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Question regarding position/displacement vectors

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data

    The position of an object at three different times is given

    At t= 4.9s
    position= (4.0,3.1,0) m

    At t= 5.4s
    position = (1.3, 7.1, 0) m

    At t= 5.9s
    position = (-0.8,12.7,0) m

    Using the best information available, what was the displacement of the bee during the time interval from 4.9 s to 4.93 s?

    The earlier parts of this question asked for the average velocity of the object between 4.9s and 5.4 s (-5.4,8,0) m/s

    And the average velocity between 4.9s and 5.9 s which is (-4.8, 9.6, 0) m/s.



    2. Relevant equations

    Displacement formula = r2 - r1 where r2 is the final position of the object and r1 is the initial position

    and also,

    r2 = r1 + (Average Velocity)(change in time)



    3. The attempt at a solution

    The way I attemped this solution is to use the average velocity between 4.9 s and 5.4 s (which was found earlier) and plug all the given values into the formula:

    r2= r1 + (Average Velocity)(Change in time)

    Which would be, (4.0,3.1,0) + ((-5.4,8,0)(0.93))
    = (-1.022, 11.1, 0) = r2

    I would then use this value for r2 into the formula for finding the displacement (r2 - r1)

    (-1.022,11.1,0) - (4.0,3.1,0)

    = (-5.022, 8, 0) m

    So the final answer would be (-5.022, 8, 0) m

    However, I'm still kind of sketchy on this solution and would just like some confirmation whether this is the right way of doing a question like this.
     
  2. jcsd
  3. Sep 15, 2009 #2
    I get a different answer ..

    Assuming the average velocity is the same for the interval 4.9 to 4.93 s
    as it is for the interval 4.9 to 5.4 s ..

    displacement for the interval 4.9 to 4.93 s =
    (average velocity)(time interval) =
    (-5.4,8,0) m/s multiplied by (4.93 s - 4.9 s)
     
  4. Sep 15, 2009 #3

    Wow mikelepore, I totally forgot about that equation for some reason. Thanks, the answer I got from your method is

    (-0.162,0.24,0)

    which makes much more sense. I just realized the magnitude for the displacement vector I had initially calculated was about 9.44 m which is almost double the magnitude of the displacement for the time interval between 4.9s and 5.4s, which doesnt seem to make sense for a question like this.

    Thanks again.
     
  5. Sep 15, 2009 #4
    Now I got the same answer as you.
     
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