#### Answer

(a) The ground speed is $39.0~m/s$
(b) The new course relative to the ground is an angle of $7.4^{\circ}$ south of west.

#### Work Step by Step

(a) We can find the west component of the ground speed:
$30.0~m/s+(10.0~m/s)~cos~30^{\circ} = 38.66~m/s$
We can find the south component of the ground speed:
$(10.0~m/s)~sin~30^{\circ} = 5.0~m/s$
We can find the ground speed:
$\sqrt{(38.66~m/s)^2+(5.0~m/s)^2} = 39.0~m/s$
The ground speed is $39.0~m/s$
(b) We can find the direction south of west:
$tan~\theta = \frac{5.0~m/s}{38.66~m/s}$
$\theta = tan^{-1}(\frac{5.0~m/s}{38.66~m/s})$
$\theta = 7.4^{\circ}$
The new course relative to the ground is an angle of $7.4^{\circ}$ south of west.