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Homework Help: Vectors Magnitudes and Angles (Respect to the X Axis)

  1. Jan 19, 2009 #1
    1. The problem statement, all variables and given/known data

    a) If Vx = 18.00 units and Vy = -15.50 units, determine the magnitude of V.
    b) Determine the angle of V with respect to the x-axis.

    2. Relevant equations



    3. The attempt at a solution

    Yeah, Part B is pretty much the only one I don't get .

    a) Answer is 23.75
    b) I'm not too sure about what the whole "respect to the x-axis" means. The angle does lie in the fourth quadrant, so I assumed the answer is to do 360 degrees - the angle (i got 40.73) and get 319.27 but that didn't work. I'm down to one try for the answers, so I have to get this one right.

    Thanks!
     
  2. jcsd
  3. Jan 19, 2009 #2
    a) If Vx = 18.00 units and Vy = -15.50 units, determine the magnitude of V.

    Magnitude = sqrt ( x^2 + y^2)

    => Magnitude = sqrt ( [(18).(18)] + [(-15.5)(-15.5)] )
    => Magnitude = sqrt (324 + 240.25)
    => Magnitude = sqrt ( 564.25)
    => Magnitude = 23.75

    With part B I get the same answer as you using:

    tan^-1 (y /x)

    tan^-1 (15.5/18)

    => Angle = 40.73 degrees
     
  4. Jan 19, 2009 #3
    Yeah, that answer is incorrect. I don't get what it means by respect to the x axis
     
  5. Jan 19, 2009 #4
    If you want to measure it with respect to the x axis it means that you start on the x - axis and move counter clockwise to your angle. I assume?
     
  6. Jan 19, 2009 #5
    Ya, that gives me 360 - 40.73 = 319.27, which was wrong :(
     
  7. Jan 19, 2009 #6
    Hmmm. What answer is given (assuming an answer is given)?
     
  8. Jan 19, 2009 #7

    rl.bhat

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    Homework Helper

    Wright the angle as -40.73 degree. Because angle in clockwise is taken as negative.
     
  9. Jan 19, 2009 #8
    Are you sure? I have one last try and I don't wanna get it wrong!
     
  10. Jan 19, 2009 #9

    rl.bhat

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    Homework Helper

    I think so.
     
  11. Jan 19, 2009 #10
    im scared to try iot :(
     
  12. Jan 20, 2009 #11
    can anyone confirm -40.73?
     
  13. Feb 16, 2009 #12
    Yes, -40.73 is what I get too.

    This means the angle is 40.73 degrees below the horizontal axis.
     
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