SOLUTION: An airplane traveling with wind flies 450 miles in 2 hours. On the return trip the plane takes 3 hours to travel the same distance. Find the speed of the airplane if the wind is st

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Question 1079094: An airplane traveling with wind flies 450 miles in 2 hours. On the return trip the plane takes 3 hours to travel the same distance. Find the speed of the airplane if the wind is still
Found 4 solutions by Fombitz, jorel1380, ikleyn, josmiceli:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
%28P%2BW%292=450
1.P%2BW=225
.
.
%28P-W%293=450
2.P-W=150
Add eqs. 1 and 2,
P%2BW%2BP-W=225%2B150
2P=375
Solve for P then use either equation to solve for W.

Answer by jorel1380(3719) About Me  (Show Source):
You can put this solution on YOUR website!
Let s be the speed of the plane in still air, and w be the speed of the wind. Then:
450/s+w=2
450/s-w=3
2s+2w=450
3s-3w=450
6s+6w=1350
6s-6w=900
12s=2250
s=187.5
w=37.5
The plane flies 187.5 mph in still air; the wind speed is 37.5 mph. ☺☺☺☺

Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
450%2F2 = u + v   (1)   is the equation for the effective speed of the plane flying with the wind.
450%2F3 = u - v   (2)   is the equation for the effective speed of the plane flying against the wind.


Where u is the speed of the airplane at no wind, and v is the rate of the wind.


Or, equivalently,

u + v = 225,          (1')
u - v = 150.          (2')

Add equations (1') and (2') (both sides). You will get

2u = 225 + 150 = 375.

Hence, u = 375%2F2 = 187.5 miles per hour.


Answer.  The speed of the plane at no wind is 187.5 mph.


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".



Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +w+ = the speed of the wind in mi/hr
Let +s+ = the speed of the plane in still air in mi/hr
-----------------------
Equation for flying with the wind
(1) +450+=+%28+s+%2B+w+%29%2A2+
Equation for flying against the wind
(2) +450+=+%28+s+-+w+%29%2A3+
----------------------------------
(1) +2s+%2B+2w+=+450+
(2) +3s+-+3w+=+450+
------------------------
(1) +s+%2B+w+=+225+
(2) +s+-+w+=+150+
Add the equations
+2s+=+375+
+s+=+187.5+
The speed of the plane in still air is 187.5 mi
----------------------
check:
(1) +450+=+%28+s+%2B+w+%29%2A2+
(1) +450+=+%28+187.5+%2B+w+%29%2A2+
(1) +225+=+187.5+%2B+w+
(1) +w+=+37.5+
and
(2) +450+=+%28+s+-+w+%29%2A3+
(2) +450+=+%28+187.5+-+w+%29%2A3+
(2) +450+=+562.5+-+3w+
(2) +3w+=+112.5+
(2) +w+=+37.5+
OK