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Direction of acceleration when direction of moving particle changes by 90 degrees?

  1. Jul 2, 2007 #1
    1. The problem statement, all variables and given/known data

    A particle is moving eastwards with a velocity of 5m/s. In 10 seconds, the velocity changes to 5 m/s northwards. What is the average acceleration time? What is the direction of accleration?


    2. Relevant equations

    a=v-u/t

    3. The attempt at a solution

    initial velocity northwards = u = 0 m/s
    final velocity northwards = v = 5 m/s
    t = 10 s
    a = v-u/t= 1/2 m/s^2

    Now initial direction = eastwards = 0 degree
    final direction = northwards = resultant = 90 degree
    therefore, using vector rules, direction of acceleration = north-west = 120 degree

    Am I right?

    Mr V
     
  2. jcsd
  3. Jul 2, 2007 #2

    bel

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    Imagine, for the sake of simplicity, that the particle was undergoing uniform circular motion, then everything will be easy.
     
  4. Jul 2, 2007 #3

    Dick

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    Don't do that. Just subtract the initial velocity vector from the final and divide by delta t (though the final division won't affect the direction). NW is ok but why 120 degrees?
     
  5. Jul 3, 2007 #4
    Oh, I just thought that if E is 0 degrees, then north west will be 120 degrees.

    Thanks a lot.

    Mr V
     
  6. Jul 3, 2007 #5

    HallsofIvy

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    No, it would be 90+ 45= 125 degrees. (Normally North is at 0 degrees and Northwest at 360- 45= 315 degrees.)
     
  7. Jul 3, 2007 #6

    HallsofIvy

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    No, it would be 90+ 45= 125 degrees. (Normally North is at 0 degrees and Northwest at 360- 45= 315 degrees.)
     
  8. Jul 3, 2007 #7
    Oh, yeah. My mistake.

    Mr V
     
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