#### Answer

(a) The average speed is $102~km/h$
(b) The average velocity is $90.8~km/h$

#### Work Step by Step

(a) We can find the distance traveled directly west:
$(108~km/h)(\frac{1}{3}~h) = 36.0~km$
We can find the distance traveled south of west:
$(90.0~km/h)(\frac{1}{6}~h) = 15.0~km$
We can find the average speed for the trip:
$average~speed = \frac{distance}{time}$
$average~speed = \frac{51.0~km}{\frac{1}{2}~h}$
$average~speed = 102~km/h$
The average speed is $102~km/h$
(b) We can find the west component of the displacement:
$36.0~km+(15.0~km)~cos~60.0^{\circ} = 43.5~km$
We can find the south component of the displacement:
$(15.0~km)~sin~60.0^{\circ} = 13.0~km$
We can find the magnitude of the displacement:
$\sqrt{(43.5~km)^2+(13.0~km)^2} = 45.4~km$
We can find the average velocity $v_{av}$:
$v_{av} = \frac{displacement}{time}$
$v_{av} = \frac{45.4~km}{\frac{1}{2}~h}$
$v_{av} = 90.8~km/h$
The average velocity is $90.8~km/h$