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Deriving constant acceleration equations

  1. Sep 27, 2011 #1
    Tired and not understanding Calculus.

    So, velocity equals dx/dt
    and acceleration is dv/dt

    But to derive the constant acceleration equations, my professor took the integral of vdt = dx, which we got from v = dx/dt.

    I follow all of the math that comes after that, but I don't understand why this integral produces equations for constant acceleration. Why are we taking the integral of velocity if acceleration is the derivative of velocity, and why does this only work for constant acceleration?

    Thanks.
     
  2. jcsd
  3. Sep 27, 2011 #2
    While I'm not quite sure of the desired equations, what you could be aiming at is like the following:
    Let a = dv/dt be the constant acceleration.
    Then integral( a dt) = integral( dv ) for suitable integral limits.
    Since a is constant it slips through the integral and
    a integral( dt ) = integral( dv )
    For an initial value problem which specifies the velocity, v_0 at time t_0, one can solve this as
    a (t - t_0) = v - v_0 => v = v_0 + a(t - t_0). Now suppose you'd like to get x into the act,
    v = dx/dt
    So,
    dx/dt = v_0 + a(t - t_0)
    Taking integrals with suitable limits,
    integral( dx ) = integral( v_0 dt ) + a ( integral( t dt ) - t_0 integral( dt ) )
    Taking x = x_0 at t = t_0
    x - x_0 = v_0 (t - t_0) + a[ (t^2/2 - t_0^2/2) - t_0 (t - t_0) ]
    => x - x_0 = (t - t_0)[v_0 + a (t - t_0)/2]
    And I hope I haven't messed up the algebra.

    Is this the kind of thing you were looking for?
     
  4. Sep 27, 2011 #3

    HallsofIvy

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    Just v= dx/dt can't "produce equations for constant acceleration" because that equation alone says nothing about constant acceleration. You can start from "acceleration= dv/dt" and write dv= adt where a is that constant acceleration. Since a is constant that's easy to integrate and gives [itex]v= at+ v_0[/itex]. NOW we can use v= dx/dt to write [itex]dx/dt= at+ v_0[/itex] so that dx= (at+v_0)dt and integrate both sides of that to get [itex]x= (a/2)t^2+ v_0t+ x_0[/itex]. Is that what your professor did?
     
  5. Sep 27, 2011 #4
    Yes, sorry, I should have been more clear. My professor substituted v0+at for v to get (v0+at)dt = dx.

    So, I see that any integration that results from this would have a as a constant, simply because a must be a constant in order to get through the integration. But I don't understand why we integrate this to get a kinematic equation. Simplified, this is basically v = dx/dt. Why do we integrate this to get kinematic equations? I don't understand the first logical leap here.
     
  6. Sep 27, 2011 #5

    cepheid

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    [tex] a(t) = a = \textrm{const.} [/tex]

    By definition:

    [tex] v(t) = \int a(t)\,dt [/tex]

    [tex] = a \int dt = at + C_1 [/tex]

    To solve for C1, consider t = 0:

    [tex] v(0) = v_0 = a*0 + C_1 [/tex]

    Hence C1 = v(0) and therefore:

    [tex] v(t) = v_0 + at [/tex]

    Now, by definition

    [tex] x(t) = \int v(t)\,dt[/tex]

    [tex] = \int (v_0 + at)dt [/tex]

    [tex] = v_0t + \frac{1}{2}at^2 + C_2 [/tex]

    Consider t = 0:

    [tex] x(0) = x_0 = C_2 [/tex]

    Therefore:

    [tex] x(t) = x_0 + v_0t + \frac{1}{2}at^2[/tex]

    Now, can you explain clearly what specifically about this derivation you don't understand?
     
  7. Sep 27, 2011 #6

    lavinia

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    A kinematic equation describes position in terms of elapsed time. The integral gives you the position,x, in terms of elapsed time,t.
     
  8. Sep 27, 2011 #7
    Noting. That was perfect, thank you. I don't know why, but I was just totally spacing out and needed it laid out in front of me. I think you made it clearer what was going on than my professor did. He wrote it out different at the start, but now I see it means the same thing, only this was more intuitive for me. Thanks a lot.
     
    Last edited: Sep 27, 2011
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