Question 933519

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



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