Question 182476: Fred paddled for 5 hours with 7km/hr current to reach a campsite. The return trip against the same current took 10 hours. Find the speed of Fred's canoe in still water?
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Fred paddled for 5 hours with 7km/hr current to reach a campsite. The return trip against the same current took 10 hours. Find the speed of Fred's canoe in still water?
----------------------------
He traveled the same distance, s km.
Call the canoe speed c, to save typing.
The time is the distance/rate
Going downstream, his rate (r) was c+7
Going upstream, his rate was c-7
The time downstream, 5 hrs = s/(c+7)
The time upstream, 10 hrs = s/(c-7)
Now there are 2 equations in 2 unknowns, so we're home free. Almost.
5 hrs = s/(c+7)
10 hrs = s/(c-7)
s = 5(c+7) = 5c+35
s = 10(c-7) = 10c-70
Since they both equal s, they equal each other
5c+35 = 10c-70
5c = 105
c = 21 km/hr (the speed of the canoe)
---------------
Tho it wasn't asked for, the distance is 140 kms. Going 21km/hr (12.6 mph) for 140 km is stepping right along. I'd check for a motor or steroids.
|
|
|
