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Math Help - A Physics question with Trig...

  1. #1
    M23
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    A Physics question with Trig...

    Hi

    I'm stumped on a physics question here, which goes:

    The vector, v, has the following components: Vx= +65.0 m/s , Vy= -120 m/s

    Calculate the magnitude and polar positive direction of the vector, v.


    (Sorry, but please assume the "v" as the v with an arrow over it, and the x and y in Vx and Vy as subscripts...)


    So, using that information, I graphed out a triangle and found the hypotenuse.



    To get the polar positive direction, I believe I need the degree of the angle that the red arrow is pointing to. But how do I find that angle?

    I've tried using tan=opp/adj, or tan=Vy/Vx, but I get a negative number as my answer. Could somebody please help me with this question?

    Thanks a lot.

    (Also, am I right in thinking that the hypotenuse is the magnitude of the vector?)
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  2. #2
    MHF Contributor
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    Quote Originally Posted by M23 View Post
    Hi

    I'm stumped on a physics question here, which goes:

    The vector, v, has the following components: Vx= +65.0 m/s , Vy= -120 m/s

    Calculate the magnitude and polar positive direction of the vector, v.


    (Sorry, but please assume the "v" as the v with an arrow over it, and the x and y in Vx and Vy as subscripts...)


    So, using that information, I graphed out a triangle and found the hypotenuse.



    To get the polar positive direction, I believe I need the degree of the angle that the red arrow is pointing to. But how do I find that angle?

    I've tried using tan=opp/adj, or tan=Vy/Vx, but I get a negative number as my answer. Could somebody please help me with this question?

    Thanks a lot.

    (Also, am I right in thinking that the hypotenuse is the magnitude of the vector?)

    yes, |v| = \sqrt{(v_x)^2 + (v_y)^2}


    for an angle in quad IV ...

    \theta = 360^\circ + \arctan\left(\frac{v_y}{v_x}\right)
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  3. #3
    M23
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    Thank you so much. I was trying the same thing, but it was not working, until I realized my calculator was in radian mode

    Thanks anyways for confirming!
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