SOLUTION: A jet takes 5 hours to travel a distance of 3800 miles against the wind. The return trip takes 4 hours with the wind. What is the rate of the jet in still air and what is the rate

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Question 1168919: A jet takes 5 hours to travel a distance of 3800 miles against the wind. The return trip takes 4 hours with the wind. What is the rate of the jet in still air and what is the rate of the wind?
Answer by ikleyn(52798)   (Show Source): You can put this solution on YOUR website!
.

Let u be the plane speed at no wind. 

Let v be the speed of the wind.



The effective speed of the plane with the wind is

    u + v =  = 760 mph.    (1)



The effective speed of the plane against the wind is

    u - v =  = 950 mph.    (2)



From equations (1) and (2), the speed of the plane at no wind is   = 855 mph.



Then from equation (1), the speed of the wind is  855-760 = 95 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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