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Graph of acceleration of a person running in an oval

  1. Sep 24, 2008 #1
    1. The problem statement, all variables and given/known data

    [​IMG]

    2. Relevant equations

    none

    3. The attempt at a solution

    I thought it was none of the above because since velocity changed from positive to negative, at some some interval, but i was wrong.
     
  2. jcsd
  3. Sep 24, 2008 #2

    danago

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    Gold Member

    Think about where the acceleration is zero and where it is non zero (remember, acceleration is a change in velocity, which includes direction). That should allow you to eliminate quite a few of them.
     
  4. Sep 24, 2008 #3
    yeah i got that part, but i dont know what happens to the acceleration at the ends. because the direction of velocity is neither horizontal or vertical. I thought acceleration is negative at some point because it cahnges from a positive direction to a negative direction. Therefore decreasing velocity? cause acceleration is v2-v1/t2-t1

    i dunno i'm kinda lost
     
  5. Sep 24, 2008 #4

    danago

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    Gold Member

    The graph show the magnitude of the acceleration, so there wont be any negative.

    What type of motion is the dancer undergoing around the bends?
     
  6. Sep 24, 2008 #5
    angular motion. I'm not that familiar with the properties of angular motion.
     
  7. Sep 24, 2008 #6
    is it choice 6 i just read about angular acceleration my text book
     
  8. Sep 24, 2008 #7
    Remember that the radial acceleration dictates the change in direction while the tangential is about the velocity. Since the velocity is constant between SR and PQ , the graph should reflect only the change everywhere else.
     
  9. Sep 24, 2008 #8
    i already established that acceleration is 0 where the straights are but im still not sure what it looks like at the curves. And the acceleration vector is towards the center of the track.
     
  10. Sep 24, 2008 #9
    CAn someone tell me whether acceleration is constant on the curves?
     
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