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In this lesson we begin our study of the solution of uniform motion word problems. We study these problems because solving them helps us develop useful.

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Presentation on theme: "In this lesson we begin our study of the solution of uniform motion word problems. We study these problems because solving them helps us develop useful."— Presentation transcript:

1 Lesson 22: Uniform motion problems, equal distances, similar triangles and proportions

2 In this lesson we begin our study of the solution of uniform motion word problems. We study these problems because solving them helps us develop useful problem-solving skills.

3 Uniform motion problems involve statements about people or things that move at a constant velocity (speed) or at an average velocity. The uniform motion problems on which we will concentrate will contain four statements about things that are equal or that differ by a specified amount.

4 Each of these statements can be turned into an equation
Each of these statements can be turned into an equation. We find that these equations are easy to write if we use four variables. In this book we will use subscripted variables. For example: R or R or R To stand for the rate that Ruby walked. W RW R

5 To solve uniform motion problems we will write equations about rate or velocity, equations about time, and equations about distances traveled. Since the distance equations are the most difficult to write, we will consider these equations to be the key equations and we will write the distance equation first.

6 We will always draw a diagram to help us write the distance equation
We will always draw a diagram to help us write the distance equation. If both people or both things travel the same distance, the diagrams will the same distance.

7 Distance = Rate x Time

8 Example: Roger made the trip on Sunday, and Judy made the same trip on Monday. Roger traveled at 12 mph. Judy traveled at 20 mph, so her time was 2 hours less than Roger’s time. How far did they travel?

9 Answer: D = D R = 12 R = 20 T + 2 = T D = 60 miles
R J R J J R

10 Example: Brier Rabbit hopped off toward the briar patch at 10 kilometers per hour at 10 a.m. At noon Brier Wolf began the chase from the same starting point. If he caught Brier Rabbit at 2 p.m. and ate him alive, how fast did he run?

11 Answer: D = D R = 10 T = 4 T = 2 R = 20 kph
R W R W

12 Example: Solve for x.

13 Answer: Scale Factor = 2 x = 20

14 HW: Lesson 22 #1-30


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