Answer:
The vector for the ball’s position is [tex]\vec{r} = (18.1t)\hat{i} + (4.20t - 4.90t^2)\hat{j} \:m[/tex]
Step-by-step explanation:
The position vector for a particle moving in the x-y plane can be written
[tex]\vec{r} = x\hat{i} + y\hat{j}[/tex]
where x, y, and [tex]\vec{r}[/tex] change with time as the particle moves while the unit vectors [tex]\hat{i}[/tex] and [tex]\hat{j}[/tex] remain constant.
We know that the x and y coordinates as functions of time are given by [tex]x = 18.1t[/tex] and [tex]y = 4.20t - 4.90t^2[/tex], where x and y are in meters and t is in seconds.
Therefore, the vector for the ball’s position is [tex]\vec{r} = (18.1t)\hat{i} + (4.20t - 4.90t^2)\hat{j} \:m[/tex]
