- #1
Alem2000
- 117
- 0
correct me if I am 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
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
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