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Math Help - please help

  1. #1
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    please help

    The compass in a light aircraft shows that it is flying due north. The air speed indicator gives a reading of 150km/h. If there is a 50km/h wind blowing from west to east, what is the ground speed of the aircraft?
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by bobby77
    The compass in a light aircraft shows that it is flying due north. The air speed indicator gives a reading of 150km/h. If there is a 50km/h wind blowing from west to east, what is the ground speed of the aircraft?
    The plane is moving 150km/h forward, it is also moving 50km/h right. If you draw out the vectors, as shown in the diagram below, then you'll see we want to calculate the hypotenuse of the triangle.

    use pythagorean theorum...
    c=\sqrt{150^2+50^2}

    c=\sqrt{22500+2500}

    c=\sqrt{25000}

    c=50\sqrt{10}\approx158.114
    Attached Thumbnails Attached Thumbnails please help-air-speed.jpg  
    Last edited by Quick; July 18th 2006 at 04:53 PM.
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  3. #3
    Super Member malaygoel's Avatar
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    It is always good to specify direction.
    Direction will be tan^{-1}(\frac{1}{3})east of north.

    Malay
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  4. #4
    MHF Contributor Quick's Avatar
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    A question by itself

    I'm just learning vectors mind you...

    would the velocity of the ground speed of the aircraft be 50\bold{i}+150\bold{j}+0\bold{k} or 50\bold{i}+0\bold{j}+150\bold{k}?
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  5. #5
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Quick
    I'm just learning vectors mind you...

    would the velocity of the ground speed of the aircraft be 50\bold{i}+150\bold{j}+0\bold{k} or 50\bold{i}+0\bold{j}+150\bold{k}?
    It depends on your coordinate system. You draw any three mutually perpendicular lines, and your coordinate system is ready(anyone of the three could be x,y and z). But for international convention, a coordinate system is universally accepted(right handed coordinate system).
    Draw a horizontal line and a line perpendicular to it(it will be obviously vertical).Label horizontal as x , vertical as y. Then z will be upwards out of the plane of the paper.
    In the question
    velocity of plane (without air)=0i+150j+0k
    velocity of air=50i+0j+0k
    resultant velocity=50i+150j+0k
    (Actually there is a method which could tell you that the cordinate system used by you is acc. to international convention or not(based on cross product of vectors))
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