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anonymous

  • one year ago

I need some help with displacement and vectors!! The displacement vectors A and B shown both have magnitudes of 3 m. The direction of vector A is theta=30 degrees. Find graphically: (a) A + B (b) A - B (c) B - A (d) A-2B Report all angles counterclockwise from the positive x-axis.

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  1. anonymous
    • one year ago
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    |dw:1440642477745:dw|

  2. jim_thompson5910
    • one year ago
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    so B is pointing directly north? and has an angle of 90 degrees?

  3. anonymous
    • one year ago
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    yep!

  4. jim_thompson5910
    • one year ago
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    one way is through the parallelogram method use the two vectors A and B to construct a parallelogram like so |dw:1440642698082:dw|

  5. jim_thompson5910
    • one year ago
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    I basically slid the vectors over to form the other missing sides

  6. jim_thompson5910
    • one year ago
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    then you draw a vector from the original tail to where the other vectors meet up along the diagonal |dw:1440642758813:dw|

  7. jim_thompson5910
    • one year ago
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    that last arrow I drew is the vector A+B

  8. anonymous
    • one year ago
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    I see. So I would solve for the diagonal to get the sum of the two vectors?

  9. jim_thompson5910
    • one year ago
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    yeah you form a parallelogram and draw the diagonal to get the sum of the two vectors

  10. jim_thompson5910
    • one year ago
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    http://hom.wdfiles.com/local--files/roberval/ParallelogramLaw2.PNG

  11. anonymous
    • one year ago
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    The diagonal would have to be 3 m, then.

  12. jim_thompson5910
    • one year ago
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    no it's going to be a bit longer

  13. anonymous
    • one year ago
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    Oh...How would I calculate that then? I understand the law, but..

  14. jim_thompson5910
    • one year ago
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    this angle is 60 degrees |dw:1440642977013:dw|

  15. jim_thompson5910
    • one year ago
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    since 60+30 = 90

  16. anonymous
    • one year ago
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    That makes sense..

  17. jim_thompson5910
    • one year ago
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    |dw:1440643005737:dw|

  18. jim_thompson5910
    • one year ago
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    |dw:1440643058820:dw|

  19. anonymous
    • one year ago
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    Ah, okay! Now I see. So then I would use the law of sines?

  20. anonymous
    • one year ago
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    Since two sides are 3 m and from focusing on the one triangle, I've found that the other two angles are 30 degrees each.

  21. jim_thompson5910
    • one year ago
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    we have this triangle |dw:1440643113916:dw|

  22. jim_thompson5910
    • one year ago
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    use the law of cosines to find x

  23. anonymous
    • one year ago
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    Whoops, sorry, that's what I meant!

  24. jim_thompson5910
    • one year ago
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    law of cosines c^2 = a^2 + b^2 - 2ab*cos(C) x^2 = 3^2 + 3^2 - 2*3*3*cos(120) x = ???

  25. anonymous
    • one year ago
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    3 times the square root of two :)

  26. anonymous
    • one year ago
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    three, sorry

  27. anonymous
    • one year ago
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    whoops. three times the square root of 3, i meant

  28. jim_thompson5910
    • one year ago
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    yes 3*sqrt(3) is the exact length of A+B

  29. anonymous
    • one year ago
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    That makes sense. How would I then start on figuring out A-B?

  30. jim_thompson5910
    • one year ago
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    A - B = A + (-B) = A + (-1*B) so you start with vector A and add on the opposite of vector B vector -B = -1*B points in the complete opposite direction of B |dw:1440643443046:dw| so in this case directly south. The magnitude or length of -B is the same as +B

  31. jim_thompson5910
    • one year ago
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    and you add A and -B just like before first form a parallelogram |dw:1440643496154:dw|

  32. jim_thompson5910
    • one year ago
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    then form the diagonal |dw:1440643523787:dw|

  33. anonymous
    • one year ago
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    I'm assuming the obtuse angles in here again are 120 degrees, like in the last part?

  34. jim_thompson5910
    • one year ago
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    my drawing is way off though geogebra is showing me this (see attached)

  35. anonymous
    • one year ago
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    Okay...so it's still the same as A + B?

  36. jim_thompson5910
    • one year ago
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    yeah more or less, you should find that angle ADF = 60 degrees from the attachment I posted above you'll use the law of cosines again c^2 = a^2 + b^2 - 2ab*cos(C) x^2 = 3^2 + 3^2 - 2*3*3*cos(60) x = ??

  37. anonymous
    • one year ago
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    Okay!! I think I got it now. Thank you!!

  38. jim_thompson5910
    • one year ago
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    you're welcome

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