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Applied Problems: Motion

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

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Applied Problems: Motion

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  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 Let’s 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 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 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 aD = RT gridwhensolving 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 GridTo 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.

  7. D=RT Grid

  8. D=RT Grid

  9. D=RT Grid 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 Pat’s distance Police Car’s distance

  11. Draw A Picture Picture 30 miles Pat’s distance Police Car’s distance

  12. Use the picture to write an equation. 30 miles Pat’s distance Police Car’s distance Police Car’s = 30 miles + Pat’s Distance Distance

  13. Now fill in the equation using our chart. 30 miles Pat’s distance Police Car’s distance Police Car’s = 30 miles + Pat’s Distance Distance 100x = 30 + 85x

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

  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.)

  16. Rate Time Distance Juan 60 x 60x Amal 72 x 72x 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. Rate Time Distance Sherry Bob 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. When finished go to the next slide to see how you did.

  18. Rate Time Distance Sherry Bob 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? Let x = the time it takes Bob to catch Sherry x + 1/2 5(x + 1/2) 5 7 x 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’s jogging distance Sherry Bob The problem is finished when Bob catches up to Sherry. Thus, their distances must equal each other. 5(x + 1/2) = 7x 5/2 = 2x 2[5/2] = 2 (2x) 5 = 4x 5x + 5/2 = 7x x = 5/4 = 1.25 hrs

  20. Now it’s 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. 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. 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: 52x Tonya 39 mi 65x Freda Do you see the equation forming from our picture?

  27. 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? 52x Tonya 39 mi 65x Freda 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. 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: START 52x 65x Freda Tonya 39 miles Do you see the equation forming from this picture?

  30. 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. START 52x 65x Tonya 39 miles Freda 52x + 65x = 39 117x=39 X= 1/3 of an hour or 1/3 of 60 min = 20 minutes

  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. 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? t 60t 2.2 - t 48(2.2 – t) 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. Distance is again computed by formula R*T = D

  33. 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? t 60t 2.2 - t 48(2.2 – t) What is the picture for this problem? 60t 48(2.2 – t) Dist 1st leg Dist 2nd leg Total dist. 120 miles Do you see the equation from the picture? 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? 60t 48(2.2 – t) Dist 1st leg Dist 2nd leg 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 don’t know the speed. We do know something about time. Let’s see how we can fill in the chart.

  36. 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? 6x 6 5 5(x + 9) The problem is over when the drummer passes Dr. John at 6 PM, so how long has Dr. John been driving? How long has the drummer been driving? What is our picture? 6x Dr. John 5x + 45 Drummer Drummer passing Dr. John

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

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