Answer
$ |\vec{v}_{avg}| =1.89$ m/s
Work Step by Step
$(c)$
We have found $\Delta\vec{r}=(-40.0m)\hat{i}+(40.0m)\hat{j}.$
Eq 4-8 gives us $ \displaystyle \vec{v}_{avg}=\frac{\Delta\vec{r}}{\Delta t}$
$ \vec{v}_{avg}$=$\displaystyle \frac{\Delta\vec{r}}{\Delta t}=\frac{(-40.0m)\hat{i}+(40.0m)\hat{j}}{30.0s}$
Its magnitude is
$ |\vec{v}_{avg}|$=$\displaystyle \frac{|\Delta\vec{r}|}{\Delta t}=\quad$ (... we found $|\Delta\vec{r}$ in part a.)
$=\displaystyle \frac{56.6m}{30.0s}$
$=1.89 m/s$
