be discovered by you through reading the problem.
Use a picture and the rate-time-distance table to help you solve the problem.

23. Homework
Complete Homework Worksheet.
Copy pages from 7th Grade text.