SOLUTION: The riverboat Hiawatha goes 30 km upstream and back in 4 hours. If the Hiawatha's speed in still water is 16 km/h, what is the speed of the current?
I set this problem up on a c
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I set this problem up on a c
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Question 370581: The riverboat Hiawatha goes 30 km upstream and back in 4 hours. If the Hiawatha's speed in still water is 16 km/h, what is the speed of the current?
I set this problem up on a chart and had 30/16-r + 30/16+r = 4 and ended up getting r=16km/h. I am not quite sure if I set this problem up right. Answer by mananth(16949) (Show Source):
You can put this solution on YOUR website! 30/(16-r)+30/(16+r)=4
30(16+r)+30(16-r)/(256-r^2)=4
480+30r+480-30r/(256-r^2)=4
960=4(256-r^2)
/4
240=256-r^2
r^2=256-240
r^2=16
Take the square root
r = 4 mph.
I think you forgot to take the square root since r^2=16
...
m.ananth@hotmail.ca
