SOLUTION: A plane traveled 514.8 miles each way to Sydney and back. The trip there was with the wind. It took 5.5 hours. The trip back was into the wind. The trip back took 11 hours. What is

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Question 1049677: A plane traveled 514.8 miles each way to Sydney and back. The trip there was with the wind. It took 5.5 hours. The trip back was into the wind. The trip back took 11 hours. What is the speed of the plane in still air?
What is the speed of the wind?
I got two equations out of this: 5.5x-5.5y=514.8, and 11x+11y=514.8. I solved the system to get 70.2 mph, and -23.4. Since speed in this sense can't be negative, does anybody know what I did wrong here?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Most likely a sign mistake, just as a guess. Possibly an interpretation mistake.

I am using variable assignments:
x, speed in absence of wind
y, speed of the wind
"Into" the wind, means "AGAINST" the wind, making the actual speed slower than in absence of any wind; and notice how the longer travel time is given for "against the wind". 
                SPEED           TIME       DISTANCE

WITHWIND        x+y              5.5       514.8

AGAINSTWIND     x-y             11         514.8


You can find what your mistake was and should be able to form the needed equations and correctly solve. The management of the language was likely the mistake; and not the otherwise written symbolic number properties steps.

Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.
A plane traveled 514.8 miles each way to Sydney and back. The trip there was with the wind. It took 5.5 hours.
The trip back was into the wind. The trip back took 11 hours. What is the speed of the plane in still air?
What is the speed of the wind?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

From the condition you have these two equations

514.8%2F5.5 = u + v,   (1)

514.8%2F11  = u - v.   (2)

The left side of the equation (1) is the speed of the plane relative the ground when flying with the wind, 
and it is the sum of the rate of the plane in still air "u" and the wind's rate "v".

The left side of the equation (2) is the speed of the plane relative the ground when flying against the wind, 
and it is the difference of the rate of the plane in still air "u" and the wind's rate "v".

Simplify the equations (1) and (2):

u + v = 93.6,    (3)
u - v = 46.8     (4)

Now add the equations (3) and (4). You will get

2u = 93.6 + 46.8 = 140.4,  or  u = 140.4%2F2 = 70.2.

Thus you just found the speed of the plane in still air. It is 70.2 miles per hour.

Now it is easy to find the speed of the wind. It is

v = 93.6 - 70.2 = 7 - 5.5 = 23.4 miles per hour.


Answer.  The speed of the plane in still water is 70. mph.  The speed of wind is 23.4 mph.

It is a typical "tailwind and headwind" word problem.

See the lessons
    - Wind and Current problems
    - 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.