Question 1026187
Moving walkways are common in airports. Drenell conducts an experiment in which he walks with a 126-foot walkway and then against the walkway. 
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It takes Drenell 18 seconds to walk the 126-foot walkway going with the walkway and it takes him 63 seconds to walk the walkway going against the walkway. What is Drenell's normal walking speed? 
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With the walkway DATA:
time = 18 sec ; dist = 126 ft ; rate = 126/18 = 7 ft/sec
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Against the walkway DATA:
time = 63 sec ; dist = 126 ft ; rate = 126/63 = 2 ft/sec
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Equation:
w = his walking speed ; s = speed of the walkalator
w + s = 7 ft/sec
w - s = 2 ft/sec
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Add and solve for "w"::
2w = 9 
walking speed = 4.5 ft/sec
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Solve for "s"::
w+s = 7 ft sec
4.5 + s = 7
s = 2.5 ft/sec (speed of the walkalator
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Cheers,
Stan H.
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