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Acceleration time graph

  1. Mar 12, 2015 #1

    Suraj M

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    1. The problem statement, all variables and given/known data
    A particle starts from rest. Its acceleration vs time graph is as shown. What will be the maximum speed?
    WIN_20150312_160506.JPG


    2. Relevant equations
    usual kinematic equations(3+)
    3. The attempt at a solution
    velocity at t=11 seconds should be max, so the area under the given graph = change in velocity
    So vf-0 = 55
    so max. velocity= 55 m/s
    But, if i try it this way:
    ##a=\frac{dv}{dt}##
    ##a.dt=dv##
    integrating..
    ##a∫_{0}^{11} dt = ∫_{0}^{v} dv##
    so ##a(11-0)=v-0##
    i know this is wrong because ##a## is not constant. But how do i include it?
    Any help is appreciated.
     
    Last edited: Mar 12, 2015
  2. jcsd
  3. Mar 12, 2015 #2

    Merlin3189

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    Since a is a function of time, ∫adt is not a∫dt

    You need to write a as a function of t then integrate it correctly.
     
  4. Mar 12, 2015 #3
    But if a particle starts from rest and its acceleration vs time graph is as shown, I ask to myself why at the intial time the initial acceleration is not null?
     
  5. Mar 12, 2015 #4

    Merlin3189

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    You could equally ask, why the question chose to set time 0 at the moment the acceleration starts.
    Seems a sensible choice. Say the acceleration is given by a spring which is compressed. It could sit in that state indefinitely, until you release the particle. Then something interesting happens, so you start measuring from that moment and call that t=0.

    To keep you happy, acceleration was 0 or null at t=-1, t=-7.5, t= -0.000000000000000000001, but rises to 10 m/sec2 (or whatever that blurred figure is) at t=0.
     
  6. Mar 12, 2015 #5

    Suraj M

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    oh yeah, i don't how i missed that..
    so ##a=10-βt## where ##β= \frac{da}{dt}##
    so ##β=\frac{10}{11}##
    so i get $$∫_{0}^{11}(10 - \frac{10}{11}t)dt =∫_{0}^{v}dv$$
    so then i get ## 10(11-0)-(\frac{10}{11}).(11^2 -0) = v##
    so ##v = 110-110=0## huh?
     
  7. Mar 12, 2015 #6

    NascentOxygen

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    Staff: Mentor

    Try this integration again. Show intermediate steps.
     
  8. Mar 12, 2015 #7

    Suraj M

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    $$∫_{0}^{11}(10dt -\frac{10}{11}tdt)=∫_{0}^{v_f}dv$$
    $$10(t_f -0) - \frac{10}{11}(\frac{t_f^2}{2}-0)=v_f-0$$
    so then substituting ##t_f=11##
    Ok got it.. thanks, sorry for the stupid mistake
     
    Last edited: Mar 12, 2015
  9. Mar 12, 2015 #8
    What is the integral of tdt? (Hint: it's not t2)

    Chet
     
  10. Mar 12, 2015 #9
    What physical law that you are aware of says that the acceleration of a particle has to be continuous with time?

    Chet
     
  11. Mar 12, 2015 #10

    Merlin3189

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    Yes. It's just your integral.
    ∫ k dt = kt , but ∫ kt dt ≠ kt2

    Try differentiating kt2 and see what you get.

    Edit - You posted the right answer while I was writing this!
     
  12. Mar 12, 2015 #11

    Suraj M

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    Thank you, i corrected it.. in post #7
    thank you for your help.
     
  13. Mar 12, 2015 #12
    Mr. Chet ,
    I wanted only say that the particle in question is defined to "start at rest" while in the graph, for how I think to interpretate it,seems to start with an initial acceleration not null. Isn't definition of rest "no acceleration"?
     
  14. Mar 12, 2015 #13
    No. Definition of "at rest" is zero velocity.

    Chet
     
  15. Mar 12, 2015 #14
    and then also the acceleration?
     
  16. Mar 12, 2015 #15
    No. Acceleration is rate of change of velocity.
     
  17. Mar 12, 2015 #16
    but then let me know how is the acceleration of a particle at rest, that is with speed null?
     
  18. Mar 12, 2015 #17
    Imagine that, at times t<0, the velocity is zero, and at times t>0, the velocity is kt, where k is a constant. So the velocity is a continuous function of time. What is the acceleration at times t < 0? What is the acceleration at times t > 0? Is the acceleration a continuous function of time at t = 0?

    Chet
     
  19. Mar 12, 2015 #18
    I've understand: before a particle moves with constant speed v, it needs to be accelerated.
    It can't move from 0 to v instantly. Also in free fall it should be an acceleration not null at the start time, 9.8 m/s^2
     
  20. Mar 12, 2015 #19

    NascentOxygen

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    Staff: Mentor

    If a stationary particle were to have no acceleration, it would forever remain at rest.
     
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