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tankster tankster
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7 years ago
Arthur and Betty start walking toward each other when they are 100 m apart. Arthur has a speed of 3.0 m/s and Betty has a speed of 2.0 m/s. Their dog, Spot, starts from Arthur's side at the same time and runs back and forth between them. By the time Arthur and Betty meet, what is Spot's displacement?
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Physics

Physics


Edition: 4th
Author:
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longhairlonghair
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#2
4 years ago
Arthur and Betty start walking toward each other when they are 613 ft apart. Arthur has a speed of 4.3 m/s and Betty has a speed of 2.7 m/s. Their dog, Spot, starts from Arthur's side at the same time and runs back and forth between them. By the time Arthur and Betty meet, what is Spot's displacement?
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#3
4 years ago
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Arthur and Betty start walking toward each other when they are 613 ft apart. Arthur has a speed of 4.3 m/s and Betty has a speed of 2.7 m/s. Their dog, Spot, starts from Arthur's side at the same time and runs back and forth between them. By the time Arthur and Betty meet, what is Spot's displacement?

When both Arthur and Betty meet, both of them walked for the same amount of time.

=> Time taken by Arthur = Time taken by Betty

=> Distance traveled by Arthur / speed of Arthur = Distance traveled by Betty / speed of Betty

=> x1 / (4.3m/s) = x2 / (2.7m/s)

=> x2 = 27 x1 / 43 - - - -equation 1

Total distance traveled by both = initial distance between both

=> x1 + x2 = x

=> x1 + x2 = 613 ft

=> x2 = 613 ft - x1 - - - - equation 2

Equating the value of x2 from equations 1 and 2,

27 x1 / 43 = 613 ft - x1

=> x1 (1 + 27/43) = 613 ft

=> x1 = 376.56 ft or 114.77 m

Spot's displacement is the shortest distance between his initial and final position. Which equals x1 as shown in the figure.

Thus, Spot's displacement = x1 = 376.56 ft or 114.77 m
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