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Vector Magnitude/Direction Problem

  1. Jan 30, 2005 #1
    Hey guys I need your help on this one. There is something I am doing wrong on this problem. Can you help me figure out what I am doing wrong?

    Problem: A golfer takes three putts to get the ball into the hole. The first putt displaces the ball 3.66m north, the second 1.83m southeast, and the third 0.91m southwest. What are (a) the magnitude and (b) the direction of the displacement needed to get the ball into the hole on the first putt?

    vector a = 3.66
    vector b = 1.83
    vector c = 0.91

    So far I have drawn a graph with vector a 3.66m directly due north. Then from the head of that vector drawn vector b 1.83m directly southeast, and from that
    I have drawn vector c 0.91m directly southwest.

    unit vector notation:
    A = (0)i + (3.66)j
    B = (1.83cos(45))i + (1.83sin(45))j
    C = (.91cos(45))i + (.91sin(45))j

    R = A + B + C
    R = 1.37i + 5.74j
    R = [1.37^2 + 5.74^2]^1/2 = 5.90

    Answers (a) 1.8m (b) 69 degress north of due east.

    What am I doing wrong to derive the answer?

    Any help is appreciated. Thanks
    Last edited: Jan 30, 2005
  2. jcsd
  3. Jan 30, 2005 #2
    Remember, North=positive and South=negative.
    So unit vector notation:
    A = (0)i + (3.66)j
    B = (1.83cos(45))i + (-1.83sin(45))j
    C = (-0.91cos(45))i + (-0.91sin(45))j

    Try that.
  4. Jan 30, 2005 #3
    Thanks man. That was such a simple error to overlook. I got the answer. :)
  5. Jan 30, 2005 #4
    Would I use the same technique as outlined above on this problem?

    Two beetles run across flat sand, starting at the same point. Beetle 1 runs 0.5m due east, then 0.80m at 30 degrees north of due east. Beetle 2 also makes two runs; the first is 1.6m at 40 degrees east of due north. What must be (a) the magnitude and (b) the direction of its second run if it is to end up at the new location of beetle 1?

    I ask this because I have attempted it and it isn't working.
    A = 0.5m
    B = 0.8m
    C = 1.6m

    A = .5i + 0j
    B = (.8cos(30))i + (.8sin(30))j
    C = (1.6cos(50))i + (1.6sin(50))j

    R = 2.35i + 1.78j
    R = [2.35^2 + 1.78^2]^1/2 = 2.95m
    arctan=(1.78/2.35)=37.14 degrees

    Answer should be: (a) 0.84m (b) 79 degrees

    What am I doing wrong here? Am I approaching it wrong? I am sure I didn't make the same mistake as last time.
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