# Rate-Time-Distance Problems Algebra Rate-Time-Distance Problems An object is in uniform motion when it moves without changing its speed, or rate. These.

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Rate-Time-Distance Problems Algebra

Rate-Time-Distance Problems An object is in uniform motion when it moves without changing its speed, or rate. These problems fall into 3 categories: Motion in opposite directions, Motion in the same direction, & Round Trip. Each is solved using a chart, a sketch, and the distance formula.

Time Speed and Distance Sunday, 28 November 2004 These quantities and found by using these formulae (rules): Distance = Time X Speed Speed = Distance ÷ Time Time = Distance ÷ Speed

Distance = Time X Speed Speed = Distance ÷ Time Time = Distance ÷ Speed D ST 1.A car travels at 30 mph for 2 hours. How far has it travelled? D = S x T  D = 30 X 2  D = 60 miles 2.A cyclist travels 45 miles in 3 hours. What is the cyclist’s speed? S = D ÷ T  S = 45 ÷ 3  S = 15 mph 3.A plane covers a distance of 1200 miles at a speed of 300 mph. How long will it take to complete this journey? T = D ÷ S  T = 1200 ÷ 300  T = 4 hours

If the time is in hours and minutes, change it to hours and decimal parts of a hour like this: 2 hours 24 minutes = 2. 4 24 ÷ 60 = 0.4

Motion in the Same Direction A helicopter leaves Central Airport and flies north at 180 mi/h. Twenty minutes later a plane leaves the airport and follows the helicopter at 330 mi/h. How long does it take the plane to overtake the helicopter. RateTimeDistance Helicopter Plane 330 180 t t + ⅓ 330t 180(t + ⅓)

RateTimeDistance Helicopter Plane 180 330 t + ⅓ t 180(t + ⅓) 330t When the plane overtakes the helicopter, the two distances are equal.

Officer Barbrady left his home at 2:15 PM and had driven 60 miles when he ran out of gas. He walked 2 miles to a gas station, where he arrived at 4:15 pm. If he drives 10 times as faster than he walks, how fast does he walk? DistanceRateTime Driving Walking r 10r 60 2

Set to Equal Office Barbrady walked at a rate of 4/mph

Two cyclists are traveling in the same direction on the same bike path. One travels at 10mph and other at 12mph. After how many hours will they be ten miles apart?

First cyclist (10mph) Second cyclist (12mph)

1 st Cyclist 10mphX10x 2 nd Cyclist 12mphX12x

1 st Cyclist 10mphX10x 2 nd Cyclist 12mphX12x THE RATE: How fast they or it is traveling at. THE DISTANCE: How far they or it has traveled. THE DISTANCE is equal to THE RATE times THE TIME. THE TIME: How long it takes to go THE DISTANCE. Since you are trying to figure out what the time is it is “X”

Going back to the picture…………………. First cyclist (10mph) Second cyclist (12mph) The first cyclist plus ten miles isequal tothe second cyclist

First CyclistSecond Cyclist Distance (miles) apart from the two cyclist

1 st Cyclist 10mph 2 nd Cyclist 12mph

Two cars leave the same town at the same time. One travels north at 60 mph and the other south at 45 mph. In how many hours will they be 420 miles apart? Bob Sherry DistanceTimeRate Car #1 Car #2 60 45 t t 60t 45t

Bob Sherry DistanceTimeRate Car #1 Car #2 60 45 t t 60t 45t In 4 hours the cars will be 420 miles apart.

Motion R T = D Two cars leave the same town at the same time. One travels north at 60 mph and the other south at 45 mph. In how many hours will they be 420 miles apart? R T =D 1 st car – (60 mph) (time) = distance 2 nd car – (45 mph) (time) = distance

Motion R T = D Two cars leave the same town at the same time. One travels north at 60 mph and the other south at 45 mph. In how many hours will they be 420 miles apart? R T =D 1 st car – (60 mph) (time) = distance 2 nd car – (45 mph) (time) = distance Let t = time Distance apart = 420 = 60t + 45t

Motion R T = D Two cars leave the same town at the same time. One travels north at 60 mph and the other south at 45 mph. In how many hours will they be 420 miles apart? 1 st car – 60 mph time (t) = distance 2 nd car – 45 mph time (t) = distance Distance apart = 420 = 60t + 45t 105t = 420 t = 4 hours

Check Motion R T = D Two cars leave the same town at the same time. One travels north at 60 mph and the other south at 45 mph. In how many hours will they be 420 miles apart? CHECKt = 4 hours 60 · 4 + 45 · 4 = 420 ?? 240 + 180 = 420OK They will be 420 miles apart in 4 hours.

Bicyclist Brent and Jane started at noon from 60 km apart and rode toward each other, meeting at 1:30 P.M. Brent’s speed was 4 km/h greater than Jane’s speed. Find their speeds. RateTimeDistance Brent Jane r r + 41.5 1.5(r + 4) 1.5r

RateTimeDistance Brent Jane r + 4 r 1.5 1.5(r + 4) 1.5r

Round Trip A ski lift carried Maria up a slope at the rate of 6 km/h, and she skied back down parallel to the lift at 34 km/h. The round trip took 30 min. How far did she ski? RateTimeDistance Up Down 34 6 t 0.5 - t 34 t 6(0.5 – t)

RateTimeDistance Up Down 6 34 0.5 - t t 6(0.5 – t) 34 t In round trip problems, the two distances are equal. Maria skied for 0.075h, or 4.5 min, for a distance of 2.55km.

Rate-Time-Distance Problems Review Additional Problems Part B

Sherry and Bob like to jog in the park. Sherry can jog at 5 mph, while Bob can jog at 7 mph. If Sherry starts 30 minutes ahead of Bob, how long will it take Bob to catch up to Sherry? Bob Sherry DistanceTimeRate 5 7 x Let x = the time it takes Bob to catch Sherry x + 1/25(x + 1/2) 7x Sherry started 30 minutes or 1/2 hour before Bob, so her time must reflect that amount. Rate is in miles per hour, time must be in hours so our units match.

When Bob catches-up with Sherry their distances will be equal. So the Equation will be sherry’s distance equals Bob’s distance. It will take Bob 1.25 hrs to catch up with Sherry. Bob Sherry DistanceTimeRate 5 7 x x + 1/25(x + 1/2) 7x

Bernadette drove 120 miles. The first part of the trip she averaged 60 mph, but on the second part of the trip she ran into some congestion and averaged 48 mph. If the total driving time was 2.2 hours, how much time did she spend at 60 mph? RateTimeDistance Part 1 Part 2 60 482.2-t t 60t 48(2.2-t)

RateTimeDistance Part 1 Part 2 60 48 2.2-t t60t 48(2.2-t) The time traveled at 60 mph was 1.2 hours.

A passenger train’s speed is 60 mi/h, and a freight train’s speed is 40 mi/h. The passenger train travels the same distance in 1.5 h less time than the freight train. How long does each train take to make the trip. DistanceTimeRate Freight train 4.5h. Passenger train 3h

Ali rode her bike to visit a friend. She traveled at 10 mi/h. While she was there, it began to rain. Her friend drove her home in a car traveling 25 mi/h. Ali took 1.5 hours longer to go to her friend’s than to return home. How many hours did it take Ali ti ride to her friend’s house? DistanceTimeRate 2.5 h

Katie rides her bike the same distance as Jill walks. Katie rides her bike 10 km/h faster than Jill walks. If it takes Katie 1 h. and Jill 3 h to travel the same distance, how fast does each travel? DistanceTimeRate Katie 15 km/ h Jill 5 km/h

At 10:00 A.M., a car leaves a house at a rate of 60 mi/h. At the same time, another car leaves the same house at a rate of 50 mi/h in the opposite direction. At what time will the car be 330 miles apart? DistanceTimeRate 1:00 P.M.

Brittney begins walking ay 3 mi/h toward the library. Her friend meets her at the halfway point and drives her the rest of the way to the library. The distance to the library is 4 miles. How many hours did Marla walk? DistanceTimeRate 2/3 h or 40 min.

Ryan begins walking towards John’s house at 3 mi/h. John leaves his house at the same time and walks towards Fred’s house on the same path at a rate of 2 mi/h. How long will it be before they meet if the distance between the houses is 4 miles? DistanceTimeRate 4/5 h. or 48 min

A train leaves the station at 6:00 P.M. traveling west at 80 mi/h. On a parallel track, a second train leaves the station 3 hours later traveling west at 100 mi/h. At what time will the second train catch up with the first? DistanceTimeRate 9:00 A.M.

It takes 1 hour longer to fly to St. Paul at 200 mi/h than it does to return at 250 mi/h. How far way is St. Paul. DistanceTimeRate 1000 mi

Motion problems use the equation D = RT where D is the distance traveled, R is the rate when traveling and T is the time spent traveling. It is helpful to use a D = RT grid when solving motion problems as shown in the following example.

Juan and Amal leave DC at the same time headed south on I-95. If Juan averages 60 mph and Amal averages 72 mph how long will it take them to be 30 miles apart? (Now would be a good time for a guess. Write yours down and try it in this table.) RateTimeDistance Juan60 x60x Amal72 x72x

x72Amal 60x x60Juan DistanceTimeRate We are looking for a time where Juan and Amal will be 30 miles apart. How do we represent the distance between the two men? 60x - 72x Or is it 72x - 60x. Which of these two would be positive? The correct equation is 72x - 60 x = 30

The purpose of the grid is to find an algebraic name for each distance. Notice that the distance 30 miles does not appear in the grid because neither Juan nor Amal traveled 30 miles. Notice also that we could use x for each time since Juan and Amal were on the road for the same amount of time. We will need to work 30 miles into the equation as follows: 72x – 60x = 30 12x = 30 x = 2.5 hrs. RateTime Distance Juan60x60x Amal72x72x

Practice Problems: Motion 1. Tonya and Freda drive away from Norfolk on the same road in the same direction. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? 2. Tonya and Freda drive away from Norfolk on the same road in opposite directions. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? Round your answer to the nearest minute.

For worked out solutions click to next slide.

1. Tonya and Freda drive away from Norfolk on the same road in the same direction. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? Let x = time it takes for them to be 39 miles apart. Construct a table to put your information in.

RateTimeDistanc e Tonya52x52x Freda65x65x Tonya and Freda drive away from Norfolk on the same road in the same direction. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? We let x stand for the time they have been driving which would be the same in this case. How far they have gone (distance) is written in the chart by using the known information and the formula R*T=D

RateTimeDistanc e Tonya52x52x Freda65x65x Tonya and Freda drive away from Norfolk on the same road in the same direction. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? Tonya and Freda are headed in the same direction so we can picture their distances as this: Tonya Freda 39 mi 52x 65x Do you see the equation forming from our picture?

RateTimeDistanc e Tonya52x52x Freda65x65x Tonya and Freda drive away from Norfolk on the same road in the same direction. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? Tonya Freda 39 mi 52x 65x One way to look at it might be to say 52x + 39 = 65x Or another way might be to say 65x – 52x = 39 Both are correct and will give you the correct answer of 3 hrs.

2. Tonya and Freda drive away from Norfolk on the same road in opposite directions. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? Round your answer to the nearest minute. Now they are going in opposite directions, but we will begin the same way by constructing our table.

RateTimeDistanc e Tonya52x52x Freda65x65x Tonya and Freda drive away from Norfolk on the same road in opposite directions. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? Round your answer to the nearest minute. As you can see the table is the same, so only the picture of the event must change. Now it looks like this: Tonya Freda 52x65x Do you see the equation forming from this picture? STARTSTART 39 miles

RateTimeDistanc e Tonya52x52x Freda65x65x Tonya and Freda drive away from Norfolk on the same road in opposite directions. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? Round your answer to the nearest minute. Tonya Freda 52x65x 52x + 65x = 39 117x=39 X= 1/3 of an hour or 1/3 of 60 min = 20 minutes STARTSTART 39 miles

3. Bernadette drove 120 miles. The first part of the trip she averaged 60 mph, but on the second part of the trip she ran into some congestion and averaged 48 mph. If the total driving time was 2.2 hours, how much time did she spend at 60 mph? The question here deals with time. Again, lets Fill in the chart.

RateTimeDistanc e 1 st leg60 2 nd leg48 Bernadette drove 120 miles. The first part of the trip she averaged 60 mph, but on the second part of the trip she ran into some congestion and averaged 48 mph. If the total driving time was 2.2 hours, how much time did she spend at 60 mph? This time totals are given for both time traveled and distance traveled. Thus totals don’t belong in the chart. Bernadette did not travel 2.2 hours at 60 mph nor did she travel 2.2 hours at 48 mph. She traveled 2.2 hours total at both of those speeds. So how do we write this in the chart. Click to observe. t 2.2 - t 60t 48(2.2 – t) Distance is again computed by formula R*T = D

RateTimeDistanc e 1 st leg60 2 nd leg48 Bernadette drove 120 miles. The first part of the trip she averaged 60 mph, but on the second part of the trip she ran into some congestion and averaged 48 mph. If the total driving time was 2.2 hours, how much time did she spend at 60 mph? What is the picture for this problem? t 2.2 - t 60t 48(2.2 – t) Do you see the equation from the picture? Dist 1 st leg Dist 2 nd leg 60t48(2.2 – t) Total dist. 120 miles 60t +48(2.2 – t) = 120

Bernadette drove 120 miles. The first part of the trip she averaged 60 mph, but on the second part of the trip she ran into some congestion and averaged 48 mph. If the total driving time was 2.2 hours, how much time did she spend at 60 mph? Dist 1 st leg Dist 2 nd leg 60t48(2.2 – t) Total dist. 120 miles 60t +48(2.2 – t) = 120 60t + 105.6 – 48t = 120 12t = 14.4 t = 1.2 hours at 60mph

4. Dr. John left New Orleans at 12 noon. His drummer left at 1:00 traveling 9 mph faster. If the drummer passed Dr. John at 6:00, what was the average speed of each? Now we don’t know the speed. We do know something about time. Let’s see how we can fill in the chart.

RateTimeDistanc e Johnx Drumm er X + 9 Dr. John left New Orleans at 12 noon. His drummer left at 1:00 traveling 9 mph faster. If the drummer passed Dr. John at 6:00, what was the average speed of each? The problem is over when the drummer passes Dr. John at 6 PM, so how long has Dr. John been driving? 6 How long has the drummer been driving? 5 What is our picture? Dr. John Drummer Drummer passing Dr. John 6x 5(x + 9) 6x 5x + 45

RateTimeDistanc e Johnx6x Drumm er X + 95(x + 9) Dr. John left New Orleans at 12 noon. His drummer left at 1:00 traveling 9 mph faster. If the drummer passed Dr. John at 6:00, what was the average speed of each? 6 5 Dr. John Drummer What equation does the picture suggest? The two distances are equal thus: 6x = 5(x + 9) 6x = 5x + 45 X = 45 mph for Dr. John X + 9 = 54 mph for the drummer

Car and Bus Motion R T = D When Michael drives his car to work, the trip takes ½ hour. When he rides the bus, it takes ¾ hour. The average speed of the bus is 12 mph less than his speed when driving. Find the distance he travels to work.

Car and Bus Motion R T = D When Michael drives his car to work, the trip takes ½ hour. When he rides the bus, it takes ¾ hour. The average speed of the bus is 12 mph less than his speed when driving. Find the distance he travels to work. Let D = distance to work

Car and Bus Motion R T = D When Michael drives his car to work, the trip takes ½ hour. When he rides the bus, it takes ¾ hour. The average speed of the bus is 12 mph less than his speed when driving. Find the distance he travels to work. Let D = distance to work R = D/T

Car and Bus Motion R T = D When Michael drives his car to work, the trip takes ½ hour. When he rides the bus, it takes ¾ hour. The average speed of the bus is 12 mph less than his speed when driving. Find the distance he travels to work. Let D = distance to work R = D/TR(car) = D / (½) R(bus) = D / (¾) Equation?

Car and Bus Motion R T = D When Michael drives his car to work, the trip takes ½ hour. When he rides the bus, it takes ¾ hour. The average speed of the bus is 12 mph less than his speed when driving. Find the distance he travels to work. Let D = distance to work R = D/TR(car) = D / (½) = 2D R(bus) = D / (¾) = 4D/3 Equation

Car and Bus Motion R T = D When Michael drives his car to work, the trip takes ½ hour. When he rides the bus, it takes ¾ hour. The average speed of the bus is 12 mph less than his speed when driving. Find the distance he travels to work. Let D = distance to work R = D/TR(car) = D / (½) = 2D R(bus) = D / (¾) = 4D/3 Equation

Check Motion R T = D When Michael drives his car to work, the trip takes ½ hour. When he rides the bus, it takes ¾ hour. The average speed of the bus is 12 mph less than his speed when driving. Find the distance he travels to work. Check D=18 The distance to work is 18 miles.

(Rate)(Time) = Distance A truck driver delivered a load to Richmond from Washington, D.C., averaging 50 mph for the trip. He picked up another load in Richmond and delivered it to NYC, averaging 55 mph on this part of the trip. The distance from Richmond to NYC is 120 miles more than the distance from Washington to Richmond. If the entire trip required a driving time of 6 hours, find the distance from Washington to Richmond.

Rate × Time = Distance A truck driver delivered a load to Richmond from Washington, D.C., averaging 50 mph for the trip. He picked up another load in Richmond and delivered it to NYC, averaging 55 mph on this part of the trip. The distance from Richmond to NYC is 120 miles more than the distance from Washington to Richmond. If the entire trip required a driving time of 6 hours, find the distance from Washington to Richmond. Let x = distance Washington to Richmond

Rate x Time = Distance A truck driver delivered a load to Richmond from Washington, D.C., averaging 50 mph for the trip. He picked up another load in Richmond and delivered it to NYC, averaging 55 mph on this part of the trip. The distance from Richmond to NYC is 120 miles more than the distance from Washington to Richmond. If the entire trip required a driving time of 6 hours, find the distance from Washington to Richmond. Let x = distance Washington to Richmond RateTimeDistance W-R50 R-NYC55

Rate x Time = Distance A truck driver delivered a load to Richmond from Washington, D.C., averaging 50 mph for the trip. He picked up another load in Richmond and delivered it to NYC, averaging 55 mph on this part of the trip. The distance from Richmond to NYC is 120 miles more than the distance from Washington to Richmond. If the entire trip required a driving time of 6 hours, find the distance from Washington to Richmond. Let x = distance Washington to Richmond RateTimeDistance W-R50x R-NYC55x + 120

Rate x Time = Distance A truck driver delivered a load to Richmond from Washington, D.C., averaging 50 mph for the trip. He picked up another load in Richmond and delivered it to NYC, averaging 55 mph on this part of the trip. The distance from Richmond to NYC is 120 miles more than the distance from Washington to Richmond. If the entire trip required a driving time of 6 hours, find the distance from Washington to Richmond. Let x = distance Washington to Richmond RateTimeDistance W-R50x R-NYC55x + 120 6 hours total

Rate x Time = Distance A truck driver delivered a load to Richmond from Washington, D.C., averaging 50 mph for the trip. He picked up another load in Richmond and delivered it to NYC, averaging 55 mph on this part of the trip. The distance from Richmond to NYC is 120 miles more than the distance from Washington to Richmond. If the entire trip required a driving time of 6 hours, find the distance from Washington to Richmond. Let x = distance Washington to Richmond RateTimeDistance W-R50x/50x R-NYC55(x+120)/55x + 120 6 hours total

Rate x Time = Distance A truck driver delivered a load to Richmond from Washington, D.C., averaging 50 mph for the trip. He picked up another load in Richmond and delivered it to NYC, averaging 55 mph on this part of the trip. The distance from Richmond to NYC is 120 miles more than the distance from Washington to Richmond. If the entire trip required a driving time of 6 hours, find the distance from Washington to Richmond. Let x = distance Washington to Richmond RateTimeDistance W-R50x/50x R-NYC55(x+120)/55x + 120 6 hours total x/50 + (x+120)/55 = 6

Rate x Time = Distance A truck driver delivered a load to Richmond from Washington, D.C., averaging 50 mph for the trip. He picked up another load in Richmond and delivered it to NYC, averaging 55 mph on this part of the trip. The distance from Richmond to NYC is 120 miles more than the distance from Washington to Richmond. If the entire trip required a driving time of 6 hours, find the distance from Washington to Richmond. Let x = distance Washington to Richmond RateTimeDistance W-R50x/50x R-NYC55(x+120)/55x + 120 6 hours total x/50 + (x+120)/55 = 6LCD = 550 11x +10x + 1200 = 3300

Rate x Time = Distance A truck driver delivered a load to Richmond from Washington, D.C., averaging 50 mph for the trip. He picked up another load in Richmond and delivered it to NYC, averaging 55 mph on this part of the trip. The distance from Richmond to NYC is 120 miles more than the distance from Washington to Richmond. If the entire trip required a driving time of 6 hours, find the distance from Washington to Richmond. Let x = distance Washington to Richmond RateTimeDistance W-R50x/50x R-NYC55(x+120)/55x + 120 6 hours total x/50 + (x+120)/55 = 6LCD = 550 11x +10x + 1200 = 3300 21x = 2100 x = 100 CHECK

Uniform Motion - Trucks 6-9Page 374 (Figure 6.1) Michele drove her empty rig 300 miles to Salina to pick up a load of cattle. When the rig was finally loaded, her average speed was 10 mph less than when the rig was empty. If the return trip took her 1 hour longer, then what was her average speed with the rig empty?

Uniform Motion - Trucks 6-9Page 374 (Figure 6.1) Michele drove her empty rig 300 miles to Salina to pick up a load of cattle. When the rig was finally loaded, her average speed was 10 mph less than when the rig was empty. If the return trip took her 1 hour longer, then what was her average speed with the rig empty? x = average empty speed

Uniform Motion - Trucks 6-9Page 374 (Figure 6.1) Michele drove her empty rig 300 miles to Salina to pick up a load of cattle. When the rig was finally loaded, her average speed was 10 mph less than when the rig was empty. If the return trip took her 1 hour longer, then what was her average speed with the rig empty? x = average empty speed RateTimeDistance To Salina Return

Uniform Motion - Trucks 6-9Page 374 (Figure 6.1) Michele drove her empty rig 300 miles to Salina to pick up a load of cattle. When the rig was finally loaded, her average speed was 10 mph less than when the rig was empty. If the return trip took her 1 hour longer, then what was her average speed with the rig empty? x = average empty speed RateTimeDistance To Salina300 Return300

Uniform Motion - Trucks 6-9Page 374 (Figure 6.1) Michele drove her empty rig 300 miles to Salina to pick up a load of cattle. When the rig was finally loaded, her average speed was 10 mph less than when the rig was empty. If the return trip took her 1 hour longer, then what was her average speed with the rig empty? x = average empty speed RateTimeDistance To Salinax 300 Returnx - 10300

Uniform Motion - Trucks 6-9Page 374 (Figure 6.1) Michele drove her empty rig 300 miles to Salina to pick up a load of cattle. When the rig was finally loaded, her average speed was 10 mph less than when the rig was empty. If the return trip took her 1 hour longer, then what was her average speed with the rig empty? x = average empty speed RateTimeDistance To Salinax 300/x300 Returnx - 10 300/(x - 10) 300 300/x + 1 = 300/(x - 10)

Uniform Motion - Trucks 6-9Page 374 (Figure 6.1) Michele drove her empty rig 300 miles to Salina to pick up a load of cattle. When the rig was finally loaded, her average speed was 10 mph less than when the rig was empty. If the return trip took her 1 hour longer, then what was her average speed with the rig empty? x = average empty speed RateTimeDistance To Salinax 300/x300 Returnx - 10 300/(x - 10) 300 300/x + 1 = 300/(x - 10) LCD = x (x - 10) 300 (x - 10) + x (x - 10) = 300 x

Uniform Motion - Trucks 6-9Page 374 (Figure 6.1) Michele drove her empty rig 300 miles to Salina to pick up a load of cattle. When the rig was finally loaded, her average speed was 10 mph less than when the rig was empty. If the return trip took her 1 hour longer, then what was her average speed with the rig empty? x = average empty speed RateTimeDistance To Salinax 300/x300 Returnx - 10 300/(x - 10) 300 300/x + 1 = 300/(x - 10) LCD = x (x - 10) 300 (x - 10) + x (x - 10) = 300 x 300 x - 3000) + x 2 – 10 x = 300 x x 2 – 10 x – 3000 = 0

Uniform Motion - Trucks 6-9Page 374 (Figure 6.1) Michele drove her empty rig 300 miles to Salina to pick up a load of cattle. When the rig was finally loaded, her average speed was 10 mph less than when the rig was empty. If the return trip took her 1 hour longer, then what was her average speed with the rig empty? x = average empty speed RateTimeDistance To Salinax 300/x300 Returnx - 10 300/(x - 10) 300 300/x + 1 = 300/(x - 10) LCD = x (x - 10) 300 (x - 10) + x (x - 10) = 300 x 300 x - 3000) + x 2 – 10 x = 300 x x 2 – 10 x – 3000 = 0 (x - 60) (x +50) =0

Uniform Motion - Trucks 6-9Page 374 (Figure 6.1) Michele drove her empty rig 300 miles to Salina to pick up a load of cattle. When the rig was finally loaded, her average speed was 10 mph less than when the rig was empty. If the return trip took her 1 hour longer, then what was her average speed with the rig empty? x = average empty speed RateTimeDistance To Salinax 300/x300 Returnx - 10 300/(x - 10) 300 300/x + 1 = 300/(x - 10) LCD = x (x - 10) 300 (x - 10) + x (x - 10) = 300 x 300 x - 3000) + x 2 – 10 x = 300 x x 2 – 10 x – 3000 = 0 (x - 60) (x + 50) =0 x = 60 x = - 50 CHECKNot Possible

Uniform Motion Driving to Florida Susan drove 1500 miles to Daytona Beach for spring break. On the way back she averaged 10 mph less, and the drive back took 5 hours longer. Find Susan’s average speed on the way to Daytona Beach.

Driving to Florida 6-10Page 355 (Figure 6.1)

Driving to Florida Susan drove 1500 miles to Daytona Beach for spring break. On the way back she averaged 10 mph less, and the drive back took 5 hours longer. Find Susan’s average speed on the way to Daytona Beach. Let = x average speed going x - 10 = average speed returning R ·T=D Goingx1500 Returning x – 101500

The Problem Two trains leave Houston at the same time, one traveling east, the other west. The first train traveled at 50 mph and the second at 40 mph. In how many hours will the trains be 405 miles apart?

Table train speed timetotal Train 1 Train 2 Rate x time = distance Rate of Train 1: 50 mph Rate of Train 2: 40 mph Time for both trains: t Total for train 1: 50t Total for train 2: 40 t 50 mph 40 mph t t 50t 40t

The Math Set trains totals to equal 405 miles Add both totals Divide both sides by 90 T = 4.5 hours This is our equation 50t+40t=405miles 90t =405miles 90 T=4.5hours 40t + 50t = 90t