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Zero velocity, Zero acceleration

  1. Feb 19, 2015 #1
    In one dimension, the acceleration of a particle can be expressed as:
    $$a = \frac{dv}{dt} = \frac{dx}{dt} \frac{dv}{dx} = v \frac{dv}{dx}$$
    Does this equation imply that if ##v = 0## then ##a = 0##?

    EDIT: I think I figured out the source of my confusion. This equation is a differential equation and I cannot just plug in values for ##v##, am I right?
     
    Last edited: Feb 19, 2015
  2. jcsd
  3. Feb 19, 2015 #2
    [strike]I don't think your 3rd step is correct.
    ##a(t)=\frac{d}{dt}v(t)=\frac{d^2}{dt^2}x(t)=\frac{d}{dt}[\frac{d}{dt} x(t)]##.[strike] disregard this comment. I made a mistake originally.
     
    Last edited: Feb 19, 2015
  4. Feb 19, 2015 #3
    It is correct. I applied the chain rule.
     
  5. Feb 19, 2015 #4
    ##a=\int v## ##dt##, so there will always be the constant of integration. V may or may not be 0 when a=0.(This is entirely determined by the value of v at t=0)
    EDIT: Its ##v=\int a## ##dt##, major mess up on my end, sorry.
     
    Last edited: Feb 19, 2015
  6. Feb 19, 2015 #5
    Ah, that's what I must've been seeing wrong.
    I knew it was something. But shouldn't your a be x, as your integrating v(t)?
     
    Last edited: Feb 19, 2015
  7. Feb 19, 2015 #6
    Please re-check your calculations.
     
  8. Feb 19, 2015 #7
    I'm in the process of doing just that.


    You are correct. I realized what my mistake was. My apologies.
     
  9. Feb 19, 2015 #8
    This is wrong if by "a" you mean acceleration. The integral of v is the displacement and not the acceleration.
     
  10. Feb 19, 2015 #9
    Crap, sorry I interchanged the terms.
    It should be ##v=\int a dt##. (How very embarrassing for me. Please excuse it as a major slip of the finger :P)
     
  11. Feb 19, 2015 #10
    It implies that only if the derivative is finite at v=0. I mean the dv/dx.
    For a simple case, like a body thrown straight up for example, the velocity is zero at the top of the trajectory but the derivative dv/dx is proportional to 1/v so it will blow up at v=0. But the product vdv/dx will have a finite, non-zero value.

    So you may have v=0 and a=0 at the same time, as discussed in the other thread. But is not a necessary condition. v=0 does not imply a=0.

    And you did not make any mistake in your derivation of the formula.
     
  12. Feb 19, 2015 #11
    I would like to take this time to point out to The OP, in case he missed my edit, that I made a mistake in my calculations the first time about, and your derivation was correct. I apologize for possibly misleading you.
     
  13. Feb 19, 2015 #12

    PeroK

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    You had them all fooled for a bit with this one. To be a little more precise: your equation holds when all the functions involved are differentiable. The problem is that dv/dx may not be differentiable at certain points. That's when the use of the chain rule breaks down. The statement of the chain rule would say "... and if v(x) is differentiable at x ..".
     
  14. Feb 19, 2015 #13
    Here's the problem:

    Suppose a particle ##P## has an acceleration ##a##. As shown above:

    $$a = v \frac{dv}{dx}$$

    Now, both ##v## and ##\frac{dv}{dx}## are functions of ##x##. I will call them ##f(x)## and ##g(x)## respectively.

    $$a = f(x) g(x)$$

    Where ##g = f'##.

    Suppose one solution to the equation ##f(x) = 0## is ##x = b##, and I wish to find the acceleration of ##P## at this particular value of ##x##.

    $$a = f(b) g(b) = 0 ⋅ g(b) = 0$$

    The reason I find this result bizarre is that I can think of many situations where a body is at instantaneous rest, but still possesses a nonzero acceleration component due to a nonzero net force at that point, so I know I must have gone wrong somewhere. Where did I go wrong?
     
  15. Feb 19, 2015 #14

    PeroK

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    I already explained that above. You assumed that v is a differentiable function of x.

    "The problem is that dv/dx may not be differentiable at certain points. That's when your use of the chain rule breaks down."
     
  16. Feb 19, 2015 #15
    So if ##v## is a differentiable function of ##x##, then ##v = 0## always implies ##a = 0##?
     
  17. Feb 19, 2015 #16

    PeroK

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    Yes, but:

    ##\frac{dv}{dx} = \frac{\frac{dv}{dt}}{\frac{dx}{dt}} = \frac{a(t)}{v(t)}##

    So, if v = 0, then dv/dx has a tendency to be undefined at that point.
     
    Last edited: Feb 20, 2015
  18. Feb 20, 2015 #17
    This is a ##\frac{0}{0}## situation, which means we cannot compute the derivative using this equation alone; we need to be given ##v## as a function of ##x##, right?
     
  19. Feb 20, 2015 #18

    PeroK

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    You're missing the point that if v = 0, then dv/dx is undefined.

    Try to find an example of a function v(t) where v(t) = 0 at some point and dv/dx exists (finite) at that point.
     
  20. Feb 22, 2015 #19
    No it does not. If v = 0, and v STAYS equal to zero, then yes, a=0. Said in more detail, if v=0 at time t, and v is still zero a moment later, then a=0. As you've said yourself, this is a differential equation, and v, a and t all represent variables. You can't just plug in numbers like that.

    To give you another example, consider two objects that have equal velocities for some instant in time. Are they accelerating in the same manner? In other words, if v1=v2, is a1 = a2?

    Answer: not unless v1 is always equal to v2. In other words, if the velocities are variables over some time, then yes the accelerations must be equal, but if they only represent numbers (or points in time), then then the accelerations are not necessarily equal.
     
  21. Feb 22, 2015 #20
    If x1=x2, does v1=v2? In other words, if two objects are next to each other at the same position next to each other, does this mean they have the same velocities?
    What if they are ALWAYS next to each other?
     
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