#1
Its been five years since I've done any physics, but the time has come where I need to be able to solve simple problems like the one below, and I can't for the life of me remember how.

If you guys could hold my hand and guide me through the process of solving a problem such as the one below it would be much appreciated.

Q: Two airports, A and B, are 600 miles apart, one aircraft (x) takes off from A, and heads towards B at 600 miles per hour, another (y) takes off from B and heads towards A at 250 miles per hour. What distance from A do they meet?

I can work out that aircraft x takes one hour to get to B, and that aircraft y takes 2 hours 24 mins, but after that I'm lost.

P.S. Is all the info required in the question? I feel a bit embarrassed that I can't even work that out...
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#2
Neither aircraft x or y takes off on time.

Passengers g and h sleep at airport a and b.
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#3
Figure how long it takes the distance between them to reach zero.

IE, at the beginning the distance is 600mi, after thirty minutes it's something like a couple hundred miles..... Distance == (X's speed * time) - (Y's speed * time). All you have to do is solve for time, amirite?

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#4
you have to rest T-a to total time ant then you could figure where they meet... i think :S
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#5
423.5

together they should cover 600 miles to meet
S = V*t
so
600*t + 250*t = 600
850*t = 600
t = 60/85

distance from A = distance that first plane covers = 60/85 * 600 = 423.529411
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Last edited by Regensgeliebte at Aug 12, 2009,
#7
Quote by MightyAl
They don't meet. Air traffic control exists to prevent aircraft meeting in-flight.

Too bad that didn't work the other day over the Hudson.

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#8
Quote by tommyt
Neither aircraft x or y takes off on time.

Passengers g and h sleep at airport a and b.


I nearly spit out the orange juice in my mouth reading this.
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#9
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Too bad that didn't work the other day over the Hudson.

Which is why math equations are solved on paper, not in the sky.
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#10
*reported for move to maths/science thread*

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#11
Quote by MightyAl
They don't meet. Air traffic control exists to prevent aircraft meeting in-flight.



this.
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#12
make the airplanes into a function of distance with respect to time i.e

aircraft x would be d = 600t (d = distance, t= time)
aircraft y would be d = -250t + 600 (it's negative 250 because it's going in the opposite direction, and the +600 is because it is starting 600 miles away)

set the two d's equal to eachother

-250t + 600 = 600t

add 250t to both sides

600 = 850t

divide 850 from both sides and it will leave you with a time in hours

That should be the correct answer and method
#13
time of A flight = time of B flight (obviously, because they both take off at the same time) (flight meaning time from take off to the "meeting")

once you got that figured out you can go with
(distance from A) / time = speed of A (600)
(distance form B) / time = speed of B (250)


we get
(distance from A) / speed of A (600) = time
(distance form B) / speed of B (250) = time)


we put them together

(distance from A) / speed of A (600) = (distance form B) / speed of B (250)

and another one
(distance fomr A) + (distance form B) = 600

now i think you just put the equasions together and figure it out..

i'm not sure, because it's been a while since i've done phisics myself but i love it.

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#16
Yeah and my goat can't ride in the same airplane as the wolf because the wolf will eat the goat!
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