Answer
$0.037\frac{m}{s},\ 31^{\circ} west\ of\ north.$
Work Step by Step
We can find the magnitude and direction of the average velocity as follows:
First of all, we find the displacement. Let the two vectors be represented by $\vec A$a and $\vec B$
$\vec{\Delta r}= \vec A+\vec B=(120m)\hat y+(-72m)\hat x$
Now the average velocity is
$\vec v_{avg}=\frac{\vec{\Delta r}}{\Delta t}=(\frac{-72m}{62min})(\frac{1min}{60s})\hat x+(\frac{120min}{62m})(\frac{1min}{60s})\hat y$
$\vec v_{avg}=(-0.091\frac{m}{s})\hat x+(0.032\frac{m}{s})\hat y$
Now the magnitude of the velocity is
$v_{avg}=\sqrt{(-0.019)^2+(0.032)^2}=0.037\frac{m}{s}$
and the direction of the velocity is given as
$\theta=tan^{-1}(\frac{120}{-72})=31^{\circ}$ west of north.
