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Homework Help: Prove magnitude of acceleration.

  1. Feb 18, 2008 #1
    Going a distance S in time T, prove at some instant the magnitude of the acceleration of the car is (4S)/(T^2)

    I have thought a lot about the problem and how I would go about proving it. Would using the Mean Value Theorem be the best way?

    I was thinking that I could take two times A and B during the trip, and according to the MVT there exists another time c such that v'(c) <= (v(b) - v(a) ) / (b - a )

    I also thought about doing an epsilon-delta proof to state that as (b - a) approaches T, (v(b) - v(a) ) / (b - a ) would approach a value that was equal to or greater than (4S)/(T^2).

    I am not sure what to do to prove this. Any suggestions would be great!
  2. jcsd
  3. Feb 18, 2008 #2


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    There is something missing in the statement of the problem. For example if the car is going at a constant speed S/T, then the acceleration will always be 0.
  4. Feb 18, 2008 #3
    The problem is given in its entirety. My thought is that a car can't go from point A to point B without ever accelerating, so initial velocity = 0. I believe I would do it using the Mean Value Theorem, to prove that there exists a mag of acceleration equal to 4S/T^2 somewhere between point A and B
    Last edited: Feb 18, 2008
  5. Feb 18, 2008 #4
  6. Feb 18, 2008 #5
    I know, I just can't seem to apply it in the context of this problem...

    I know that a(t)=v'(t) and there is some point is [0,S] where (S - 0 ) / (T - 0 ) = (4S) / (T^2 )

    I know the magnitude of acceleration is the slope of the tangent at a point on a speed-time graph.

    Any other hints for setting up a proof using the MV Theorm? Or maybe a better way? I can't have someone work it all out for me, I just ask for any hints or suggestions.

  7. Feb 18, 2008 #6
    Are we assuming it starts from rest? How about just integrate the acceleration given and show that if the acceleration is less then that value then there is not enough acceleration to cover the distance in time.
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