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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) = x0(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
    The average velocity is 1 and since the position function is continuous (since its differentiable) and v(0) = 0 = v(1), there must be some time at which the velocity is greater than 1 (for otherwise the average would be less than 1). Therefore L > 1.

    I take it that v' is the derivative of the velocity function, i.e. the acceleration?
     
  4. May 2, 2008 #3
    Thnkx very much, but how do u know that The average velocity is 1??
     
  5. May 2, 2008 #4

    exk

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    L=1 simply by taking the slope of a straight line that connects the 2 endpoints.
     
  6. May 2, 2008 #5

    HallsofIvy

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    The car went distance 1 in time 1: 1/1= average speed!
     
  7. May 6, 2008 #6
    How u jumped from "The average velocity is 1 and since the position function is continuous (since its differentiable) and v(0) = 0 = v(1)" To "L > 1."

    Is it a rule ?Give me the sorce u got that from
    If not ,how u profed it??
     
  8. May 9, 2008 #7
    tHESE R my trials
    pls tell me wt I must do??
     

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  9. May 10, 2008 #8

    HallsofIvy

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    Actually, it is NOT true that L> 1. If the car moved with constant velocity, then L would be 1. The average speed is 1. If that speed is not constant, then there must be sometime when the velocity was less than 1, some time when the velocity was greater than 1. What is true is that [itex]]L\ge 1[/itex].
     
  10. May 10, 2008 #9

    Redbelly98

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    The velocity is zero at the beginning and end of the trip, which is less than the average. So it must be greater than the average for some portion of the trip.
     
  11. May 10, 2008 #10
    How I can complete the question??
    I know that L>1
    I think I must verify that the opposite of l v’ l > L^2/(L-1) is wrong
    pls help me today?
     
  12. May 10, 2008 #11

    HallsofIvy

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    Thanks. I didn't even think of that!!
     
  13. May 10, 2008 #12

    HallsofIvy

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    Yes, that's a good idea! Notice that | |. You must take both acceleration and deceleration into account. Suppose the acceleration were never greater than L^2/(L-1). In fact, in order to make it simpler, suppose the acceleration were always equal to L^2/(L-1). How long would it take the car to reach speed L? If that is larger than 1/2, the car has less than 1/2 second left to go from L to 0. Assuming the deceleration is never less than -L^2/(L-1), how long will it take the car to go from speed L to 0?
     
  14. May 10, 2008 #13

    Redbelly98

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    I tried (and failed) in an attempt to contradict the statement we're supposed to prove. Choosing a constant acceleration for 0<t<0.5, and the negative of that acceleration for 1/2<t<1, I got the following.

    In order to have x=1 at t=1:

    a = 4, t<0<1/2
    a = -4, 1/2<0<1

    v = 4t for t<0<1/2
    v = 4(1-t) for 1/2<t<1

    So L = 2 (maximum of v, occurs at t=1/2)

    And |v'| = |a| is equal to L^2/(L-1) = 4, never greater than it.

    But ... in this example v is not differentiable at t=1/2, so this doesn't disprove the statement after all.

    p.s on a side note, I notice that it was "necessary to solve quickly" this problem more than 1 week ago. When is/was the actual deadline for this?

    I prefer it when thread titles are actually descriptive of the question being asked. Too many threads titled "URGENT" or "solve quickly" or "very very important" can be trying on peoples' goodwill and patience.
     
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