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A jet plane flying with the wind went 1780 mi in 4 h. Against the wind, the plane could only fly 1520 mi in the same amount of time.
Find the rate of the plane in calm air and the rate of the wind.
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Let "u" be the speed of the plane at no wind (=same as in calm air), in mph.
Let "v" be the speed of wind.
Then the effective speed of the plane flying WITH the wind is (u+v) mph (relative to the ground).
the effective speed of the plane flying AGAINST the wind is (u-v) mph (relative to the ground).
According to the condition,
the plane effective speed is = 445 mph flying with the wind,
and effective speed is = 380 mph flying against the wind.
It gives you two equations
u + v = 445, (1)
u - v = 380. (2)
Add the two equations (both sides). You will get
2u = 445 + 380 = 825. Hence, u = = 412.5.
Thus the speed of the plane at no wind is 412.5 mph.
Next, from the equation (1) v = 445 - u = 445 - 412.5 = 32.5 mph.
Thus the speed of the wind is 32.5 mph.
Answer. The speed of the plane at no wind is 412.5 mph.
The speed of the wind is 32.5 mph.
Check. The speed of the plane with the wind is 412.5 + 32.5 = 445 mph, and the flight time is = 4 hours.
The speed of the plane against the wind is 412.5 - 32.5 = 380 mph, and the flight time is = 4 hours.
Checks !
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.
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".