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Question on fundemental theorem of calc.

  1. Sep 3, 2006 #1
    If a particle traverse distance 1 in time, beginning and ending at rest, then prove that at some point in the interval it must have been subjected to an accleration equal to 4 or more.

    i have a feeling that this question in order to answer it you need to use the intermediate theorem of integral calculus and the fundemental theorem of calcs, my problem is i dont know how to formulate the answer.
     
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  3. Sep 3, 2006 #2

    StatusX

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    If you know the total distance and time, you know the average velocity. The mean value theorem tells you it must actually have had this velocity at some point. If it had this velocity at time t1, then you know the average acceleration on [0,t1] and on [t1,tf]. etc.
     
  4. Sep 3, 2006 #3
    but then isn't my answer depneds on the way i divide the interval, i mean i could find the accleration greater than 5, 6 etc.
    i mean if [0,1/5] [1/5,2/5] [2/5,3/5] [3/5,4/5] [4/5,1] then the avg velocity is 1 in one of these intervals and thus the avg accleration is greater than 5.
     
  5. Sep 3, 2006 #4

    StatusX

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    Just divide it into two intervals. If one is too long to give you a high enough average acceleration, look at the other one.
     
  6. Sep 3, 2006 #5

    0rthodontist

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    No--because you might know the velocity to be 1 at some point in an interval, but you don't know the velocity at the endpoints of the interval. It could have zero acceleration in any particular interval that has velocity 1, if the velocity is also 1 at the endpoints of the interval.

    StatusX, I may be doing it wrong, but when I try your method I can only get a minimum acceleration of 2. There is a geometric argument:

    Let ABC be the triangle where A = (0, 0), B = (1/2, 2), C = (1, 0). If the acceleration is less than 4, the velocity may never reach a point on or above the top edges of ABC--it can be verified algebraically using the MVT that this would yield an acceleration whose absolute value is greater than or equal to 4. But since the triangle's area is exactly 1, if the velocity is uniformly less than its top edges, the integral of the velocity must be less than 1; contradiction.
     
    Last edited: Sep 3, 2006
  7. Sep 3, 2006 #6
    i dont understand statusx why should i divide the interval into only 2 subintervals cant i divide it any way that i want?
    and in some subinterval i would get the velocity 1
     
  8. Sep 4, 2006 #7

    0rthodontist

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    I believe StatusX's point is this:
    --Divide the interval (0, 1) into (0, x) and (x, 1) where v(x) = 1, if v is the velocity function. You know this can be done because of the mean value theorem and the fact that the position function is continuous and passes through the points <0, 0> and <1, 1>.
    --Depending on whether x is closer to 0 or to 1, at least one of (0, x) and (x, 1) will have time duration less than or equal to one half (x <= 1/2 or 1-x <= 1/2).
    --By the MVT, the acceleration exactly equals one of 1/x or -1/(1-x) over that interval, for an absolute value of at least 2.

    This depends on your knowledge that v(0) = 0, v(x) = 1, v(1) = 0. If you divided up (0, 1) into an arbitrary number of intervals, you wouldn't know the velocity at all of the endpoints.
     
    Last edited: Sep 4, 2006
  9. Sep 4, 2006 #8
    but then you get that the accleration is greater or equals 2 and not greater or equals 4.
     
  10. Sep 4, 2006 #9

    StatusX

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    Right, sorry, my method doesn't give a good enough bound. Try orthodontist's idea. Try integrating the velocity from 0 to 1/2, keeping in mind you start with v=0 and must have |v'|<4. You should be able to show the integral is less than 1/2, and likewise if you integrate from 1 back to 1/2.
     
    Last edited: Sep 4, 2006
  11. Sep 4, 2006 #10

    HallsofIvy

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    Am I missing something? "a particle traverse distance 1 in time". In what time? In order to know the acceleration at any point, you are going to have to know at least the average time and that requires knowing the time in which the particle took to "traverse distance 1".
     
  12. Sep 5, 2006 #11
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