.
Let u be the speed of the tugboat in still water (in miles per hour).
Let v e the speed of the current.
Then the speed of the tugboat downstream is (u+v) miles per hour.
The speed of the tugboat upstream is (u-v) miles per hour.
From the other side, the speed downstream is = 12 mph. (it is the formula D =R*T, written in the form R = )
(Here R is the rate moving downstream.)
The speed upstream is = 8 hours. (Again, it is the formula D =R*T, written in the form R = )
(Here R is the rate moving upstream.)
So you have these two equations
u + v = 12, (1)
u - v = 8. (2).
Add the two equations. You will get
2u = 12+8 = 20 ====> u = = 10.
It is the speed of the tugboat in still water.
Now the speed of the current, from the equation (1), is v = 12-u = 12 - 10 = 2 mph.
Answer. The speed of the tugboat in still water is 10 mph.
The speed of the current is 2 mph.
Solved.
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It is a typical and standard Upstream and Downstream trip word problem.
You can find many similar fully solved problems on upstream and downstream round trips with detailed solutions in lessons
- Wind and Current problems
- More problems on upstream and downstream round trips
- Wind and Current problems solvable by quadratic equations
- Unpowered raft floating downstream along a river
- Selected problems from the archive on the boat floating Upstream and Downstream
in this site.
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".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.