Question 1113978: Any help on this question would be greatly appreciated. The assignment is due on April 9, 2018. If you can help before then, please do. Thank you so much!
A boat traveling upstream goes 70 kilometers in 7 hours. If the return trip takes only 5 hours, what is the speed (in kilometers per hour) of the boat in still water?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39623)   (Show Source):
Answer by ikleyn(52832)  (Show Source):
You can put this solution on YOUR website! .
Let u be the boat speed in still water (in kilometers per hour, km/h).
Let v be the rate of the current, in km/h.
Then the boat's effective speed downstream is u+v km/h,
while the boat's effective speed upstream is u-v km/h.
// It is first major point you need understand and use in this sort of problems.
Now, "speed" equation for boat floating upstream is
 = u - v (1) (speed upstream = the distance divided by time upstream)
Next, "speed" equation for boat floating downstream is
 = u + v (2) (speed downstream = the distance divided by time downstream)
// It is the second major point in solving such problems: you must understand and write these equation automatically !
Simplify equations (1) and (2)
u - v = 10 (3)
u + v = 14 (4)
Now add equations (3) and (4) to eliminate "v". You will get
2u = 10 + 14 = 24 ====> u =  = 12.
Thus you just found the boat' speed in still water. It is 12 kilometers per hour.
At this point, you just answered the problem's question.
For completeness, you can find the current rate from eq(4) v = 14 - u = 14 - 12 = 2 km/h.
Answer. The boat' speed in still water is 12 kilometers per hour.
The current speed is 2 km/h.
Solved. I advise you to make the check on your own.
After making the check, you will understand the problem and the solution MUCH better.
When solving such problems, you inevitably will reduce the problem to solving equations of the form (3), (4).
The method of their solution I demonstrated to you is the standard version of the Elimination method.
It is the simplest way/method solving such equations.
And this method solving equations is the third major point of the whole procedure . . .
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It is a typical and standard Upstream and Downstream round 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.
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The lessons to learn from this solution :
1. Expressions u+v and u-v for the effective downstream and upstream speeds.
2. "Speed" equations.
3. The solution method to system of equation.
If you know these major points - you know how to solve it !
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