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Slowest particle for decreasing acceleration

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data
    I need to find the acceleration, a(t), for which a particle moves on a straight line of a length D.
    V(0)=0
    V(final, that is at x = D) = U
    a is always positive, but NON-INCREASING!!!
    What is a(t) for which the particle reaches the end of the line the slowest as possible?


    2. Relevant equations
    Calculus Variational


    3. The attempt at a solution
    Obviously I tried to address it with functional derivative with Lagrange constraints, but
    I failed.
     
  2. jcsd
  3. Apr 30, 2013 #2

    mfb

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    I would guess the best a(t), and show that deviations from that are worse.
     
  4. Apr 30, 2013 #3
    The question does not make any sense."Slowest" is about the speed. However, the end speed is fixed and is U. So it cannot be slower or faster, it can only be U.

    Does the question really mean that the total time taken to go from 0 to D, with V(0) = 0 and V(D) = U, must be maximized?
     
  5. Apr 30, 2013 #4

    mfb

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    "Slowest" could mean the average speed. This is identical to the interpretation you proposed, and the way I interpreted the problem.
     
  6. Apr 30, 2013 #5
    I think you are giving out too much :)
     
  7. May 1, 2013 #6
    Slowest

    That means that it took the object the longest period of time to reach the end.
    In other words, the time that took for the object to get to the end is to be maximised.
     
  8. May 1, 2013 #7
    Guess it must involve variational calculus.
    I'm desperate... :-(
     
  9. May 1, 2013 #8

    mfb

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    You could consider a few examples first, and try to find the optimal solution (without proof). This might give a hint how to prove it, too.
     
  10. May 1, 2013 #9
    Then you need to express total time in terms of v(x). For that, consider how long it takes the particle to go through some very small displacement Δx.
     
  11. May 1, 2013 #10
    Yes, I tried.
    but I couldn't derive something valueable
     
  12. May 1, 2013 #11
    Show what you have tried. By the rules of the forum, you have to demonstrate an attempt, otherwise we can't help you.
     
  13. May 1, 2013 #12
    All integrals boundaries are (0,D) and relating to x.
    v is a function of x.

    T=∫1/v
    ∫v'=U --> ∫[v'-U/D]=0
    Adding Lagrange Multiplier:
    T=∫[1/v-λ(v'-U/D)]
    That functional needs be maximised. Using EL:
    v''=-1/(λv^2)
    Solution to this isnt simple, and moreover seems not to be correct.
     
  14. May 1, 2013 #13
    First of all, T=∫1/v is incorrect. It must be T=∫1/v dx. Second, I do not see where you use the requirement that a is positive and not-increasing.

    Finally, your Euler-Lagrange equation is not even correct.
     
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