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Help me understand kinematics lecture on derivation of laws of motion

  1. Sep 4, 2010 #1
    I was checking online videos for lectures on kinematics and I watched one made by yale university but there was some part of it that I didn't understand when he talks about calculus derivation of Vf^2=Vi^2 + 2ax
    here is the video

    the derivation starts from about 1:08:45
    I didn't understand the part of canceling the dts even though he explained it When it can be allowed and when not? and why does it have to be vanishingly small as he says on 1:11:00? and why does he say in 1:12:14 if delta x is small like 1 second what does that mean? is it distance or time?
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Sep 4, 2010 #2


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    In essence what you are doing is integrating both sides w.r.t. t. So when he has d/dt(v2/2)=a(dx/dt)

    and he 'cancels' the 'dt's it becomes like this:

    [tex]\int \frac{d}{dt}(\frac{v^2}{2})dt = \int a \frac{dx}{dt}dt[/tex]

    'a' is constant so you get:

    [tex]\int \frac{d}{dt}(\frac{v^2}{2})dt = a \int \frac{dx}{dt}dt[/tex]

    The limits of integration would just be from final to initial.
    which will give him

    [tex] \frac{v^2}{2} \right| _{v_i} ^{v_f} = ax \right|_{x_0} ^{x}[/tex]

    If you differentiate something and then integrate it back ,you will get the something. Which will give you the result.

    That is the 'proper' way to do it. The canceling he did just happens to work since differentials are defined by the limit as it goes to zero.

    So a= dv/dt = lim(Δt→0) Δv/Δt
  4. Sep 4, 2010 #3
    ok but why does it only work when delta t is vanishingly small? as he said? and it wont work if it's one second?
  5. Sep 4, 2010 #4


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    Because dv/dt = lim(Δt→0) Δv/Δt this means that you will only get the actual value of dv/dt iff Δt is infinitesimally small (or tending to zero).
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