You can put this solution on YOUR website! Bill rowed 30 km upstream against a 2 km/h current and back again in 8 total hours. How fast can he row in still water?
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Let s = his speed in still water
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Then
(s+2) = speed with the current
(s-2) = speed against the current
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Write a time equation: time = dist/speed; (dist given as 30):
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time up + time back = 8 + = 8
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Eliminate the denominators, mult eq by (s-2)(s+2), resulting in:
30(s+2) + 30(s-2) = 8(s+2)(s-2)
30s + 60 + 30s - 60 = 8(s^2 - 4); FOILed
60s = 8s^2 - 32
0 = 8s^2 - 60s - 32
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A quadratic equation:
8s^2 - 60s - 32 = 0
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Simplify, divide eq by 4
2s^2 - 15x - 8
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Factors to:
(2s + 1)(s - 8) = 0
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s = + 8 km/hr in still water is the solution we want
:
:
Check:
30/6 + 30/10 =
5 + 3 = 8
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Make sense to you?
