Explanation It is given that the bicyclists are 20 miles away from each other and travel 10 miles per hour each towards each other. The fly flies at a speed of 15 miles per hour. Let the fly meet the second bicyclist after t h. The distance covered by fly in t h is 15 miles per hour × t h = 15 t miles In this time t h, the second bicyclists would have covered 10 t miles. The total distance is 20 miles so that the value of t becomes 15 t + 10 t = 20 25 t = 20 t = 20 25 t = 4 5 In this time, the first bicyclist covers a distance of 10 mph × 4 5 h = 8 m . Now, the fly travels from second bicyclist to the first, thus the first bicyclist and the fly travels in a distance of 20 − 2 × 8 = 4 m. Let the fly meet the first bicyclist after t h. The distance covered by fly in t s is 15 miles per hour × t h = 15 t miles In this time t h, the second bicyclists would have covered 10 t miles. The total distance is 4 miles so that the value of t becomes 15 t + 10 t = 4 25 t = 4 t = 4 25 t = 4 5 2 In this time, the second bicyclist covers a distance of 10 mph × 4 5 2 h = 8 5 m . Again, the fly travels from first bicyclist to the second, thus the second bicyclist and the fly travels in a distance of 20 − 2 × 8 5 = 4 5 m. Let the fly meet the second bicyclist after t h. The distance covered by fly in t s is 15 miles per hour × t h = 15 t miles . In this time t h, the second bicyclists would have covered 10 t miles

Calculus With Applications, Books a la Carte Plus MyLab Math Package (11th Edition)

11th Edition

ISBN: 9780133886849

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 12.4, Problem 36E

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To solve: The distance covered by the fly until it is trapped between the wheels by both ways.

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Chapter 12.4, Problem 35E

Chapter 12.4, Problem 37E

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Calculus With Applications, Books a la Carte Plus MyLab Math Package (11th Edition)

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