#### Answer

(a) The average velocity is $68.5~km/h$ at an angle of $12.5^{\circ}$ north of east.
(b) The average velocity is $68.5~km/h$ at an angle of $12.5^{\circ}$ south of west.

#### Work Step by Step

(a) We can find the x-component of the displacement:
$\Delta x = (80.0~km/h)(\frac{45}{60}~h)+(60.0~km/h)(\frac{1}{2}~h)~cos~38.0^{\circ}$
$\Delta x = 83.64~km$
We can find the y-component of the displacement:
$\Delta y = (60.0~km/h)(\frac{1}{2}~h)~sin~38.0^{\circ}$
$\Delta y = 18.47~km$
We can find the magnitude of the displacement:
$\sqrt{(83.64~km)^2+(18.47~km)^2} = 85.66~km$
We can find the average velocity:
$v_{ave} = \frac{displacement}{time}$
$v_{ave} = \frac{85.66~km}{1.25~h}$
$v_{ave} = 68.5~km/h$
We can find the angle $\theta$ north of east from the starting point:
$tan~\theta = \frac{18.47}{83.64}$
$\theta = tan^{-1}(\frac{18.47}{83.64})$
$\theta = 12.5^{\circ}$
The average velocity is $68.5~km/h$ at an angle of $12.5^{\circ}$ north of east.
(b) Since the magnitude of the displacement is the same and the time is the same, the magnitude of the average velocity is the same. Since the displacement is in the opposite direction, the displacement is at an angle of $12.5^{\circ}$ south of west.
The average velocity is $68.5~km/h$ at an angle of $12.5^{\circ}$ south of west.