Download presentation

Presentation is loading. Please wait.

Published byIsaiah Heckstall Modified over 3 years ago

1
Applied Problems: Motion By Dr. Marcia L.Tharp Dr. Julia Arnold

2
Applied Problems: Motion Motion problems use the equation D = RT where D is the distance traveled. R is the rate of travel. T is the time spent traveling.

3
Applied Problems: Motion Motion Problem 1 Lets suppose that you drive your car 120 miles in 2 hours than we can find the rate you traveled by using the distance equation. D = RT 60 = R Since 120 is a distance we let D = 120 Since the time you traveled is 2 hours we let T= 2. Placing this in the equation D=RT we have 120 = R 2 120 = R 2 2 So your rate is 60 miles per hour.

4
Motion Problem 2 Pat has been caught speeding by the airplane patrol. He is doing 85miles per hour. A police car who is 30 miles behind him can do 100 miles per hour to catch him. How long will it take the police car to catch Pat if they continue in the same direction at this speed? Print this page.

5
It is helpful to use a D = RT grid when solving motion problems as shown in the following example. The purpose of the grid is to find an algebraic name for each distance.

6
D=RT Grid To use this grid fill in the rate and time for each driver. We will let x = the time it takes for the police car to catch Pat. Then multiply the rate by the time to get the distance traveled. RateTimeDistance Pat Police Car

7
D=RT Grid RateTimeDistance Pat85 Police Car 100

8
D=RT Grid RateTimeDistance Pat85X Police Car 100X

9
D=RT Grid RateTimeDistance Pat85X85x Police Car 100X100x Notice that neither the policeman or Pat are represented by the 30 miles. We will use this later.

10
Next Draw A Picture Picture 30 miles Pats distance Police Cars distance

11
Draw A Picture Picture 30 miles Pats distance Police Cars distance

12
Use the picture to write an equation. 30 milesPats distance Police Cars distance Police Cars = 30 miles + Pats Distance

13
Now fill in the equation using our chart. 30 milesPats distance Police Cars distance Police Cars = 30 miles + Pats Distance 100x = 30 + 85x RateTimeDistance Pat 85X85x Police Car 100X100x

14
Next solve this equation. 100x = 30 + 85x So it took the police car 2 hours to catch Pat. 100x – 85x = 30 + 85x – 85x 15x = 30 15 15 x = 2 hours Subtract 85 x from both sides. Divide by 15 on each side.

15
Problem 3 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

16
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

17
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? Draw a grid like the one below and fill in what you know about Sherry and Bob. Bob Sherry DistanceTimeRate When finished go to the next slide to see how you did.

18
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.

19
Sherry Sherrys jogging distance Bob The problem is finished when Bob catches up to Sherry. Thus, their distances must equal each other. 5(x + 1/2) = 7x 5x + 5/2 = 7x 5/2 = 2x 2[5/2] = 2 (2x) 5 = 4x x = 5/4 = 1.25 hrs

20
Now its time to see what You can do.

21
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.

22
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? 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?

23
For worked out solutions click to next slide.

24
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.

25
RateTimeDistance 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

26
RateTimeDistance 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?

27
RateTimeDistance 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.

28
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.

29
RateTimeDistance 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

30
RateTimeDistance 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

31
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.

32
RateTimeDistance 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 dont 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

33
RateTimeDistance 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

34
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

35
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 dont know the speed. We do know something about time. Lets see how we can fill in the chart.

36
RateTimeDistance Johnx Drumme r 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

37
RateTimeDistance Johnx6x Drumme r 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

38
Back to Word Problem Menu Back to Math 03 Modules Now go on to the Geometry Lessons.

Similar presentations

OK

MT 8.3: Working with Rate, Time, and Distance The formula that has to be remembered for this section is… R ● T = D (Rate x Time = Distance) There are four.

MT 8.3: Working with Rate, Time, and Distance The formula that has to be remembered for this section is… R ● T = D (Rate x Time = Distance) There are four.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on schottky diode voltage Themes free download ppt on pollution Kids ppt on batteries Download ppt on motivational thing Ppt on isobars and isotopes of hydrogen Ppt on mars orbiter mission Ppt on stock market in india Ppt on famous indian entrepreneurs in america Ppt on layered network architecture Ppt on nature and human beings