Answer
The average velocity with the least magnitude is for the displacement ending at point $C$.
$| \vec{v}_{avg,AC}|=0.0083\mathrm{m}/\mathrm{s}.$
Work Step by Step
The $(x,y)$ coordinates of the points are
$A=(15$ m$, -15$ m$), t_{A}=0,$
$B=(30$ m$, -45$ m$), t_{B}=300\mathrm{s}$
$C=(20$ m$, -15$ m$)$ , $t_{C}=600\mathrm{s}$
$D=(45$ m$,45$ m$)$ , $t_{D}= 900\mathrm{s}$.
Use formulas:
$\left[\begin{array}{lll}
\Delta\vec{r}=\vec{r}_{2}-\vec{r}_{1} & (4- 2) & \text{Displacement} \\
\vec{v}_{avg}=\frac{\Delta\vec{r}}{\Delta t} & (4- 8) & \text{average velocity}
\end{array}\right]$
(a)
$\Delta\vec{r}_{AB}=\vec{r}_{B}-\vec{r}_{A}=(30m-15m)\hat{i}+(-45m-(-15m)\hat{j}\\=(15m)\hat{i}+(-30m)\hat{j} $
$ \displaystyle \vec{v}_{avg,AB}=\frac{\Delta\vec{r}_{AB}}{\Delta t}=\frac{(15m)\hat{i}+(-30m)\hat{j} }{300s}, $
$| \displaystyle \vec{v}_{avg,AB}|=\frac{\sqrt{(15m)+(-30m)} }{300s}=0.11$ m/s
$\Delta\vec{r}_{AC}=\vec{r}_{C}-\vec{r}_{A}=(20m-15m)\hat{i}+(-15m-(-15m)\hat{j}=(5m)\hat{i}$
$| \displaystyle \vec{v}_{avg,AC}|=\frac{5m }{600s}=0.0083$ m/s
$\Delta\vec{r}_{AD}=\vec{r}_{D}-\vec{r}_{A}=(45m-15m)\hat{i}+(45m-(-15m)\hat{j}=(30m)\hat{i}+(60m)\hat{j} $
$| \displaystyle \vec{v}_{avg,AD}|=\frac{\sqrt{(30m)+(60m)} }{900s}=0.07$ m/s
The average velocity with the least magnitude is for the displacement ending at point $C$.
$| \vec{v}_{avg,AC}|=0.0083\mathrm{m}/\mathrm{s}.$
