Question 453533: A child is running on a moving sidewalk in an airport. When he runs against the sidewalk's motion, he travels 48 feet in 12 seconds. When he runs with the sidewalk's motion, he travels 120 feet in the same amount of time. What is the rate of the child on a still sidewalk and what is the rate of the moving sidewalk?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A child is running on a moving sidewalk in an airport. When he runs against the sidewalk's motion, he travels 48 feet in 12 seconds. When he runs with the sidewalk's motion, he travels 120 feet in the same amount of time. What is the rate of the child on a still sidewalk and what is the rate of the moving sidewalk?
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Against-motion DATA:
distance = 48 ft ; time = 12 sec ; rate = d/t = 4 ft/sec
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With-motion DATA:
distance = 120 ft; time = 12 sec ; rate = d/t = 10 ft/sec
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Equations:
Let rate of child be "c"
Let rate of sidewalk be "s"
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c + s = 10 ft/sec
c - s = 4 ft/sec
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Add and solve for "c":
2c = 14
c = 7 ft/sec (child's rate)
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Solve for "s":
c+s = 10
7 + s = 10
s = 3 ft/sec (sidewald rate)
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Cheers,
Stan H.
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