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Rowing with the current, a canoeist paddled 10 mi in 2 h. Against the current, the canoeist could paddle only 6 mi in the same amount of time.
Find the rate of the canoest in calm water and the rate of the current.
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Let u be the the canoe rate in still water, in mph.
Let v be the rate of the current.
Then your governing equations are
= u + v (1) (the rate paddling with the stream)
= u - v (2) (the rate paddling against the srteam)
Simplify:
u + v = 5, (3)
u - v = 3. (4)
To solve, add the equations (3) and (4). You will get
2u = 5 + 3 ---> 2u = 8 ---> u = 4.
Thus you found the canoe rate in still water. It is 4 mph.
Then from (3) v = 5 - u = 5-4 = 1.
Answer. the canoe rate in still water is 4 mph. The rate of the current is 1 mph.
It is a typical Upstream and Downstream round trips word problem.
You can find similar fully solved similar problems on upstream and downstream round trips with detailed solutions in the lessons
- Wind and Current problems
- More problems on upstream and downstream round trips
- Selected problems from the archive on the boat floating Upstream and Downstream
Read them attentively and learn how to solve this type of problems once and for all.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this textbook under the section "Word problems", the topic "Travel and Distance problems".