SOLUTION: Hope you can help me! :)
When Neil was 2 miles upstream from camp on a canoe trip, he passed a log floating downstream with the current. He paddled upstream for 1 more hour and
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Question 934682: Hope you can help me! :)
When Neil was 2 miles upstream from camp on a canoe trip, he passed a log floating downstream with the current. He paddled upstream for 1 more hour and then returned to camp just as the log arrived. What was the rate of the current?
Found 2 solutions by mananth, Alan3354:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
let rate of canoe be x mph
rate of current = y
Time taken by log to reach starting point = 2/y.....................(1)
he canoes for 1 hour at speed of (x-y) upstream
so distance he paddles = x-y
Return
downstream speed = (x+y)
Time taken to reach starting point = time to reach the crossing point + time taken for 2 miles
=(x-y)/(x+y) +2/(x+y)
2/y = 1+(x-y)/(x+y) +2/(x+y)
2/y = (x+y+x-y+2)/(x+y)
2/y = (2x+2)/(x+y)
2/y = 2(x+2)/(x+y)
1/y =(x+1)/(x+y)
x+y = xy+y
x=xy
y=1
rate of current = 1mph
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
When Neil was 2 miles upstream from camp on a canoe trip, he passed a log floating downstream with the current. He paddled upstream for 1 more hour and then returned to camp just as the log arrived. What was the rate of the current?
-------------------
Neil and the log are moving in the same medium, the water, so the time moving away from the log is the same as the time to get back to it.
1 hour away from it and 1 hour back to it.
During that time, the log moved 2 miles.
--> current is 1 mi/hr
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