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Flying with the wind, a plane can travel 1100 miles in 4 hours.
Against the wind, the plane requires 5 hours to fly the same distance.
Find the speed of the plane in calm air and the speed of the wind, both in miles per hour.
As needed, round your answers to the nearest tenth
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The effective speed with the wind is = 275 mph,
and it is equal to the sum of the plane speed at no wind, "u", and the speed of the wind, "v"
u + v = 275. (1)
The effective speed against the wind is = 220 mph,
and it is equal to the DIFFERENCE of the plane speed at no wind and the speed of the wind
u - v = 220. (2)
Thus you have these two equations, (1) and (2) for two unknowns, u and v.
To solve the system, add the equations. You will get then
2u = 275 + 220 = 495
u = 495/2 = 247.5 miles per hour.
Then from equation (1),
v = 275 - 247.5 = 27.5 miles per hour.
ANSWER. The speed of the plane at no wind is 247.5 miles per hour.
The speed of the wind is 27.5 miles per hour.
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.