.
Let v be the rate of the jet stream, in miles per hour.
Then the effective speed of the plane flying with the wind is (520+v) mph, and the time of the flight with tailwind is hours.
The effective speed of the plane flying against the wind is (520-v) mph, and the time of the flight with headwind is hours.
The time difference is of an hour, which gives you an equation
- = .
To solve it, multiply both sides by 2*(520-v)*(520+v). You will get
2*1600*(520+v) - 2*1600*(520-v) = (520-v)*(520+v)
2*1600*v + 2*1600*v = 520^2 - v^2
v^2 + 6400v = 520^2
v^2 + 2*3200 + 3200^2 = 520^2 + 3200^2
= 10510400
v + 3200 = +/-
v = - 3200 + 3241.975 = 41.975.
Check. - = 0.5. ! Correct !
Answer. The rate of the jet stream was 41.975 mph.
Solved.
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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, where you will find other similar solved problems with detailed explanations.
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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I agree with the @Alan's note that the geographical settings in this problem are FAR from to be perfect.