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Challanging Problem It is nessecary to solve quicly

  1. Apr 30, 2008 #1
    Challanging Problem " It is nessecary to solve quicly"

    I have a problem :
    A car moves along the real line from x = 0 at t = 0 to x = 1 at t = 1, with
    differentiable position function x(t) and differentiable velocity function v(t) = x’(t).
    The car
    begins and ends the trip at a standstill; that is v = 0 at both the beginning and the end of
    the trip. Let L be the maximum velocity attained during the trip. Prove that at some time
    between the beginning and end of the trip, l v’ l > L^2/(L-1).

    Can you verify that L > 1 ???

    Thankx
     
  2. jcsd
  3. Apr 30, 2008 #2

    nicksauce

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    You can show L > 1 using the Mean Value Theorem
     
  4. May 2, 2008 #3
    By mean value theorem I can find that L>0
     
  5. May 2, 2008 #4

    nicksauce

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    The MVT would say that there exists c in t = [0,1] such that
    f'(c) = (x(1) - x(0)) / (t1 - t0) = (1-0) / (1-0) = 1.
     
  6. May 9, 2008 #5
    That is my solution
    How I can compleat it??
     

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  7. May 10, 2008 #6
    You are from UOS
    right?????
     
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