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Thread: Airplane vector when descending

  1. #1
    Junior Member B9766's Avatar
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    Airplane vector when descending

    I have a question from my chapter on vectors:

    Find the component form of the vector that represents the velocity of an airplane descending at a speed of 100 mph at and angle of $30^\circ$ below the horizontal.

    For the angle of descent I calculated $0^\circ - 30^\circ = -30^\circ = 330^\circ$

    The answer at the back of the book calculated the vector using $180^\circ + 30^\circ = 210^\circ$ so my answer was wrong.

    But, wouldn't that vector have the airplane going backward? It seems to me the x component of the vector would be positive while the y component would be negative, putting it in Quadrant IV.

    Where is my thinking wrong on this?

    BTW, my answer was $v=\langle 50\sqrt{3}, -50 \rangle$ while the textbook says it should be $v=\langle -50\sqrt{3}, -50 \rangle$
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  2. #2
    MHF Contributor
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    Re: Airplane vector when descending

    If this is the entire problem it's awful.

    You don't seem to be given any information on the x and y directions of motion. Is it just assumed to be along the positive x axis?

    Assuming that's the case. You have positive velocity in the x direction and negative velocity in the z direction.

    Using the numbers given and your ability to visualize you should be able to see that

    $v_x = 100 \cos(30^\circ)=50\sqrt{3}$

    $v_z = -100 \sin(30^\circ)=-50$

    and thus the velocity vector $v = (50\sqrt{3},~-50)$ which appears to be exactly what you got.

    Apparently the plane is travelling along the negative x axis, though why you'd know this I have no idea.
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  3. #3
    Junior Member B9766's Avatar
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    Re: Airplane vector when descending

    I agree Romsek. I should have put quotes around the question. That's exactly how it appears in the textbook.

    I'll report it as an error and see what happens. I sure wouldn't want to be on any airplane they were piloting.
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