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An airplane takes 5 hours to travel a distance of 2800 miles with wind. The return trip takes 5 hours and 36 minutes against the wind.
Find the speed of the plane in still air, and the speed of the wind. (Hint: Convert all time to hours.)
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5 hours and 36 minutes = hours = hours = hours.
Now, the effective velocity of the plane with the wind is = 560 miles per hour (I divided the distance by time to get the speed).
It is the sum of the plane speed at no wind "u" and the speed of wind "v".
It gives you your first equation
u + v = 560.
Next, the effective velocity of the plane flying against the wind is = 500 miles per hour (I divided the distance by time to get the speed).
It is the difference of the plane speed at no wind "u" and the speed of the wind "v".
It gives you your second equation
u - v = 500.
Write these two equations together
u + v = 560, (1)
u - v = 500. (2)
Now ADD the two equations (both sides). You will get
2u = 560 + 500 = 1060. Hence, u = 530 miles per hour.
Thus you just found the speed of the plane at no wind. It is 530 miles per hour.
Now from equation (1) v = 560 - 530 = 30.
So, 30 miles per hour is the speed of the wind.
Answer. The speed of the plane at no wind is 530 miles per hour.
The speed of wind is 30 miles per hour.
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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.
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".