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More fun with vectors

  1. Feb 13, 2005 #1
    A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 2m, y = 1.5m, and has velocity Vo =(3.3m/s)i + (-7m/s)j. The accelerationis given by a = (6m/s^2) i + (5.5 m/s^2)j. What is the x component of velocity afer 1.5s?

    im stuck
     
  2. jcsd
  3. Feb 13, 2005 #2
    A constant acceleration "a" implies the velocity is "a*t" where t is the time elapsed from 0.
     
  4. Feb 13, 2005 #3

    dextercioby

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    Put it as it should be,in vector form
    [tex] \Delta\vec{v}=\vec{a} \Delta t[/tex]

    Now project on the Ox axis and make a simple multiplication.

    Daniel.
     
  5. Feb 13, 2005 #4
    First off, do you understand the problem? Can you draw it? do you know what it means that a = (6m/s^2) i + (5.5 m/s^2)j. What have you tried to do so far?
     
  6. Feb 13, 2005 #5

    what is the Ox axis?
     
  7. Feb 13, 2005 #6

    dextercioby

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    What do you mean?You're given that the motion takes place in the xy plane,so it's not difficult to imagine the 2 mutually perpendicular Ox & Oy axis...?

    Daniel.
     
  8. Feb 13, 2005 #7
    i mean what is O
     
  9. Feb 13, 2005 #8

    dextercioby

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    O is the origin of the coordinate system,or if you want to,the point in which the 2 mutually perpendicular axis meet...

    However,this is a useless detail for this problem...

    Daniel.
     
  10. Feb 13, 2005 #9
    so when i get (Delta V)i subtract the initial velocity of the x component (3.5 m/s) from it and that will be the x component of velocity after 1.5 seconds.
     
  11. Feb 13, 2005 #10

    dextercioby

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    I've given you the equation already in post #3.I've explained what i meant about "Ox projection" and now i'm asking you to interpret the scalar equality
    [tex] \Delta v_{x}=a_{x} \Delta t [/itex]

    in a correct manner.

    Daniel.
     
  12. Feb 13, 2005 #11
    ok i think i got it now. it would just be the acceleration of the i componet multiplied by the change in time (1.5s). thanx
     
  13. Feb 13, 2005 #12

    dextercioby

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    That will be "Delta v",to compute the final velocity (component "x") u'll have to add the initial value...

    Daniel.
     
  14. Feb 13, 2005 #13
    curious...i did like you said, but im still not getting the correct answer. are u sure that is the correct formulae??
     
  15. Feb 13, 2005 #14
    okay thats where I went wrong, i dont know why i was subtracting. u were right i am supposed to add them to find the total final velocity
     
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