MultiStep Problems with DistanceRateTime In this lesson we will work on using the distanceratetime equation to solve problems that involve several steps. In the distanceratetime lesson from the geometry module, you may have learned how to use the distanceratetime equation. If you want a gentler refresher on this please review that lesson first. Recall the main formula that relates distance (how far one has gone), rate (how fast one is going), and time (how long the journey took). If d represents the distance, r represents the rate, and t represents the time, then d = rt By dividing both sides by r, the equation becomes
d and by dividing both sides of the first equation by t, the equation becomes
d These three equation will be fundamentally important for this lesson. Here is an example. Example 1 Roberto can run 200 meters in 25 seconds. At this rate, how many meters can he run in 80 seconds?
Solution We can use the distance rate time equation. First we know that Roberto's distance d is 200 meters when his time t is 25 seconds. We are given d and t, so we can find r using the third equation
d Plugging in 200 for d and 25 for t gives
200 Now that we have the rate, we can find the distance (number of meters) when the time is 80 seconds. We have d = rt = (8)(80) = 640 We can conclude that Roberto can run 640 meters in 80 seconds.
This problem is typical of CAHSEE problems in that the distance and time are given in one of the sentences. The strategy is to use this information and the r = d / t equation to find the rate. Then use this rate to either find the distance or the time. Here is another example. Imani has noticed that it took her 30 minutes to travel 15 miles on her way to work. If the total distance to work is 20 miles, how many more minutes will it take her to get to work? Solution First we can use the fact that it took t = 30 minutes to travel d = 15 miles to find her rate r. Use the equation
d and plug in 15 for d and 30 for t gives
15 1 Next, since she has already traveled 15 miles on the way to work and the total distance is 20 miles, we can subtract to find the remaining distance. Remaining Distance = Total Distance  Distance already traveled = 20  15 = 5 Now we find the time remaining by using the equation
d Plug in d = 5 and r = 1/2 to get
5 Now recall that dividing fractions is the same as multiplying by the reciprocal. We get
5 2 = 10 It will take Imani 10 more minutes to get to work.
Now try a couple by yourself. If you want to see the answers, put your mouse on the yellow rectangle and the answer will appear. Exercise 1 The Mendenhall glacier has retreated 700 feet in the last six years. If this rate continues, how many feet will the glacier retreat during the next 15 years?
Exercise 2 While riding the Red Line at maximum speed, Juanita noticed that she had traveled 7 miles in 6 minutes. If she rides the Red Line at maximum speed for an additional 9 minutes, what is the total distance that Juanita will have traveled?
Some problems involve putting together distances or times for two separate trips or one trip that involves two different rates. For this type of problem, we will want to calculate the distances or times for each trip separately and then add them up when we are finished. Here is an example. Example 3 Angelica and Mary teamed together for a relay race. Angelica ran 9 feet per second and Mary ran 10 feet per second. If Angelica ran for 120 seconds and then Mary ran for 80 seconds, what was the total distance they covered? Solution Here we are given that Angelica ran 9 feet per second for 120 seconds and that Mary ran 10 feet per second for 80 seconds. We first compute their separate distances and then add them together. We use d = rt For Angelica's distance d, we plug in her rate r = 9 and her time t = 12. Angelica's d = (9)(120) = 1080 For Mary's distance d, we plug in her rate r = 10 and her time t = 8. Mary's d = (10)(80) = 800 Now for the total distance we add the two distances. Total Distance = Angelica's Distance + Mary's Distance = 1080 + 800 = 1880 We can conclude that the total distance that they covered was 1880 feet. Now one by yourself. If you want to see the answers, put your mouse on the yellow rectangle and the answer will appear. Exercise 3 Steve rides his bike at 16 miles per hour on the level road and 12 miles per hour on the hills. If Steve took a bike ride that involved 24 miles of level roads and 6 miles of hills, how long did the trip take him?
Once we understand how to solve distanceratetime problems, we can stretch our understanding to include distances that are not measured with a ruler but rather as an amount or number of something completed. Here are a few examples that illustrate this idea. Example 4 Carlos' car has full gas tank that holds 18 gallons of gasoline. If after 3 hours of driving, the car used 12 gallons, how many more hours can Carlos drive until he empties the gas tank? Solution This example is very similar to Example 2 from this lesson. Here we can think of the gallons used up as the distance. That is, the distance was 12 gallons for the first part of the trip and the time was 3 hours. The rate in gallons per hour was
d Plugging in 12 for d and 3 for t gives
12 Since we want to find how many more hours Carlos can drive, we reason that he has already used up 12 gallons of gasoline. To determine how many gallons are left, we subtract the amount used up from the tank size: Gallons Left = Total Gallons  Gallons Used = 18  12 = 6 So the rate (in gallons per hour) is r = 4 and the distance (in gallons) is d = 6. We can now use the second of the three distanceratetime equations to get
d
6
3 = 1.5 We can conclude that Steve can drive for 1.5 more hours before the tank is empty. Here is another example that is a variation of the distanceratetime equation. Example 5 Margarita manages a theater where there are 504 patrons who need to be seated. She notices that in the first 15 minutes that the doors are open, 270 patrons are able to be seated. How much more time will it take to seat the rest of the patrons? Solution This Example is just like Example 4 but instead of gallons of gasoline, we are working with theater patrons. Here we can think of the theater patrons who have been served as the distance. That is the distance was 270 patrons when the time was 15 minutes. The rate in patrons per minute is
d Plugging in 270 for d and 15 for t gives
270 Since we want to find how much more time it will take for the rest of the patrons to be seated, we subtract the number already seated from the total number of patrons to be seated: Patrons Left = Total Patrons  Patrons Already Seated = 504  270 = 234 So the rate (in patrons per minute) is r = 18 and the distance (in patrons) is d = 234. We can use the second of the three distance rate time equations to get
d
234 = 13 We can conclude that it will take 13 more minutes to get all of the patrons seated.
Now try a couple by yourself. If you want to see the answers, put your mouse on the yellow rectangle and the answer will appear. Exercise 4 Carlos is reading a 462 page book for his Language Arts class. It took him 6 days to read the first 252 pages. If he continues reading at the same rate, how many more days will it take him to complete the book?
Exercise 5 A postal worker's route includes delivery to 532 residences. It took the postal worker 3 hours to deliver to the first 228 residences. If the postal worker continues at the same rate, how many more hours will it take the postal worker to complete the route?
