To find the average velocity vector $\vec{v}_{av}$ from $t = 0$ to $t = 5s,$ we use the formula:$\\ \vec{v}_{av} = \frac{\Delta \vec{r}}{\Delta t} \\$where $\Delta \vec{r}$ is the change in position vector and $\Delta t$ is the change in time.$\\$Given positions:$\\$• At $t = 0,$ position $(x_0, y_0) = (2m, 3m) \\$• At $t = 5s,$ position $(x_5, y_5) = (13m, 14m) \\$Calculate the change in position:$\\ \Delta \vec{r} = (x_5 - x_0)\hat{i} + (y_5 - y_0)\hat{j} \\ \Delta \vec{r} = (13m - 2m)\hat{i} + (14m - 3m)\hat{j} \\ \Delta \vec{r} = 11\hat{i} + 11\hat{j} \\$Calculate the change in time:$\\ \Delta t = 5s - 0s = 5s \\$Substitute into the average velocity formula:$\\ \vec{v}_{av} = \frac{11\hat{i} + 11\hat{j}}{5} \\ \vec{v}_{av} = \frac{11}{5} \hat{i} + \frac{11}{5} \hat{j} \\$Thus, the average velocity vector is $\frac{11}{5} (\hat{i} + \hat{j}). \\$This corresponds to Option 4.