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Homework Help: Direction of a vector after I have already found magnitude

  1. May 24, 2012 #1
    1. The problem statement, all variables and given/known data
    The x component of vector is -16.2 m and the y component is +54.6 m. (a) What is the magnitude of ? (b) What is the angle (in radians) between the direction of and the positive direction of x?





    2. Relevant equations
    magnitude of vector is sqrt of a^2+b^2
    direction of vector is tan theta= ay/ax


    3. The attempt at a solution
    I got A part correct by finding the magnitude using sqrt of a^2+b^2= 56.95m but i can't seem to get b part. i did inverse tan of 56.4m/-16.2m and got -1.28 radians and its wrong. HELP!!!
     
  2. jcsd
  3. May 24, 2012 #2
    You did 56.4/-16.2.
    Your given info, however states that the y component is +54.6 m.

    So:
    inverse tan (54.6/-16.2) is your answer.
     
  4. May 25, 2012 #3
    That was a typo. I have been calculating inverse tan of 54.6/-16.2 and get -1.28 which is wrong according to the automatic grader
     
  5. May 25, 2012 #4

    gneill

    User Avatar

    Staff: Mentor

    The inverse tangent function cannot "tell" whether the negative sign is to be associated with the numerator or denominator in its argument; ##\frac{-y}{x}## is indistinguishable from ##\frac{y}{-x}##. It will return a value that is in the range -90 ≤ θ ≤ +90 degrees.

    You need to apply some smarts regarding the signs of the given components in order to place the resulting vector in the correct quadrant; You may have to add or subtract 180° (##\pi## radians) from the value it gives you in order to "shift" the quadrant.

    Alternatively, if your calculator has an atan2(y,x) function, then it'll handle the signs automatically. It may also have a "rectangular to polar" conversion capability which will both find the magnitude and the proper angle for you in one step.
     
  6. May 25, 2012 #5
    oh ok so since -16.5 x and 54.6 y puts it in the 2nd quadrant the angle has to be between 90 and 180. -73.47 is in the 4th quadrant so by adding 180 to that and converting it to radians i get 1.855 radians which is correct!
     
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