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Algebra 1 Section 3.7.

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Presentation on theme: "Algebra 1 Section 3.7."— Presentation transcript:

1 Algebra 1 Section 3.7

2 Uniform Motion Uniform motion problems involve one or more objects traveling at constant speed. Formula: d = rt

3 Be sure to use the right units!
d = rt rate × time = distance r t = d 50 45 3 t 5 = 150 50t 225 Be sure to use the right units!

4 Example 1 Both groups travel the same distance.
The high school travels at 5 mi/hr for an “unknown” time [x]. The junior high travels at 3 mi/hr for one hour longer than the high school [x + 1].

5 Since the distances are equal:
Example 1 r t = d junior high high school 3 5 t + 1 t = 3(t + 1) 5t Since the distances are equal: 3(t + 1) = 5t

6 Example 1 3(t + 1) = 5t 3t + 3 = 5t 3 = 2t t = 1½
The high-school class needs 1½ hours to catch up with the junior-high class.

7 Solving Uniform Motion Problems
Read the problem carefully. Plan a strategy. Solve an equation. Check.

8 For both trains, t = 3 hours
Example 2 Northbound train rate = r + 5 Total distance: 333 mi. Mattoon Southbound train rate = r For both trains, t = 3 hours

9 Since total distance is 333 mi:
Example 2 r t = d northbound southbound r + 5 r 3 = 3(r + 5) 3r Since total distance is 333 mi: 3(r + 5) + 3r = 333

10 Example 2 3(r + 5) + 3r = 333 3r + 15 + 3r = 333 6r + 15 = 333

11 Example 2 r = 53 The southbound train travels at a rate [r ] of 53 mi/hr. The northbound train travels at a rate [r + 5] of 58 mi/hr.

12 Example 3 Both runners run for the same amount of time [t ].
Jon travels 400 m more than Paul. Jon’s distance Paul’s distance 400

13 Since Jon’s distance is 400 m greater than Paul’s distance:
Example 3 r t = d Jon Paul 375 325 t = 375t 325t Since Jon’s distance is 400 m greater than Paul’s distance: 375t = 325t + 400

14 Example 3 375t = 325t + 400 50t = 400 t = 8 minutes

15 Example 3 t = 8 minutes Both runners ran for 8 minutes.
Jon ran 375(8) = 3000 m, which equals 7.5 laps. Paul ran 325(8) = 2600 m, which equals 6.5 laps.

16 Homework: pp


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