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Against the wind a commercial airline in South America flew 540 miles in 3 hours. With a tailwind the return trip took 2.5 hours.
What was the speed of the plane in still air? What was the speed of the wind?
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Let u be the speed of the plane (not the airline!) in miles per hour with no wind.
Let v be the speed of wind.
Then
= u - v (1) ( = the speed of the plane flying against the wind )
= u + v (2) ( = the speed of the plane flying with the wind )
Simplify (1) and (2):
u - v = 180, (1')
u + v = 216 (2')
Now add (1') and (2') (both sides). You will get
2u = 180 + 216 = 396 ---> u = = 198 mph.
Thus the speed of the plane with no wind is 198 mph.
Next, from (2') v = 216 - u = 216 - 198 = 18 mph.
The speed of the wind is 18 mph.
Solved.
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".
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.