SOLUTION: at 9am a boy rides his bike due north from campus at 16 mph. two hours later another boy rides his bike due east from the same point, also at 16 mph. when will they be 160 miles ap
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Question 787497: at 9am a boy rides his bike due north from campus at 16 mph. two hours later another boy rides his bike due east from the same point, also at 16 mph. when will they be 160 miles apart? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! at 9am a boy rides his bike due north from campus at 16 mph. two hours later another boy rides his bike due east from the same point, also at 16 mph. when will they be 160 miles apart?
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North-boy DATA:
rate = 16 mph ; time = x hrs. ; distance = 16x miles
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East-boy DATA:
rate = 16 mph ; time = x-2 hrs. ; distance = 16(x-2) miles
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Use Pythagoras to solve for "x":
(16x)^2 + (16(x-2))^2 = 160^2
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256x^2 + 256(x^2-4x+4) = 25600
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x^2 + x^2-4x+4 = 100
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2x^2 - 4x - 96 = 0
x^2 - 2x - 48 = 0
Factor:
(x-8)(x+6) = 0
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Positive solution:
x = 8 hours
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Answer: 9am + 8 hrs = 5pm
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Cheers,
Stan H.
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