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Math Help - Rate and Speed

  1. #1
    Newbie
    Joined
    Oct 2007
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    Rate and Speed

    Greetings, I believe this is possibly the correct place to put this, I am sorry if I am wrong.
    I was wondering if I could get help on a couple of problems that have pestered me.

    There are two, but I think if I can get the first one the second should be a breeze.

    Question: A pilot can travel 400 miles with the wind in the same amount of time as 336 miles against the wind. Find the speed of the wind if the pilot's speed in still air is 230 miles per hour.

    So here is how I set it up with the quation D=RT.
    D R T
    400 x
    336 x
    230 x

    Quite honestly, I am very bad at word problems like this; I am hoping I can get a step in the right direction, if at all possible. I know that to find Rate, it would be R=D/T, however I don't know that one is going faster or slower, so I would probably set it up as
    400/x=336/x, which doesn't make sense.
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  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
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    Hello, Artimid!

    A pilot can travel 400 miles with the wind in the same amount of time
    as 336 miles against the wind.
    Find the speed of the wind if the pilot's speed in still air is 230 miles per hour.
    We have: . \text{[Distance]} \;=\;\text{[Speed]} \times \text{[Time]}\quad\Rightarrow\quad T \:=\:\frac{D}{S}

    Let w = speed of the wind.

    Flying with the wind, the plane's speed is: 230 + w mph.
    To fly 400 miles, it takes: . \frac{400}{230 + w} hours.

    Against the wind, the plane's speed is: 230 -w mph.
    To fly 336 miles, it takes: . \frac{336}{230 - w} hours.

    Since these times are equal, our equation is: . \frac{400}{230 + w} \:=\:\frac{336}{230-w}

    Go for it!

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  3. #3
    Newbie
    Joined
    Oct 2007
    Posts
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    Aha!

    I was so caught up in my x value, that I didn't think at all to add w. I see how that works now. I appreciate the help.
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