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Math Help - Word Problem I have never seen before

  1. #1
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    Word Problem I have never seen before

    I know this word problem involves systems of equations, but I have never seen one similar to this before, and I don't have the slightest idea where to start. I can solve it when it's set up, I just don't know how to set it up.

    ~~~~~~~~~~

    Two planes leave Pittsburgh and Philadelphia at the same time, each going to the other city. One plane flies 25 mph faster than the other. Find the air speed of each plane if the cities are 275 miles apart and the planes pass each other after 40 minutes of flying time.

    ~~~~~~~~~

    Thank you for any and all help.
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  2. #2
    Moo
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    Hello,

    Let x be the speed of the first plane (mph), the slowest, say he's leaving from Philadelphia (P)
    The speed of the second plane will be x+25

    Let t=40min=2/3 hour be the time the planes pass each other.

    Let d the distance from P to the point the planes pass each other. The distance from this point of meeting to Pittsburgh (T) is 275-d.



    We know that the first plane went from P to the meeting point in 40 min (because they meet 40 min after their departure), at the speed x.

    So x=\frac{d}{\frac{2}{3}}

    (you have to put the time in term of hours, because the speed is in term of hours)

    The second plane went from T to the meeting point (which is 275-d distance) in 40 min too, at the speed x+25.

    So x+25=\frac{275-d}{\frac{2}{3}}


    Now, you have to solve for x, given the system :

    \begin{Bmatrix}<br />
x=\frac{d}{\frac{2}{3}} \\<br />
\\<br />
x+25=\frac{275-d}{\frac{2}{3}}<br />
\end{Bmatrix}<br />

    Can you make that ? Did you understand the way i used to come up with these equations ?
    Last edited by Moo; March 30th 2008 at 09:42 AM. Reason: too many latex errors oO
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  3. #3
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    hmm.. lets see

    let the speed of one of the plane to be V1 and the other V2 where V2 is the faster plane:

    V1 = V1
    V2 = V1 +25

    given they cross each other after a period, t = 40 minutes = 0.666... hours

    using s=ut +(1/2)at^2 we can calculate the speed of one of the plane
    also since they are both moving at a constant speed(no acceleration mentioned in the question), a= 0

    therefore the equation simplifies to

    s = ut

    now the total distance beteen both cities is 275 and the planes flies from each cities. this means they will both in total travelled 275 miles when they cross each other

    distance travelled by the first plane, S1
    S1 = V1t

    distance travelled by the second plane, S2
    S2 = V2t
    = (V1+25)t --------(1)

    total distance travelled by both planes S(total) = 275

    therefore,

    S1+S2 = S(total)
    V1(2/3) + (V1+25)(2/3) = 275

    solving for V1 we get 193.75 mph

    since V2 = V1 +25
    V2 = 218.75 mph
    Last edited by Danshader; March 30th 2008 at 09:35 AM.
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  4. #4
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    I am completely lost.

    Can someone break down the steps to getting the problem set up? (use x and y for the unknown variables please.)
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  5. #5
    Moo
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    Well, i can hardly see how to explain more .
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  6. #6
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    Quote Originally Posted by mathgeek777 View Post
    Two planes leave Pittsburgh and Philadelphia at the same time, each going to the other city. One plane flies 25 mph faster than the other. Find the air speed of each plane if the cities are 275 miles apart and the planes pass each other after 40 minutes of flying time.
    Let me try to explain it a little more clearly. My method is similar to Moo's, but written without fractions in the denominators of fractions, etc.

    Let the slower plane have speed x. So, as above, the faster plane has speed x + 25.

    They pass each other after 40 minutes, so the time here is 2/3. (Since our speed is in miles per hour, our time has to be in miles per hour.)

    Then, since d = r x t, the first (slower) plane travels \frac {2}{3} * x = \frac{2x}{3}.

    The faster plane travels for the same time, but goes further, since it is faster. It's distance, also equal to rate x time, is (x + 25) * \frac{2}{3} = \frac{2x+50}{3}.

    Now, we know that total distance traveled is 275 miles (imagine, for example, that one travels 100 miles. The other MUST have traveled 175 miles if they're going to meet.)

    So add up the two distances and set them equation to 275.

     \frac{2x}{3} + \frac{2x+50}{3} = 275.

    If you want to get rid of the denominators, multiply both sides by 3, giving:

    2x + 2x + 50 = 825.

    Solve for x, and remember that it is the slower speed of the two planes.

    This problem doesn't require two variables or anything really complex - it just requires understanding d = r x t, really.

    Does this make sense?
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