Answer
$59^{\circ}$ north of east, $4.9\frac{m}{s}$
Work Step by Step
The magnitude and direction of the average velocity can be determined as:
$\theta=tan^{-1}(\frac{r_y}{r_x})=tan^{\frac{2500ft}{1500ft}}=59^{\circ}north\space of\space east$
and the magnitude of the displacement is
$r=\sqrt{(r_x)^2+(r_y)^2}$
$r=\sqrt{(1500ft)^2+(2500ft)^2}=2900ft\times 0.305\frac{m}{ft}=890m$
Now the magnitude of the average velocity is given as
$v_{avg}=\frac{\Delta r}{\Delta t}$
We plug in the known values to obtain:
$v_{avg}=\frac{890}{3.0min\times 60\frac{s}{min}}=4.9\frac{m}{s}$
