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**Distance, Speed and Time**

Lesson 4.2.3

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**Distance, Speed and Time**

Lesson 4.2.3 Distance, Speed and Time California Standard: Algebra and Functions 4.2 Solve multistep problems involving rate, average speed, distance, and time or a direct variation. What it means for you: You’ll learn the formula for speed, and how to use it to solve problems. Key words: speed distance time formula

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**Distance, Speed and Time**

Lesson 4.2.3 Distance, Speed and Time Speed is a rate — it’s the distance you travel per unit of time. 55 miles per hour is the speed limit on some roads. If you drive steadily at this speed, you’ll travel 55 miles every hour. There’s a formula that links speed, distance, and time — and you’re going to use it in this Lesson.

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**Distance, Speed and Time**

Lesson 4.2.3 Distance, Speed and Time Speed is a Rate Speed is a rate. It is the distance traveled in a certain amount of time. 2 hours 10 miles per hour 20 miles Speed can be measured in lots of different units, such as miles per hour, meters per second, inches per minute... distance time speed = The formula for speed is:

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**Distance, Speed and Time**

Lesson 4.2.3 Distance, Speed and Time Example 1 Gila walked 6 miles in 8 hours. What was Gila’s average speed? Solution Use the formula, and substitute in the values from the question. distance time speed = = 6 miles 8 hours = (6 ÷ 8) miles per hour = 0.75 miles per hour Gila’s average speed was 0.75 miles per hour. Solution follows…

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**Distance, Speed and Time**

Lesson 4.2.3 Distance, Speed and Time Rearrange the Equation to Find Other Unknowns You can rearrange the speed formula, and use it to find distance or time. distance time speed = To change the equation into an equation that gives distance in terms of speed and time, multiply both sides of the equation by time. distance × time time speed × time = distance = speed × time You can find the equation for time in terms of speed and distance in a similar way.

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**Distance, Speed and Time**

Lesson 4.2.3 Distance, Speed and Time Example 2 Alyssa runs for 0.5 hours at a speed of 11 kilometers per hour. How far does she run? Solution Use the formula for distance, and substitute the values for speed and time. Distance = speed × time = 11 kilometers per hour × 0.5 hours = 5.5 kilometers Solution follows…

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**Distance, Speed and Time**

Lesson 4.2.3 Distance, Speed and Time Example 3 Andy is planning a walk. He walks at an average speed of 3 miles per hour, and plans to cover 15 miles. How long should his walk take him? Solution You need to rearrange the speed formula. distance = speed × time speed × time speed distance = Divide both sides by speed Solution continues… Solution follows…

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**Distance, Speed and Time**

Lesson 4.2.3 Distance, Speed and Time Example 3 Andy is planning a walk. He walks at an average speed of 3 miles per hour, and plans to cover 15 miles. How long should his walk take him? Solution (continued) Now you can use the formula to answer the question: distance speed time = 15 miles 3 mph = = (15 ÷ 3) hours = 5 hours Andy’s walk should take him 5 hours.

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**Distance, Speed and Time**

Lesson 4.2.3 Distance, Speed and Time Guided Practice 1. Juan ran in a marathon that was 26 miles long. If his time was 4 hours, what was his average speed? 2. Moesha goes to school every day by bike. The journey is 6 miles long, and takes her 0.6 hours. What is her average speed? 3. Monica travels 6 miles to work at a speed of 30 miles per hour. How long does the journey take her each morning? Speed = distance ÷ time = 26 ÷ 4 = 6.5 miles per hour Speed = distance ÷ time = 6 ÷ 0.6 = 10 miles per hour Time = distance ÷ speed = 6 ÷ 30 = 0.2 hours or 12 minutes Solution follows…

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**Distance, Speed and Time**

Lesson 4.2.3 Distance, Speed and Time Guided Practice Josh has been walking for 5 hours at a speed of 4 miles per hour. 4. His walk is 22 miles long. How far does he have left to walk? 5. How much longer will he take if he continues at the same speed? Find how far Josh has already walked. Distance = speed × time = 4 × 5 = 20 miles So Josh has 22 – 20 = 2 miles left to walk Time = distance ÷ speed = 2 ÷ 4 = 0.5 hours or 30 minutes Solution follows…

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**Distance, Speed and Time**

Lesson 4.2.3 Distance, Speed and Time Example 4 On a three-hour bike ride, a cyclist rode 58 miles. The first two hours were downhill, so the cyclist rode 5 miles per hour quicker than she did for the last hour. a) What was her speed for the first two hours? b) What was her speed for the last hour? Solution Let the cyclist’s speed for the first two hours be (x + 5) mi/h. So her speed for the last hour = x miles per hour. You need to write an equation using the information given. Solution continues… Solution follows…

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**Distance, Speed and Time**

Lesson 4.2.3 Distance, Speed and Time Example 4 On a three-hour bike ride, a cyclist rode 58 miles. The first two hours were downhill, so the cyclist rode 5 miles per hour quicker than she did for the last hour. a) What was her speed for the first two hours? b) What was her speed for the last hour? Solution (continued) Total distance = distance traveled in first two hours distance traveled in last hour 58 = (x + 5) × 2 x × 1 distance = speed × time 58 = 2x + 10 x Solution continues…

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**Distance, Speed and Time**

Lesson 4.2.3 Distance, Speed and Time Example 4 On a three-hour bike ride, a cyclist rode 58 miles. The first two hours were downhill, so the cyclist rode 5 miles per hour quicker than she did for the last hour. a) What was her speed for the first two hours? b) What was her speed for the last hour? Solution (continued) Solve the equation to find x: 58 = 2x x 58 = 3x + 10 Þ 48 = 3x Þ x = 16 a) The speed for the first two hours was (x + 5) = = 21 mi/h b) So the speed for the last hour was x = 16 mi/h

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**Distance, Speed and Time**

Lesson 4.2.3 Distance, Speed and Time Guided Practice 6. Train A travels 20 mi/h faster than Train B. Train A takes 3 hours to go between two cities, and Train B takes 4 hours to travel the same distance. How fast does each train travel? Let d = distance between the two cities Train A speed = d ÷ 3 Train B speed = d ÷ 4 Train A speed = Train B speed + 20 mi/h d ÷ 3 = (d ÷ 4) d = 3d Þ d = 240 mi Train A speed = 240 ÷ 3 = 80 mi/h, Train B speed = 240 ÷ 4 = 60 mi/h Solution follows…

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**Distance, Speed and Time**

Lesson 4.2.3 Distance, Speed and Time Independent Practice 1. A mouse ran at a speed of 3 meters per second for 30 seconds. How far did it travel in this time? 2. A slug crawls at 70 inches per hour. How long will it take it to crawl 630 inches? 3. A shark swims at 7 miles per hour for 2 hours, and then at 9 miles per hour for 3 hours. How far does it travel altogether? 90 meters 9 hours 41 miles Solution follows…

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**Distance, Speed and Time**

Lesson 4.2.3 Distance, Speed and Time Independent Practice 4. Bike J moves at a rate of x miles per hour for 2 hours. Bike K travels at 0.5x miles per hour for 4 hours. Which bike travels the furthest? 5. On a two-day journey, you travel 500 miles in total. On the first day you travel for 5 hours at an average speed of 60 mi/h. On the second day you travel for 4 hours. What’s your average speed for these 4 hours? Both travel the same distance 50 mi/h Solution follows…

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**Distance, Speed and Time**

Lesson 4.2.3 Distance, Speed and Time Round Up You need to remember the formula for speed. distance time speed = If you know this, you can rearrange it to figure out the formulas for distance and time when you need them — so that’s two less things to remember.

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Speed, Velocity, and Acceleration. Speed Speed: how fast something is moving Ex 1: A racecar travels at 90 miles per hour (mi/hr) Ex 2: A runner runs.

Speed, Velocity, and Acceleration. Speed Speed: how fast something is moving Ex 1: A racecar travels at 90 miles per hour (mi/hr) Ex 2: A runner runs.

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