Deriving the formula for distance

  • Thread starter DarthRoni
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  • #1
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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.
 

Answers and Replies

  • #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.
 
  • #3
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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 ?
 
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  • #4
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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
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.
 
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  • #7
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Thanks guys !
 

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