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Final Exam Review: Motion of an Airplane

  1. May 5, 2012 #1
    This is a question from an old test that I took this semester. My grader did not input anything, so I am left to trying to figure out by myself if my answers are correct or not.

    Question:

    The motion of an airplane leaving from miami can be described using a cartesian coordinate system with due East coincendet with the positive x-direction and due Noth with the positrive y-direction. The position vector r of the airplane as a function of time from take-off can then be written as:

    vector r = at(i) + (Bt - yt^3)(j)

    [a.] What are the units of the three constants?

    This one was relatively simple. I found that the units were: a = m/s. B = m/s. and y = m/s^3.

    [b.] Find the time(s) when the motion of the airplane is due NE.

    By motion I am guessing they mean velocity,

    So Vx = a and Vy = B - 3yt^2.

    Because it is exactly NE, these two components should be equal to one another, giving Vx = Vy

    ∴ a = B - 3yt^2,

    t = [(a - B)/3y]^1/2.

    The negative answer for t can be discarded.

    [c.] Find the plane's position when the motion is due NE.

    Position means the angle.

    So theta = tan^-1(ry/rx)

    Which is tan^-1([Bt - yt^3]/at)

    -----------------------------------------------------------------------------------------

    Please help me confirm whether I am wrong or right as my final is on Monday and I need to do extremely well in order to pass this course

    Thanks.
     
  2. jcsd
  3. May 6, 2012 #2

    haruspex

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    I would have thought position meant position in 2 dimensions, probably in Cartesian co-ordinates.
     
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