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Bike Vectors

  1. Jan 22, 2008 #1
    [SOLVED] Bike Vectors

    1. The problem statement, all variables and given/known data
    A student bikes to school by traveling first d_N = 1.00 \rm {miles} north, then d_W = 0.400 \rm {miles} west, and finally d_S = 0.100 \rm {miles} south.

    If a bird were to start out from the origin (where the student starts) and fly directly (in a straight line) to the school, what distance d_b would the bird cover?


    2. Relevant equations



    3. The attempt at a solution

    What I have attempted with this problem was to graph the vectors making north going positive on the y-axis. I then went left from the end of of the first point and finally down. I then drew a vector from the initial point to the final point. I know adding the vectors doesn't make sense because the distance should be shorter. I'm not sure which vectors to add or subtract or if that is even the proper way to handle this. I'm not even sure if I was correct in graphing the vectors. I understand how to get the resultant vector when I have angles and can use sine and cosine. Is it possible to find the angles for this problem? This is online question and I'm really lost on how to proceed without just blindly guessing. I know I will never find the answer that way.
     
  2. jcsd
  3. Jan 23, 2008 #2

    Kurdt

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    You sound like you've graphed it properly. You can just add the vectors. Remember that south is in the opposite direction to north and is thus the same as the north vector multiplied by -1.
     
  4. Jan 23, 2008 #3
    So therefore you are saying that I would add them as 1.00N+.4+(-.1)? This gives me and answer of 1.30 which I have already tried and it was returned as incorrect. Am I just understanding wrong? The hint given is: You cannot simply add the different components of a vector to find the magnitude. Try to draw a vector diagram of the distances traveled by the student and use geometry to find d_b.
     
  5. Jan 23, 2008 #4

    Kurdt

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    Adding the vectors gives you the final position vector. You then need to find the magnitude of that position vector. Have you covered that before?
     
  6. Jan 23, 2008 #5
    Well I don't think we have covered that in class, but I may just not be connecting what your saying to something we've done. What we have done in class as far as finding vectors is to either find the resultant vector by either using the tail to tip method or by using vector components. I tried using the tail to tip method because the problem doesn't give any angles to be able to use the component method. Is it possible to apply either one of those methods? Does this problem have more to do with products of vectors? Do I need to find an angle between the vectors?
     
    Last edited: Jan 23, 2008
  7. Jan 24, 2008 #6

    Kurdt

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    You don't need an angle since you can work out the components of the final vector. The magnitude of that vector s given by Pythagoras' theorem.

    [tex] |\mathbf{v}| = \sqrt{x^2+y^2} [/tex]
     
  8. Jan 24, 2008 #7
    Well I figured it out now and that is extremely simple. I kind of feel like I wasted your time, but thanks for you help.
     
  9. Jan 24, 2008 #8

    Kurdt

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    Course you haven't. :smile:
     
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