SOLUTION: I am truly confused by word problems. Please help. This is the question.
A twin-engined aircraft can fly 500 miles from city A to city B in 4 hours with the wind and make the re
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Question 159281: I am truly confused by word problems. Please help. This is the question.
A twin-engined aircraft can fly 500 miles from city A to city B in 4 hours with the wind and make the return trip in 8 hours against the wind. What is the speed of the wind?
I tried setting this up in a graph and got completely lost. I know this is probably a simple problem but I have worked on it for over an hour now and have gotten myself completely confused. Distance problems confound me. I appreciate your help.
Found 2 solutions by Alan3354, jojo14344:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
I am truly confused by word problems. Please help. This is the question.
A twin-engined aircraft can fly 500 miles from city A to city B in 4 hours with the wind and make the return trip in 8 hours against the wind. What is the speed of the wind?
I tried setting this up in a graph and got completely lost. I know this is probably a simple problem but I have worked on it for over an hour now and have gotten myself completely confused. Distance problems confound me. I appreciate your help.
****************************************
A lot of people tense up when they hear "word problems." The trick (if you wanna call it that) is to get the info you need to solve the problem. Humans talk using words, but we work problems with numbers and symbols.
----------------
For this specific problem:
The ground speed of an aircraft is its speed, plus or minus the wind speed.
In this case, it goes from A to B (I've never been to either) in 4 hours with the wind speed added to its speed, and it takes 8 hours with the wind speed subtracted.
So, going there the ground speed is the aircraft speed + the wind speed.
Going back, it's the aircraft speed - the wind speed.
--------------------------
To save typing, use some symbols. Call A the speed of the aircraft, and W the wind speed.
-------------
The time to move from one place to another is determined by 2 things:
How far it is, and
How fast we move. If it's twice as far, and we move at the same rate, it will take twice as long (twice as much time). right?
If the distance is the same, and we go 3 times as fast, it will take 1/3 of the time. See that?
---------------
OK, back to the problem. The distance (call it s, don't ask me why it's s, but it's commonly used and you will see it again) s = 500 miles.
The ground speed going is the airspeed of the plane (call it A) plus the wind speed (call it W).
So, 500/(A + W) = 4 hours
***********************
Going back,
500/(A - W) = 8 hours
**************
Now, instead of a problem with words, we have 2 equations with 2 unknowns.
500/(A + W) = 4
500/(A - W) = 8
My dog can solve that. Well, maybe not, but he can do tricks for food.
----------------
500/(A + W) = 4
500/(A - W) = 8
--------------- Now it's just math.
500 = 4*(A + W)
500 = 8*(A - W)
-----------------
4A + 4W = 500 eqn 1
8A - 8W = 500 eqn 2
Multiply eqn 1 by 2
8A + 8W = 1000
8A - 8W = 500 eqn 2
Add the 2 eqns
16A = 1500
A = 1500/16 (the aircraft's speed)
Sub into either eqn, I'll use eqn 1
4A + 4W = 500 eqn 1
4*(1500/16) + 8W = 500
375 + 8W = 500
8W = 125
W = 125/8, the wind speed, what was asked for.
Answer by jojo14344(1513) (Show Source): You can put this solution on YOUR website!
Remember: }, ----> working eqn
According to our working eqn:
1st trip:
-----------------------------> eqn 1
2nd trip (going back):
, see negative (-)->"against"
-------------> eqn 2
In eqn 1, we get:
-------> eqn 3, and substitute in eqn 2:
, SPEED OF WIND
For the aircraft, go back eqn 3,
To check, go back eqn 1 or 2:
via eqn 1:
via eqn 2:
Thank you,
Jojo
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