Question 933519: Two kilometers upstream from his starting point, a canoeist passed a log floating in the river's current. After paddling upstream for one more hour, he paddled back and reached his starting point just as the log arrived. Find the speed of the current.
LOG
---------->(2km)
PERSON
<---------
-------------------->
This is the only question that I can't understand out of all the practice problems my teacher gave me,please help!
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website!
Two kilometers upstream from his starting point, a canoeist passed a log floating in the river's current.
rate of log = rate of current =x
After paddling upstream for one more hour,
Let rate of canoe = y
time upstream = 1 hour
so distance upstream from the crossing point = y-x
Time taken by log to go 2 km to reach starting point = time taken by canoe to go downstream to starting point
2/x = (y-x)/(y-x) + (y-x+2)/(y+x)
2/x = 1+(y-x+2)/(y+x)
2/x = (y+x+y-x+2)/(y+x)
2/x= 2(y+1)/(y+x)
1/x = y+1/(y+x)
x+y =xy+x
y=xy
x=1
rate of current = 1km/h
m.ananth@hotmail.ca
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