1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    User Avatar
    Staff Emeritus
    Science Advisor

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook