Given data:
* The x-coordinate of the position at time t is,
[tex]x=t^2[/tex]
* The y-coordinate of the position at time t is,
[tex]y=t^3-2t[/tex]
Solution:
The position vector at time t is,
[tex]\vec{r}=t^2i+(t^3-2t)j[/tex]
The velocity vector at time t is,
[tex]\vec{v}=\frac{\vec{dr}}{dt}[/tex]
Substituting the known values,
[tex]\begin{gathered} \vec{v}=\frac{d}{dt}(t^2i+(t^3-2t)j) \\ \vec{v}=2ti+(3t^2-2)j \end{gathered}[/tex]
The acceleration vector at time t is,
[tex]\vec{a}=\frac{d}{dt}(\vec{v})[/tex]
Substituting the known values,
[tex]\vec{a}=\frac{d}{\differentialDt t}(2ti+(3t^2-2)j)[/tex]
