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Rational Expressions Chapter 10

Rational Expressions Chapter 10. Distance – Rate - Time. T. Q. W. H. d = r t. Motion Problems. Motion problems involve distance, time and rate . The equation that links these concepts is called The Distance Formula:. miles kilometers meters feet inches. miles/hour

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Rational Expressions Chapter 10

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  1. Rational ExpressionsChapter 10 Distance – Rate - Time T Q W H

  2. d = r t Motion Problems Motion problems involve distance, time and rate. The equation that links these concepts is called The Distance Formula: • miles • kilometers • meters • feet • inches • miles/hour • km/min. • m/s • ft./sec. • inches/sec. • hours • minutes • seconds • days • years d = distance r = rate t = time

  3. d = r t Motion Problems Motion problems involve distance, time and rate. The equation that links these concepts is called The Distance Formula: • miles/hour • km/min. • m/s • ft./sec. • inches/sec. d = distance t = time r = rate You’ll need to be comfortable with using and manipulating the distance formula.

  4. Rational ExpressionsChapter 10 Distance – Rate - Time

  5. Here is a strategy to solve motion problems 1. Read the problem (three times) picking out key information 2. Draw a diagram of what is happening. 3. Make a table that relates distance, rate and time. 4. Use the table to write an equation. 5. Answer the question – include the units!

  6. Example 1 A bicyclist travels 8 miles per hour faster than a walker. The cyclist travels 54 miles in the same time it takes the walker to walk 18 miles. Find their speeds. x = speed of walker x + 8 = speed of cyclist 18 miles 54 miles Walker 18 Biker 54

  7. Example 1 A bicyclist travels 8 miles per hour faster than a walker. The cyclist travels 54 miles in the same time it takes the walker to walk 18 miles. Find their speeds. Time Walker Time Biker Walker 18 Biker 54

  8. Example 1 A bicyclist travels 8 miles per hour faster than a walker. The cyclist travels 54 miles in the same time it takes the walker to walk 18 miles. Find their speeds.

  9. # 2 One car travels 20 mph faster than the other car. While the faster car goes 240 miles the other car travels 180 miles. Find their speeds. 180 240 x = speed of red (slower) car Slow car 180 x 240 x + 20 Fast car

  10. # 2 One car travels 20 mph faster than the other car. While the faster car goes 240 miles the other car travels 180 miles. Find their speeds. Time Slow Car Time Fast Car x = speed of red (slower) car Slow car 180 x 240 x + 20 Fast car

  11. # 2 One car travels 20 mph faster than the other car. While the faster car goes 240 miles the other car travels 180 miles. Find their speeds.

  12. Wind Example 3 An airplane can fly at a speed of 240 m.p.h. in calm air. It takes the same time to fly 1080 miles with the wind as it does 840 miles against the wind. Find the speed of the wind. speed in “calm air” = 240 d = 1080 miles w = ? d = 840 miles

  13. Example 3 An airplane can fly at a speed of 240 m.p.h. in calm air. It takes the same time to fly 1080 miles with the wind as it does 840 miles against the wind. Find the speed of the wind. Let w = speed of wind With 1080 240 + w Against 240 - w 840 Time against wind Time with wind

  14. Homework #1 The speed of a freight train is 14 km/h slower than the speed of a passenger train. The freight train travels 330 km in the same time that it takes the passenger train to travel 400 km. Find the speed of each train. #2 One car travels 40 km/h faster than another. While one travels 150 km, the other goes 350 km. Find their speeds. A lab tested two high-speed trains. One travels 40 km/h faster than the other train. While one train travels 70 km, the other travels 60 km. Find their speeds. #3 #4 A person traveled 120 miles in one direction. The return trip was accomplished at double the speed and took 3 hours less time. Find the speed going

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