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Reading acceleration vs. time graphs

  1. Jun 24, 2007 #1

    exi

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    1. The problem statement, all variables and given/known data

    [​IMG]

    (The particle starts from rest at the origin.)

    1: What is the velocity at t = 5s?

    2: What is the position at t = 5s?

    3. The attempt at a solution

    I'm thinking -2 m/s velocity and -8 m position at t = 5s, but would like someone to double-check my understanding of reading velocity and position from this graph. The negative acceleration at origin combined with that instantaneous change at t = 1 have me a bit unsure of myself.
     
  2. jcsd
  3. Jun 24, 2007 #2

    Astronuc

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    Staff: Mentor

    The zero acceleration simply means that the speed is constant after 1 sec, i.e. there is no further acceleration.

    So the maximum speed (magnitude of velocity) is achieved at 1 sec.

    Starting from rest with a negative acceleration, this would imply acceleration in the negative direction. Normally negative acceleration would imply a deceleration in the + direction.
     
  4. Jun 24, 2007 #3

    exi

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    yes sir, so does my thinking at t = 5 sound correct, or am I a bit off?
     
  5. Jun 24, 2007 #4

    Astronuc

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    Staff: Mentor

    What is the velocity at 1 sec? It is constant thereafter.
     
  6. Jun 24, 2007 #5

    exi

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    Either -1 or -2 m/s. That change at t=1 is what's making me a little unsure as to whether there's an extra increment of acceleration.

    So velocity at t=5 is either -1 or -2, and position is... I'm still guessing -8 based on what I mentioned in the original post, but that depends on what I'm asking here, as well.
     
  7. Jun 24, 2007 #6

    Astronuc

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    Staff: Mentor

    Well, for 1 s, the acceleration is constant at -1 m/s2.

    v = a*t = -1 m/s2 * 1 s = ?

    At t = 1 s, the acceleration drops to zero. That is a discontinuity. Since a = 0, v is constant and equal to v(t= 1s).
     
  8. Jun 24, 2007 #7

    exi

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    Okay, so at t = 5, velocity is -1 m/s.

    Oh wow, I was way off on that. Am I correct in using:

    [tex]\Delta x = V_ot + \frac{at^2}{2}[/tex]

    twice, once for t(0,1) and t(1,5), adding the results, and getting -4.5m for position @ t=5s?
     
  9. Jun 24, 2007 #8

    Astronuc

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    Staff: Mentor

    The problem can be broken into two time intervals, yes. 0 to 1s, with constant acceleration, the 1 to 5s, at constant velocity.

    For the first part of the problem, vo = 0, a = -1 m/s2, over the time interval 0 to 1s.

    For the second part of the problem, vo = - 1 m/s, and a = 0, for t = 1 s to 5 s.

    and -4.5m for position @ t=5s is correct.
     
    Last edited: Jun 24, 2007
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