SOLUTION: when Marvin was 2 km. upstream from where he started his canoe trip, he passed a log floating with the current. After paddling upstream for one more hour, he paddled back and reach

Algebra ->  College  -> Linear Algebra -> SOLUTION: when Marvin was 2 km. upstream from where he started his canoe trip, he passed a log floating with the current. After paddling upstream for one more hour, he paddled back and reach      Log On


   

Question 549930: when Marvin was 2 km. upstream from where he started his canoe trip, he passed a log floating with the current. After paddling upstream for one more hour, he paddled back and reached his starting point just as the log arrived.
How fast was the current flowing?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
when Marvin was 2 km. upstream from where he started his canoe trip, he passed a log floating with the current. After paddling upstream for one more hour, he paddled back and reached his starting point just as the log arrived.
How fast was the current flowing?
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Draw the picture:
Upstream DATA:
time = 1Downstream DATA:
distance = p-c+2 km ; rate = p+c km/h ; time = (p-c+2)/(p+c) hrs
---------
Log DATA:
rate = c km/h ; distance 2 km ; time = d/r = 2/c hrs
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Equation:
time up + time down = 2/c hrs
1 hr + (p-c+2)/(p+c) hr = 2/c hr
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Multiplthru by c(p+c) to get:
c(p+c) + c(p-c+2) = 2(p+c)
-----
cp + c^2 + cp-c^2+2c = 2p+2c
----
2cp = 2p
cp = p
current speed is 1 km/hr
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Cheers,
Stan H.
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