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A plane travels at a speed of 205 MPH in still air. Flying with the wind, the plane is clocked over a distance of 950 miles.
Flying against the wind, it takes 2 hours longer to complete the return trip. what was the speed of the wind ?
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Let v = the wind speed (in miles per hour), now unknown.
The the tailwind speed = 205 + v miles per hour.
Headwind speed = 205 - v miles per hour.
"Time equation"
- = 2 hours.
950*(205+v) - 950*(205-v) = 2*(205^2-v^2)
2*950*v = 2*(205^2-v^2),
950v = 205^2 - v^2
v^2 + 950v - 205^2 = 0
= = .
Only positive root makes sense: v = = 42.35 mph.
Answer. The speed of the wind is 42.35 mph.
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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.
In these lessons you will find the detailed solutions of many similar problems.
Consider them as samples. Read them attentively.
In this way you will learn how to solve similar 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.