devanlevin
Nov6-08, 02:13 PM
given the acceleration
a_{x}=6t^{2}
a_{y}=-\frac{2PI}{3}cos(\frac{2PI}{3}t)
knowing that the starting point was (0,0) -- \vec{r}_{0}=0
and that the starting velocity was \vec{V}_{0}=\hat{-x}
find the velocity vector and placement vector,
what i did was, to get from \vec{a} to\vec{v}, integral on a_{x} and a_{y}
\inta_{x}dx=v_{x}=6\frac{t^{3}}{3}+v_{x_{0}}[/tex =2t^{3}-1
\inta_{y}dx=v_{y}=-\frac{2PI}{3}*\frac{sin\frac{2PI*t}{3}}{\frac{2PI} {3}}+v[tex]_{y_{0}}[/tex =-sin\frac{2PI}{3}t+0
which works out fine, \vec{V}=(2t^{3}-1,-sin\frac{2PI}{3}t) which in the case where t=0s is equal to -\hat{x} and is true to the facts
but
to get from \vec{v} to\vec{r}, -- integral on v_{x} and v_{y}
\intvx=2\frac{t^{4}}{4}-t+x_{0}=\frac{1}{2}t^{4}-t
\intvy=-([tex]\frac{-cos\frac{2PI*t}{3}}{\frac{2PI}{3}}+y_{0}=\frac{3}{ 2PI}cos(\frac{2PI}{3}t)
in this case, when t=0s, x_{0} is really 0 and true to the facts but y_{0} is \frac{2PI}{3} (since cos(0)=1) can anyone see where i have gone wrong?
a_{x}=6t^{2}
a_{y}=-\frac{2PI}{3}cos(\frac{2PI}{3}t)
knowing that the starting point was (0,0) -- \vec{r}_{0}=0
and that the starting velocity was \vec{V}_{0}=\hat{-x}
find the velocity vector and placement vector,
what i did was, to get from \vec{a} to\vec{v}, integral on a_{x} and a_{y}
\inta_{x}dx=v_{x}=6\frac{t^{3}}{3}+v_{x_{0}}[/tex =2t^{3}-1
\inta_{y}dx=v_{y}=-\frac{2PI}{3}*\frac{sin\frac{2PI*t}{3}}{\frac{2PI} {3}}+v[tex]_{y_{0}}[/tex =-sin\frac{2PI}{3}t+0
which works out fine, \vec{V}=(2t^{3}-1,-sin\frac{2PI}{3}t) which in the case where t=0s is equal to -\hat{x} and is true to the facts
but
to get from \vec{v} to\vec{r}, -- integral on v_{x} and v_{y}
\intvx=2\frac{t^{4}}{4}-t+x_{0}=\frac{1}{2}t^{4}-t
\intvy=-([tex]\frac{-cos\frac{2PI*t}{3}}{\frac{2PI}{3}}+y_{0}=\frac{3}{ 2PI}cos(\frac{2PI}{3}t)
in this case, when t=0s, x_{0} is really 0 and true to the facts but y_{0} is \frac{2PI}{3} (since cos(0)=1) can anyone see where i have gone wrong?