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**Rational Expressions Chapter 10**

Distance – Rate - Time H W Q T

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**Motion Problems d = r t The Distance Formula:**

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. hours minutes seconds days years miles kilometers meters feet inches d = distance d = r t r = rate t = time

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**Motion Problems d = r t The Distance Formula:**

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 d = r t t = time r = rate You’ll need to be comfortable with using and manipulating the distance formula.

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**Rational Expressions Chapter 10**

Distance – Rate - Time

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**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. Distance = rate x time 5. Answer the question – include the units!

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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 Distance = rate x time Walker 18 Biker 54

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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 Distance = rate x time Walker 18 Biker 54

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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.

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# 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 Distance = rate x time Slow car 180 x 240 x + 20 Fast car

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**180 240 Time Time Slow Fast Car Car x x + 20**

# 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 Distance = rate x time Slow car 180 x 240 x + 20 Fast car

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# 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.

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**An airplane can fly at a speed of 240 m.p.h. in calm air. **

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 = ? Wind d = 840 miles

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**An airplane can fly at a speed of 240 m.p.h. in calm air. **

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 Distance = rate x time With 1080 240 + w Against 840 240 - w Time with wind Time against wind

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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. #3 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. #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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1.5 “Problem Solving Strategies” Examples: 1. A train travels between Boston and Washington, a distance of 457 miles. The trip takes 6.5 hours. What is.

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