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Vectors of watermelon seeds

  1. Jan 30, 2008 #1
    1. The problem statement, all variables and given/known data
    A watermelon seed has the following coordinates: x = -3.1 m, y = 2.7 m, and z = 0 m. Find its position vector as (a) a magnitude and (b) an angle relative to the positive direction of the x axis. If the seed is moved to the xyz coordinates (6.6 m, 0 m, 0 m), what is its displacement as (c) a magnitude and (d) an angle relative to the positive direction of the x axis?

    2. Relevant equations

    adding vectors

    3. The attempt at a solution
    I got A and B to equal 4.1 m and 138.812 degrees respectively, then I got stuck on C and D
  2. jcsd
  3. Jan 30, 2008 #2
    draw a vector diagram
  4. Jan 30, 2008 #3
    Right, so the diagram that would need to be used is 9.7 for the x axis by adding the -3.1 and 6.6 together? and then you use the 2.7 with the 9.7 to get the magnitude? Which i got to be 10.07
  5. Jan 30, 2008 #4
    Then for the angle...You take the inverse sin of (2.7/10.07) which i got to be 15.55. Would I subtract that from 90 then to get the degrees?
  6. Jan 30, 2008 #5
    are you using rx= ax +bx, ry= ay + by, r = (rx^2 + ry^2)^(1/2), and tan^-1(ry/rx)
  7. Jan 30, 2008 #6
    i used Pythagorean theorem to get the 10.07 (which is correct according to the program) and then i used sinx=(2.7/10.07) Then take the inverse. Now would i subtract that answer (15.55 degrees) from 180?
  8. Jan 30, 2008 #7
    ok, you are doing a different problem then I had in mind, that sound right to me subtract the answer (15.55 degrees) from 180
  9. Jan 30, 2008 #8
    are you solved this, if so change the name
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