Question 1119713: Flying with the wind, a plane traveled 420 miles in 3 hours. Flying against the wind, the plane traveled the same distance in 5 hours. Find the rate of the plane in calm air and the rate of the wind.
rate of plane ? mph
rate of wind ? mph
Found 3 solutions by josgarithmetic, ikleyn, Alan3354: Answer by josgarithmetic(39628) (Show Source): Answer by ikleyn(52867) (Show Source):
You can put this solution on YOUR website! .
Let u = rate of the plane in calm air, in miles per hour;
v = rate of wind.
Then the effective speed flying with the wind is = 140 mph.
It is the sum u + v, so your first equation is
u + v = 140. (1)
The effective speed flying against the wind is = 84 mph.
It is the difference u - v, so your second equation is
u - v = 84. (2)
Thus you have the system of two linear equations (1) and (2).
To solve it, add the equations. You will get
2u = 140 + 84 = 224 ====> u = 224/2 = 112.
Then from equation (1) v = 140 - 112 = 28.
Answer. The rate of the plane in calm air is 112 miles per hour.
The rate of the wind is 28 miles per hour.
Check. Check the solution on your own.
After doing it, you will understand the solution much better.
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.
Answer by Alan3354(69443) (Show Source):
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