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Constant acceleration proof, dv/dt=a

  1. Apr 18, 2010 #1
    My book has stumped me on something that I thought should be simple... perhaps I'm just not getting it.

    Section - Constant Acceleration

    It states the following:

    "When acceleration is constant, the average acceleration and instantaneous acceleration are equal and we can write...

    a = aavg = (v-v0)/(t-0)

    Here v0 is the velocity at time t = 0 and v is the velocity at any later time t. We can recast this equation as:

    v = v0 + at (Eq. 2-11)

    As a check, note that this equation reduces to v = v0 for t = 0, as it must. As a further check, take the derivative of Eq. 2-11. Doing so yields dv/dt = a, which is the definition of a."

    I'm not catching the last (bold) part. The derivative of equation 2-11 being equal to a.

    When I work out the derivative (in my mind) is should work like this:

    dv/dt = at
    =a't + at'
    =1(t) + a(1)
    = t + a

    My mind is leaning towards the idea that the book is trying say that dv/dt = a, at t = 0. Is that what the book is trying to tell me? Because I think I'm just confusing myself.

    Any proof of this would probably help me out.

    Thanks, prior!
  2. jcsd
  3. Apr 18, 2010 #2

    D H

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    What is the derivative of f(t)=2 with respect to t? The derivative of f(t)=c, where c is some constant? Here, a is just another constant.
  4. Apr 18, 2010 #3
    Yeah I'm not really sure why you think this:

    dv/dt = at
    =a't + at'
    =1(t) + a(1)
    = t + a


    Here's how you should see it. Equation 2-11 shows v as a function of time with initial velocity v0. In other words,

    v(t) = v = v0 + at

    Take the first derivative with respect to time. v0 is a constant and derivative of a constant is zero. a is also a constant and the derivative of a constant times a variable is the constant itself. So...

    v'(t) = a
  5. Apr 18, 2010 #4
    That did it, I was thinking of at as being the product of two variables... instead of say, dv/dt = 6t = 6 for example... now THAT I can understand.

    Thanks for the help, it's been a few years...
  6. Apr 18, 2010 #5
    Ah I see so you were using the product rule to get = t + a

    Well, good that you understand now :)
  7. Apr 18, 2010 #6
    Funny actually, now that I think about it, I'm in the CONSTANT ACCELERATION section... CONSTANT... doh!

    Thanks again!
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