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Vector analysis question on acceleration

  1. Feb 21, 2009 #1
    1. The problem statement, all variables and given/known data

    A moving particle reaches its max. speed at the instant t = 3. (Before and after 3, its speed is less.) It follows that the particle's acceleration is 0 at the instant t = 3... Show that this is FALSE

    2. Relevant equations

    v = dR/dt
    a = d^2 R / dt^2

    3. The attempt at a solution

    How do I show this is false? The derivative of velocity is acceleration so I would think it's true and is indeed 0. This is on a chapter for vector analysis on acceleration and curvature.
     
  2. jcsd
  3. Feb 21, 2009 #2

    gabbagabbahey

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    If the particle's acceleration were 0 at t=3, why would it slow down?:wink:
     
  4. Feb 21, 2009 #3
    But the acceleration has to be 0 because at this point the speed is maximum... so on one side there should be negative acceleration and on the other side positive.
     
  5. Feb 21, 2009 #4

    gabbagabbahey

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    No, if the acceleration at t=3 were zero, then if you measured the speed of the particle a very short time later it would be unchanged.

    It's true that [tex]\frac{d}{dt} ||\vec{v}||=0[/tex] at t=3, but that doesn't necessarily mean [tex]||\vec{a}||=0[/tex] at t=3.

    This rests on the fact that [tex]\vec{v}[/tex] is a vector, it has both magnitude and direction and just because its magnitude isn't changing doesn't mean it's direction can't be changing. If it's direction is changing, then [tex]\vec{a}=\frac{d\vec{v}}{dt}\neq0[/tex] :wink:

    There is an expression for [itex]\vec{a}[/itex] in terms of its tangential and normal components that you should know (it involves curvature), use that to prove that the acceleration is non-zero at t=3!:smile:
     
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