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An airplane takes 3 hours to travel a distance of 1800 miles with the wind. The return trip takes 4 hours against the wind.
Find the speed of the plane in still air and the speed of the wind.
I had a question about how would I have to set up this problem in order to find both my speeds?
I have been stuck on this question for hours please, help!
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= u + v (1) is the equation for the flight with the wind.
(Here "u" is the airspeed of the airplane with NO wind,
"v" is the wind speed)
= u - v (2) is the equation for the flight against the wind.
Simplify equations (1) and (2) and write them as a system:
u + v = 600, (1')
u - v = 450. (2')
Now add the two equations (1') and (2'). You will get
2u = 1050 ---> u = = 525 miles per hour.
Having this, you can easily determine "v" from (1'):
v = 600 - 525 = 75 miles per hour.
Answer. The plane airspeed is 525 mph. The wind speed is 75 mph.
It is a typical "tailwind and headwind" word problem.
See the lessons
- Wind and Current problems
- Wind and Current problems solvable by quadratic equations
- Selected problems from the archive on a plane flying with and against the wind
in this site.
You will find the detailed solutions of many similar problems there.
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".
If after reading my solution you still have a question "why the equations (1) and (2) have this form ?",
then read the lessons above. They contain the detailed answer to this question.