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Deriving the formula for distance

  1. Oct 14, 2013 #1
    I am currently redoing grade 11 physics and they have introduced the following formula,
    [tex]\Delta d = v_1\Delta t + \frac{1}{2}a\Delta t^2[/tex]
    I am trying to find this equation using calculus.
    So from what I understand, [itex]\Delta d[/itex] is the area under the curve for [itex]v(\Delta t)[/itex].
    We can define [itex]v(\Delta t) = v_1 + a\Delta t[/itex] where [itex]\Delta t = t_2 - t_1[/itex]
    I can take the integral of [itex]v(\Delta t)[/itex] to find [itex]\Delta d[/itex] call it [itex]D(\Delta t)[/itex]. Where, [itex]D'(\Delta t) = v(\Delta t)[/itex].
    [tex]D(\Delta t) = \int v(\Delta t)\ d\Delta t \implies D(\Delta t) = \int (v_1 + a\Delta t)d\Delta t\implies D(\Delta t) = v_1\Delta t + \frac{1}{2}a\Delta t^2[/tex]

    and suppose I know [itex]v(t)[/itex] could I say:
    [tex]\Delta d = \int_{t_1}^{t_2}v(t)dt = P(t_2) - P(t_1)\ where\ P'(t) = v(t)[/tex]
    [tex]\Delta d = D(\Delta t) \implies D(\Delta t) = P(t_2) - P(t_1)[/tex]

    Is this all correct ? Any feedback would be appreciated.
     
  2. jcsd
  3. Oct 14, 2013 #2

    UltrafastPED

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    We define velocity as v=dx/dt, and acceleration as a=dv/dt.

    Start with the constant acceleration, a. One integration gives the change in velocity as v=at; the second gives the change in distance as d=1/2 at^2.

    Then add in the expressions for an initial velocity, and an initial displacement - this gives the general result for constant acceleration.
     
  4. Oct 14, 2013 #3
    So are you saying since [itex] a = \frac{\Delta v}{\Delta t}\ and\ v = \frac{\Delta d}{\Delta t}[/itex]
    [tex]\Delta v = \int (a)dt = at[/tex]
    [tex]\int (\Delta v) dt = \int (at)dt = \frac{1}{2}at^2[/tex]
    and since [itex]v_1[/itex] is constant
    [tex]\int (v_1)dt = v_1t[/tex]
    and since [itex]v(t) = v_1[/itex] when [itex]t = 0[/itex], we have
    [tex]\int v(t)dt = \int (v_1)dt + \int (\Delta v)dt[/tex]
    [tex]\implies \int v(t)dt = v_1t + \frac{1}{2}at^2[/tex]

    Is this what you mean? Where does initial displacement come into play? How would I proceed to use definite integrals ?
    [tex]\int_{t_1}^{t_2}v(t)dt = (v_1t_2 + \frac{1}{2}at_2^2) -(v_1t_1 + \frac{1}{2}at_1^2)[/tex]
    does this work ?
     
    Last edited: Oct 14, 2013
  5. Oct 14, 2013 #4
    well I suppose all we did was define v(t) and took the integral. I'm still not sure what you mean by adding initial displacement.
     
  6. Oct 14, 2013 #5

    UltrafastPED

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  7. Oct 14, 2013 #6

    jtbell

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    Here's another way to do it besides the one in the link UltrafastPED gave. Integrate the acceleration as an indefinite integral, so you add a constant of integration:
    $$v=\int {a dt}\\
    v = at + C$$

    To find the constant of integration, you have to specify an "initial condition", namely that ##v = v_0## at t = 0:
    $$v_0 = a \cdot 0 + C\\
    C = v_0$$

    Therefore ##v = at + v_0##

    You can probably do the second integration to get x, with the initial condition ##x = x_0## at t = 0.
     
  8. Oct 14, 2013 #7
    Thanks guys !
     
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