1.10 Solving Linear Equations - Distance, Rate and Time

1 Linear Equations - distance , rate and TimeObjective: Solve distance problems by creating and solvinga application of Linear Equations can be found in distance problems. Whensolving distance problems we will use the relationshiprt=dor rate (speed) timestime equals distance . For example, if a person were to travel30 mph for 4 find the total distance we would multiply rate times time or(30)(4) = person travel a distance of 120 miles. The problems we will be Solving herewill be a few more steps than described above. So to keep the information in theproblem organized we will use a table. An example of the basicstructure of thetable is blow:RateTimeDistancePerson 1 Person 2 Table of distance ProblemThe third column, distance , will always be filled in by multiplying the rate andtime columns together. If we are given a total distance of both persons or trips wewill put this information below the distance column. We willnow use this table toset up and solve the following example1 Example joggers start from opposite ends of an 8 mile course running towards eachother.

2 One jogger is running at a rate of 4 mph, and the other isrunning at arate of 6 mph. After how long will the joggers meet?RateTimeDistanceJogger1 Jogger2 The basic table for the joggers,one and twoRateTimeDistanceJogger14 Jogger26We are given the rates for each are added to the tableRateTimeDistanceJogger14tJogger26tW e only know they both start and end at thesame use the variabletfor both timesRateTimeDistanceJogger14t4tJogger26 t6tThe distance column is filled in by multiplyingrate by time8We havetotal Distance, 8miles,under distance4t+ 6t= 8 The distance column gives equation by adding10t= 8 Combine like terms,4t+ 6t1010 Divide both sides by 10t=45 Our solution fort,45hour(48 minutes)As the example illustrates, once the table is filled in, the equation to solve is veryeasy to find. This same process can be seen in the following exampleExample and Fred start from the same point and walk in opposite directions. Bobwalks 2 miles per hour faster than Fred.

3 After 3 hours they are30 miles fast did each walk?RateTimeDistanceBob3 Fred3 The basic table with given times filled inBoth traveled3hours2 RateTimeDistanceBobr+ 23 Fredr3 Bob walks2mph faster than FredWe know nothing about Fred,so userfor his rateBob isr+ 2,showing2mph fasterRateTimeDistanceBobr+ 233r+ 6 Fredr33rDistance column is filled in by multiplying rate sure to distribute the3(r+ 2)for distanceis put under distance3r+ 6 + 3r=30 The distance columns is our equation,by adding6r+ 6 =30 Combine like terms3r+ 3r 6 6 Subtract6from both sides6r=24 The variable is multiplied by666 Divide both sides by6r= 4 Our solution forrRateBob4 + 2 = 6 Fred4To answer the question completely we plug4in forrin the traveled6miles per hour andFred traveled4mphSome problems will require us to do a bit of work before we can just fill in thecells. One example of this is if we are given a total time, rather than the indi-vidual times like we had in the previous example.

4 If we are given total time wewill write this above the time column, usetfor the first person s time, and makea subtraction problem, Total t, for the second person s time. This is shown inthe next exampleExample campers left their campsite by canoe and paddled downstream at an averagespeed of 12 mph. They turned around and paddled back upstreamat an averagerate of 4 mph. The total trip took 1 hour. After how much time did the campersturn around downstream?RateTimeDistanceDown12Up4 Basic table for down and upstreamGiven rates are filled in1 Total time is put above time columnRateTimeDistanceDown12tUp41 tAs we have the total time,in the first time we havet,the second time becomes the subtraction,total t3 RateTimeDistanceDown12t12tUp41 t4 4t= distance column is found by multiplying rateby sure to distribute4(1 t) they cover thesame distance ,=is put after the down distance12t= 4 4tWith equal sign, distance colum is equation+ 4t+ 4tAdd4ttobothsidessovariableisonlyonones ide16t= 4 Variable is multiplied by 161616 Divide both sides by 16t=14 Our solution,turn around after14hr(15 min)Another type of a distance problem where we do some work is when one personcatches up with another.

5 Here a slower person has a head startand the fasterperson is trying to catch up with him or her and we want to know how long itwill take the fast person to do this. Our startegy for this problem will be to usetfor the faster person s time, and add amount of time the head start was to get theslower person s time. This is shown in the next leaves his house traveling 2 miles per hour. Joy leaves 6hours later to catchup with him traveling 8 miles per hour. How long will it take her to catch upwith him?RateTimeDistanceMike2 Joy8 Basic table for Mike and JoyThe given rates are filled inRateTimeDistanceMike2t+ 6 Joy8tJoy,the faster person,we usetfor timeMike stime ist+ 6showing his6hour head startRateTimeDistanceMike2t+ 62t+12 Joy8t8t= distance column is found by multiplying the rateby sure to distribute the2(t+ 6)for MikeAs they cover thesame distance ,=is put afterMike sdistance2t+12= 8tNow the distance column is the equation 2t 2tSubtract2tfrom both sides12= 6tThe variable is multiplied by666 Divide both sides by62 =tOur solution fort,she catches him after2hours4 World View Note:The 10,000 race is the longest standard track event.

6 10,000meters is approximately miles. The current (at the time of printing) worldrecord for this race is held by Ethiopian Kenenisa Bekele with a time of 26 min-utes, second. That is a rate of miles per hour!As these example have shown, using the table can help keep allthe given informa-tion organized, help fill in the cells, and help find the equation we will solve. Thefinal example clearly illustrates a 130 mile trip a car travled at an average speed of 55 mph andthen reducedits speed to 40 mph for the remainder of the trip . The trip hours. Forhow long did the car travel 40 mph?RateTimeDistanceFast55 Slow40 Basic table for fast and slow speedsThe given rates are filled time is put above the time tAs we have total time,the first time we havetThe second time is the subtraction t100 40tDistance column is found by multiplying rateby sure to distribute 40( t)for slow130 Total distanceis put under distance55t+100 40t=130 Thedistancecolumngivesourequationbyaddin g15t+100=130 Combine like terms 55t 40t 100 100 Subtract 100 from both sides15t=30 The variable is multiplied by 301515 Divide both sides by 15t= 2 Our solution 2 = answer the question we plug2in fortThe car traveled 40 mph for hours(30 minutes)Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative CommonsAttribution Unported License.

7 ( ) Practice - distance , Rate, and Time Problems1. A is 60 miles from B. An automobile at A starts for B at the rate of 20 milesan hour at the same time that an automobile at B starts for A at the rate of25 miles an hour. How long will it be before the automobiles meet?2. Two automobiles are 276 miles apart and start at the same time to traveltoward each other. They travel at rates differing by 5 miles per hour. If theymeet after 6 hours, find the rate of Two trains travel toward each other from points which are 195 miles travel at rate of 25 and 40 miles an hour respectively. Ifthey start at thesame time, how soon will they meet?4. A and B start toward each other at the same time from points 150 miles A went at the rate of 20 miles an hour, at what rate must B travel if theymeet in 5 hours?5. A passenger and a freight train start toward each other at the same time fromtwo points 300 miles apart. If the rate of the passenger trainexceeds the rateof the freight train by 15 miles per hour, and they meet after 4hours, whatmust the rate of each be?

8 6. Two automobiles started at the same time from a point, but traveled inopposite directions. Their rates were 25 and 35 miles per hour how many hours were they 180 miles apart?7. A man having ten hours at his disposal made an excursion, riding out at therate of 10 miles an hour and returning on foot, at the rate of 3 miles an the distance he A man walks at the rate of 4 miles per hour. How far can he walkinto thecountry and ride back on a trolley that travels at the rate of 20 miles per hour,if he must be back home 3 hours from the time he started?9. A boy rides away from home in an automobile at the rate of 28 miles an hourand walks back at the rate of 4 miles an hour. The round trip requires 2 far does he ride?10. A motorboat leaves a harbor and travels at an average speed of 15 mphtoward an island. The average speed on the return trip was 10 mph. How farwas the island from the harbor if the total trip took 5 hours?

9 11. A family drove to a resort at an average speed of 30 mph and later returnedover the same road at an average speed of 50 mph. Find the distance to theresort if the total driving time was 8 As part of his flight trainging, a student pilot was required to fly to an airportand then return. The average speed to the airport was 90 mph, and theaverage speed returning was 120 mph. Find the distance between the twoairports if the total flying time was 7 A, who travels 4 miles an hour starts from a certain place 2hours in advanceof B, who travels 5 miles an hour in the same direction. How many hoursmust B travel to overtake A?14. A man travels 5 miles an hour. After traveling for 6 hours another man startsat the same place, following at the rate of 8 miles an hour. When will thesecond man overtake the first?15. A motorboat leaves a harbor and travels at an average speed of 8 mph towarda small island. Two hours later a cabin cruiser leaves the same harbor andtravels at an average speed of 16 mph toward the same island.

10 In how manyhours after the cabin cruiser leaves will the cabin cuiser bealongside themotorboat?16. A long distance runner started on a course running at an average speed of 6mph. One hour later, a second runner began the same course at an averagespeed of 8 mph. How long after the second runner started will the secondrunner overtake the first runner?17. A car traveling at 48 mph overtakes a cyclist who, riding at 12 mph, has hada 3 hour head start. How far from the starting point does the car overtakethe cyclist?18. A jet plane traveling at 600 mph overtakes a propeller-driven plane which hashad a 2 hour head start. The propeller-driven plane is traveling at 200 far from the starting point does the jet overtake the propeller-driven7plane?19. Two men are traveling in opposite directions at the rate of 20 and 30 miles anhour at the same time and from the same place. In how many hourswill theybe 300 miles apart?20. Running at an average rate of 8 m/s, a sprinter ran to the end of a track andthen jogged back to the starting point at an average rate of 3 m/s.