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YumYum247

  • one year ago

someone please help me?!?!?!?!?

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  1. YumYum247
    • one year ago
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  2. YumYum247
    • one year ago
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    How do i figure out the distance and displacement???? :(

  3. YumYum247
    • one year ago
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    but before that please check if my drawing is ok???? :)

  4. YumYum247
    • one year ago
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    @Michele_Laino

  5. YumYum247
    • one year ago
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    @dan815

  6. dan815
    • one year ago
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    wow u went all out with ur measuring lol

  7. YumYum247
    • one year ago
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    dan you are prally that missing brother i always wanted.... :")

  8. YumYum247
    • one year ago
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    when your marks are dangling by a single branch.....you have to XD

  9. arindameducationusc
    • one year ago
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    Displacement= Final position-initial position

  10. arindameducationusc
    • one year ago
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    for distance, just add 14+6+12+2

  11. YumYum247
    • one year ago
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    but dear there are directions that are opposite to each other and even more weird.... E, N, NE, NE

  12. arindameducationusc
    • one year ago
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    Distance doesnot depend on direction! Only magnitude

  13. YumYum247
    • one year ago
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    how do i subtract out the two vector direction from a one dimentional???

  14. arindameducationusc
    • one year ago
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    Ya, displacement is little complicated here.... give me some time....

  15. arindameducationusc
    • one year ago
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    For distance, just add (simple)

  16. YumYum247
    • one year ago
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    sorry i was referring to displacement......!!!

  17. YumYum247
    • one year ago
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    :D

  18. arindameducationusc
    • one year ago
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    okay....

  19. YumYum247
    • one year ago
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    so what's up with displacement???? ?XD

  20. YumYum247
    • one year ago
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    can someone please check my drawing if i've make it right so we can proceed with the question????

  21. arindameducationusc
    • one year ago
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    Trignometric functions cannot be used here... So, If you have taken accurate scale drawings, The method is absolutely correct.

  22. YumYum247
    • one year ago
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    that's right trig can't be used here since it's not a right triangle....

  23. arindameducationusc
    • one year ago
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    This is a pure scaling and drawing problem... Yup you are right..... Yes because no 90 degrees

  24. arindameducationusc
    • one year ago
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    Good job!

  25. YumYum247
    • one year ago
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    XD

  26. YumYum247
    • one year ago
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    the thing is that there are three triangles in this problem...XD

  27. arindameducationusc
    • one year ago
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    If protector is allowed in this question, taking angles and taking length of the distance gives answer.... which you have done....

  28. YumYum247
    • one year ago
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    displacement is my main concern here.....

  29. arindameducationusc
    • one year ago
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    Ofcourse. See the idea is after taking all those angles and carefully going the path, take the final point and the initial point(in this case 0) and take the difference. Got it?

  30. YumYum247
    • one year ago
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    @joeyboy111

  31. anonymous
    • one year ago
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    the question was answered why did you tag me?

  32. YumYum247
    • one year ago
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    @joeyboy111 how do i find the displacement????

  33. anonymous
    • one year ago
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    take the final point and the initial point(in this case 0) and take the difference

  34. YumYum247
    • one year ago
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    what!!!! :O

  35. johnweldon1993
    • one year ago
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    Displacement, Hmm, I know above we said trig cannot be used...but I cannot see why not? The trip looks like |dw:1438831247053:dw| If we think about "How far total East did he travel?" and "How far total North did he travel" Then we just have 1 giant right triangle with 2 sides known East...14 + 12sin(15) + 2sin(65) = ? North...6 + 12cos(15) + 2cos(65) = ? It's late so let me know if this is totally wrong

  36. johnweldon1993
    • one year ago
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    Also yes I know my 15 degree angle and 65 degree angled vectors should be swapped...but the equations remain the same 15 should be much more "north" and the 65 should be much more "east"

  37. YumYum247
    • one year ago
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    now worries about the angles, just tell me once i have the followings figured out, then i make a triangle and solve for the hypothenuse ? East...14 + 12sin(15) + 2sin(65) = ? North...6 + 12cos(15) + 2cos(65) = ?

  38. YumYum247
    • one year ago
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    i meant right triagle****

  39. arindameducationusc
    • one year ago
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    @johnweldon1993 for 12km, the angle should be 75

  40. arindameducationusc
    • one year ago
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    and for 2km, angle should be 25

  41. arindameducationusc
    • one year ago
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    Yes trig can be used.. that was cool... @johnweldon1993

  42. johnweldon1993
    • one year ago
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    Well the angles I'm using North 15 degrees east...I'm interpreting as it would be north...but instead its just 15 degrees east of north Thus the reason I'm labeling |dw:1438832198665:dw|

  43. johnweldon1993
    • one year ago
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    So for that reasoning, is why I'm using "sin" as the function for 'east' and "cos" for 'north' using those angles

  44. arindameducationusc
    • one year ago
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    @johnweldon1993 Can I give my argument here if you dont mind?

  45. johnweldon1993
    • one year ago
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    Yeah absolutely, that's why I said It's late it could be wrong. Always open to criticism

  46. arindameducationusc
    • one year ago
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    |dw:1438833553890:dw| See the black line I have drawn, As per the definition of displacement, we need that measurement. So, we need the length of the hypotenuse, i.e14^2+6^2

  47. johnweldon1993
    • one year ago
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    Correct

  48. arindameducationusc
    • one year ago
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    So, as per the displacement definition, if you see @YumYum247 's diagram. The triangles are not in 90 degrees. only the first hypotenuse can be used, not others as per the definition. What do you say @johnweldon1993 ?

  49. johnweldon1993
    • one year ago
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    Do you agree that every vector has both a 'x' and 'y' "and z irrelevant here" component?

  50. YumYum247
    • one year ago
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    oh my god, so do you use the x component and y component way????? i know how do to that way....

  51. arindameducationusc
    • one year ago
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    See the red line here, I guess this should be the displacement, ofcourse they have component x and y..... give me one minute ........ hmmmm

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  52. johnweldon1993
    • one year ago
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    That should INDEED be the displacement My approach is to find how far over "x" and how far up "y" in order to find that red line being the hypotenuse \[\large Displacement = Red Line = \text{How far over}^2 + \text{How far up}^2\]

  53. arindameducationusc
    • one year ago
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    yes.....

  54. YumYum247
    • one year ago
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    Thank you bunch of lumps of flesh and pile of dead skin cells, i really appreciated you help.... :")

  55. johnweldon1993
    • one year ago
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    haha no problem...let us know what you end up with!

  56. YumYum247
    • one year ago
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    sure will. :")

  57. arindameducationusc
    • one year ago
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    We had a blackout... Sorry couldnot reply k, bye....

  58. YumYum247
    • one year ago
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  59. YumYum247
    • one year ago
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  60. YumYum247
    • one year ago
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    can someone please check my work, i had to figure out the average velocity and average speed too, see if i've done it correct. :")

  61. YumYum247
    • one year ago
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