## College Physics (4th Edition)

(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.
(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.