Solving Rate-Time-Distance Problems - The Learning Lab at HFCC

HFCC Math Lab Intermediate Algebra - 13SOLVING RATE-TIME-DISTANCE PROBLEMSThe variables involved in a motion problem are distance (d), rate (r), and time (t). These variables aredrelated by the equation d rt , which can be solved for any one of the three variables d rt , r , ortdt . Ordinarily, one of these three forms of the equation will be used in setting up a motion problem.rThe first step in any application problem is to read and reread the statement of the problem until it isfully understood. In addition, in a motion problem all the times given should be in the same units, andall the distances should be in the same units. The rate units must be compatible with the time anddistance units. For example, if the rates are in miles/hour, then the distances must be in miles and thetimes must be in hours.In almost all motion problems the data can be placed in a table in the format shown here:ABRate Time DistanceA and B may be categories such as two individuals, two vehicles, or two types of terrain.Steps to a SolutionDefine the variable (the unknown). This will involve one of the columns of the table.Introduce any given data into the table. Often this will complete another column.Use d rt ,rdt , ortdrto complete the table.Write the equation and solve. For this step each problem is different. We generally look forsome statement of relationship not yet used in the setup involving time, rate, or distance.Ex 1: A man walks from point A to point B at a pace of 5 miles per hour. He returns at the rate of 41miles per hour. If the total time for his round trip is 2 hours, then what is the distance from A4A to BB to Ato B?We let d represent the distance between A and B, with the first row of the table representing hisinitial trip and the second row representing his return trip:Rate Time DistanceddRevised 04/10 1

We now introduce the given data. We are given both rates. (We omit the units, which arecompatible, from the table.)Rate Time DistanceA to B 5 dB to A 4 dSincetdr, we now complete the table.Rate Time DistanceA to B 5d5dB to A 4d4dWe now write an equation. In this case the equation comes from the statement in the problem1 9that the total time is 2 hours. Thus we have4 4(time from A to B) + (time from B to A) = 9 4d d 95 4 44d 5d 45(Multiply each side by the LCD, 20.)9d45d5The distance between the points is thus 5 miles.Ex 2: Two planes leave an airport at the same time and travel in opposite directions. If one planeaverages 440 miles per hour and the other 560 miles per hour, then in how long will the planesbe 2500 miles apart?We let t be the time of travel, since that is what we are trying to find. Each row represents therate, time, and distance of one of the planes.Rate Time DistancePlane 1tRevised 04/10 2

Plane 2tThe rates are given, so we enter those into the table:Rate Time DistancePlane 1 440 tPlane 2 560 tWe use the relation drt to complete the table:Rate Time DistancePlane 1 440 t 440tPlane 2 560 t 560tThe equation comes from the fact that when the planes are 2500 miles apart, the distancestraveled by the planes will total 2500 miles:440t 560t2500t2.5The time required is thus 2.5 hours.Ex 3: Two planes leave an airport at the same time and travel in the same direction. If one planeaverages 440 miles per hour and the other 560 miles per hour, then in how long will the planesbe 2500 miles apart?We note that the table to be used for this example will be the same as the table used in Example2:Rate Time DistancePlane 1 440 t 440tPlane 2 560 t 560tOur equation changes, though:(faster plane’s distance) – (slower plane’s distance) = 2500 miles560t 440t2500t2500 520120 6The time required would thus have been520 6hours, or 20 hours 50 minutes.Revised 04/10 3

Ex 4: A car was driven 280 miles. If it had gone 5 miles per hour faster, the trip would have beenshorter by an hour. At speed was the car driven?We will let r represent the rate at which the car was actually driven. On the hypothetical trip, thecar was driven at r 5 miles per hour.Rate Time DistanceActualrHypothetical r 5We are given the distance the car was driven. The distances would be the same for the actualand the hypothetical trips:Rate Time DistanceActual r 280Hypothetical r 5280We use the relationtdrto complete the table:Rate Time DistanceActualr280r280Hypothetical r 5280r 5280The equation is derived from the statement that the time for the hypothetical trip is 1 hour lessthan the time for the actual trip:(time for actual trip) 1(time for hypothetical trip)280 2801r r 5280 r 5 r r 5 280r (Multiply each side by the LCD, r r 5 .)2280r 1400 r 5r 280r20 r 5r1400r 40 r 35 0(Factor. If the quadratic polynomial were notr 40 or r 35factorable, we would use the quadratic formula.)In this application, r 40 is meaningless.Revised 04/10 4

ExercisesThe speed of the car on the (actual) trip was thus 35 miles per hour.Solutions to odd-numbered exercises and answers to even-numbered exercises appear at the end.1. Two airplanes pass each other going in opposite directions. One travels at a speed of 270 milesper hour; the other travels at 210 miles per hour. How much time elapses after they pass eachother before they are 1200 miles apart?2. A pair of hikers, 14 miles apart, begin at the same time to hike toward each other. If one walksat a rate that is 1 mile/hour faster than the other, and if they meet two hours later, then how fast isthe slower hiker walking?3. One cyclist pedals at a rate 6 miles/hour faster than the other. At those speeds one covers thesame distance in three hours that the other covers in four hours. What is the speed of the fastercyclist?4. At 4:00p two vehicles leave towns 470 miles apart. They travel toward each other, one at anaverage speed that is 20 miles/hour faster than the other. If the automobiles pass at 9:00p, thenwhat is the average speed of the faster vehicle?5. A motorboat travels 20 miles/hour in still water. If the boat travels 62 miles upstream in thesame time that it takes to travel 98 miles downstream, then what is the rate of the current?6. A boat goes 50 km downstream in the same time that it takes to go 30 km upstream. The speedof the stream is 3 km/hr. Find the speed of the boat in still water.7. At noon a Pinto sets out westward at a steady rate from a point X on a narrow, one-lane road. At2:00p, a Prius moving westward along the same road reaches point X and suddenly accelerates toa speed 80 miles/hour faster than that of the Pinto. At 3:00p these cars “meet.” How far fromthe point X has each traveled at that time?8. A speeding automobile is traveling 55 miles/hour, and a highway trooper, 2 miles behind, ischasing it at a speed of 70 miles/hour. How long will it take the trooper to overtake the speedingcar if they both maintain these speeds?9. A driver averages a secondary-road speed of 40 miles/hour and a turnpike speed of 70miles/hour. If a trip of 160 miles over these roads took 150 minutes, then how many miles weredriven on secondary roads?10. A man bikes at a rate that is 6 miles/hour faster than the rate at which he jogs. After 30 minutesjogging and 90 minutes biking, the man has covered 21 miles. Over what distance was the manjogging?Revised 04/10 5

Answers and Solutions1.Rate Time DistancePlane 1 270 t 270tPlane 2 210 t 210t(dist. traveled by Plane 1 after they pass) + (dist. traveled then by Plane 2) = 1200270t 210t1200t 2.5hours2. 3 miles/hour3.Rate Time DistanceCyclist 1 r 3 3rCyclist 2 r 64 4 r 6(distance traveled by Cyclist 1) = (distance traveled by Cyclist 2)3r4 r 63r 4r24r 24 miles/hour4. 57 miles/hour5. We note that when traveling upstream the boat will travel at its still-water speed minus the speedof the current (which works against it), and we note that when traveling downstream the boatwill travel at its still-water speed plus the speed of the current (which helps it along). Here we’llneed a variable, say c, for the speed of the current.Rate Time DistanceUpstream 20 c6220 c62Downstream 20 c9820 c98(time upstream) = (time downstream)62 9820 c 20 c62 20 c 98 20160c720c 4.5 miles/hour6. 12 km/hrRevised 04/10 6c

7. One method of solution:8.215 hr (= 8 minutes) Rate Time DistancePriusd1d 1 dPintod33 d(rate at which Prius travels) = (rate at which Pinto travels) + 80 miles/hourdd3803d d 240d 120The cars meet 120 miles west of point X.An alternate solution method:Rate Time DistancePrius r 801 r 80Pinto r 3 3r(distance traveled by Prius) = (distance traveled by Pinto)r 80 3rr 40We now know that the rate at which the Pinto traveled is 40 miles/hour. But we are asked for thedistance that it and the Prius have traveled from the point X. We substitute 40 for r in eitherdistance expression: 3r becomes 3 40 120.The cars meet 120 miles west of point X.Revised 04/10 7

9. We need compatible units. Since the rates are given in miles/hour, we convert 150 minutes to1 52 2 2hours.One method of solution:Rate Time DistanceSecondaryd40dRoads40160 dTurnpike 70160 d70(time on secondary roads) + (time on turnpike) = 5 2d 160 d 540 70 27d4 160 d 700d20(Multiply each side by the LCD, 280.)SecondaryRoadsThus 20 miles were driven on secondary roads.An alternate solution method:Turnpike 70Rate Time Distance40 t 40t(distance on secondary roads) + (distance on turnpike) = 160 miles540t70 t 160240t 175 70t16015 30tt1hour252t570 2tThe time spent on secondary roads is thus a half-hour. We are asked for the number of milesdriven on secondary roads, though. We substitute the value of t into the expression representingthe distance driven on secondary roads:Revised 04/10 8

10. 3 miles140 20 miles driven on secondary roads2You can get additional instruction and practice by going to the following websites:http://www.purplemath.com/modules/distance.htmThis website gives eight additional examples.http://www.algebralab.org/Word/Word.aspx?file=Algebra_DistanceRateTimeI.xmlThis website gives an additional example and four practice problems.Revised 04/10 9