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Kinematics - Acceleration equation confustion

  1. Jul 13, 2012 #1
    Kinematics -- Acceleration equation confustion

    I have been following a derivation and can't quite wrap my head round what is happening. I don't know if the person who wrote it has made a mistake somewhere or I am making the mistake.

    I am informed that

    [tex]a = \frac{dv}{dt}[/tex]

    Fine. Then I am told

    [tex]a= \frac{dv}{dx}\frac{dx}{dt}[/tex]

    (which is odd, one moment it is dv/dt now it seems to be dv/dx)

    and that I should convince myself that this is the acceleration through a cross multiplication.... but I did it and I got

    [tex]\frac{dvdt}{dx dx}[/tex]

    This surely isn't acceleration, have I made a mistake somewhere? And if the person I am following is wrong, what is the correct way to state this...?

    They further said that

    [tex]\frac{dx}{dt}[/tex]

    is just velocity, so they end up with

    [tex]a = v \frac{dv}{dx}[/tex]

    Can someone help?
     
  2. jcsd
  3. Jul 13, 2012 #2
    Re: Kinematics

    No.

    The acceleration a does NOT seem to be dv/dx.

    It would seem that a is dv/dx if we have a = dv/dx and we do not have that because we have instead that

    a = (dv/dx)(dx/dt).
     
  4. Jul 13, 2012 #3

    ehild

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    Re: Kinematics

    Assume that the acceleration is function of the position x: a=a(x(t)), a composite function of t. You get the derivative with respect to t if you derive with respect to x first then derive x with respect to t.

    da(x(t))/dt=(da/dx)*(dx/dt).

    ehild
     
  5. Jul 13, 2012 #4
    Re: Kinematics

    Hi help1please,
    you just messed up your cross multiplication
    dv/dx * dx/dt is not dvdt/dxdx but dvdx/dtdx as you can see, you could just as well simplify by the 'artificially' injected dx/dx (=1) and get back to a=dv/dt

    Cheers...
     
  6. Jul 13, 2012 #5

    CAF123

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    Re: Kinematics

    Acceleration is (dv/dt) but by application of the chain rule this can be written as a = (dv/dx)(dx/dt) because we are effectively cancelling out the dx top and bottom.
    (I say effectively because dx is really a differential operator)

    In your last eqn, (dx/dt) = v so we get a = v (dv/dx)

    This sort of chain rule application can be important if you want to find the velocity of a particle as a function of position, acceleration as function of velocity etc.., although this requires some extra mathematics.(Separable diff eqns)
     
  7. Jul 13, 2012 #6

    CWatters

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    Re: Kinematics

    You probably understand it by now but..

    a = dv/dt

    Multiply top and bottom by dx

    = (dv/dt)(dx/dx)

    rearrange

    = (dv/dx)(dx/dt)
     
  8. Jul 13, 2012 #7
    Re: Kinematics

    That's simply some manipulation (multiplying numerator and denominator by same thing), scratch out dx from numerator and denominator, you get the same expression as before. :wink:
     
  9. Jul 16, 2012 #8
    Re: Kinematics

    I didn't mess up anything, it was work I was following and noticed something was wrong, that's all. But thanks!
     
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