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Integrating gravity(g)= -9.8m/s

  1. Sep 25, 2004 #1
    correct me if im wrong....

    By integrating gravity(g)= -9.8m/s....you get the motion equations with constant acceleration

    I didnt know how to set the limits of the integral...LATEX IS TUFF
    [tex]d\vec{a}=(-g)dt[/tex] 1

    [tex]\int_{\vec{V}_0}^{\vec{V}}dv=\int_{t_0}^{t_1}{-g}dt[/tex] 2

    [tex]\Delta{\vec{v}}=(-g)t[/tex] 3

    [tex]\vec{v}dt=d\vec{r}[/tex] 4

    [tex]\d\vec{r}=(\vec{v}_0+(-g)t)dt[/tex] 5

    [tex]\intd\vec{r}=\int(\vec{v}_0+(-g)tdt[/tex] 6

    [tex]\Delta\vec{r}=\vec{v}_0t+1/2(-g)t^2[/tex] 7

    [tex]\Delta{\vec{v}}=\int{\vec{v}_0}dt[/tex] 8


    HOLD ON A SEC IM TRYN TO LATEX...Can anyone prove the force equations?

    HI merons dad
     
    Last edited: Sep 25, 2004
  2. jcsd
  3. Sep 25, 2004 #2

    Tide

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    Are you talking about Newton the physicist or Newton the mathematician?

    Physics: force = mass times acceleration or change in momentum per unit time

    acceleration = time rate of change of velocity

    Mathematics: [itex]a = \frac {dv}{dt}[/itex]
     
  4. Sep 25, 2004 #3
    Gravity

    Did I make some mistakes...?..i think i fixed them all. Can anyone show how to derive Torqu?

    [tex]\Delta{\vec{v}}=(-g)t[/tex] 3

    [tex]\vec{v}dt=d\vec{r}[/tex] 4

    [tex]d\vec{r}=(\vec{v}_0+(-g)t)dt[/tex] 5

    [tex]\intd\vec{r}=\int(\vec{v}_0+(-g)tdt[/tex] 6

    [tex]\Delta\vec{r}=\vec{v}_0t+1/2(-g)t^2[/tex] 7

    [tex]\Delta{\vec{v}}=\int{\vec{v}_0}dt[/tex] 8
     
    Last edited: Sep 25, 2004
  5. Sep 25, 2004 #4
    Oooops I ment to put that on the physics posts not calculus :frown: :frown:
     
  6. Oct 30, 2004 #5
    how did he figure it out..experiment?
     
    Last edited: Oct 30, 2004
  7. Nov 1, 2004 #6
    No English!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
     
  8. Nov 1, 2004 #7

    Gokul43201

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    Alem, there appears to be a problem with your line #8.
     
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