Answer
$|\Delta\vec{r}|=56.6$ m
Work Step by Step
Define a coordinate system.
Let the flagpole be the origin,
let the +x direction be due east,
and +y due north.
The formulas available to us :
$\left[\begin{array}{lll}
\vec{r}=x\hat{i}+y\hat{j}+z\hat{k} & (4- 1) & \text{Position vector} \\
\Delta\vec{r}=\vec{r}_{2}-\vec{r}_{1} & (4- 2) & \text{Displacement} \\
\vec{v}_{avg}=\frac{\Delta\vec{r}}{\Delta t} & (4- 8) & \text{average velocity} \\
\vec{a}_{avg}=\frac{\vec{v}_{2}-\vec{v}_{1}}{\Delta t}=\frac{\Delta\vec{v}}{\Delta t} & (4- 15) & \text{average acceleration}
\end{array}\right]$
The text now gives us
$\vec{r_{1}}=(40.0m)\hat{i} ,\qquad \vec{v}_{1}=(-10.0m/s)\hat{j}$
$\vec{r}_{2}=(40.0m)\hat{j} ,\qquad \vec{v_{2}}=(10.0m/s)\hat{i}.$
$t=30.0s$
(a)
Eq. 4-2 gives us the displacement
$\Delta\vec{r}=\vec{r_{2}}-\vec{r_{1}}=(-40.0m)\hat{i}+(40.0m)\hat{j}$
Magnitude:
$|\Delta\vec{r}|=\sqrt{(-40.0\mathrm{m})^{2}+(40.0\mathrm{m})^{2}}=56.6$ m
