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Suppose that the wind is blowing at approximately 18 miles per hour. You want to make a round-trip flight (on the same day) of 1,295 miles
(in each direction). Flying against the wind, the trip takes about 11 minutes longer than flying with the wind.
Find the speed of your plane in still air, and determine how long each leg of the flight actually takes.
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Let "u" be the speed of the plane in still air, in miles per hour (the correct term is "airspeed").
Then the speed of the plane with the wind is (u+18) mph,
while the speed of the plane against the wind is (u-18) mph.
The time spent for the flight against the wind is hours.
The time spent for the flight with the wind is hours.
The difference of the two times is 11 minutes = hour.
It gives you an equation
- = .
To solve it, multiply the both sides of the equation by 60*(u-18)*(u+18).
You will get a quadratic equation for "u".
From this point please complete the assignment on your own.
It is a typical "tailwind and headwind round trip" word problem.
The way I described here is the standard way for solving such problems.
You can find the detailed solutions of many similar problems in 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.
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".
The way how the "josgarithmetic" explains the solution, is the way to "nowhere".
Nobody and never solves such problems in his way.
Simply ignore his writing.