## College Physics (4th Edition)

(a) $average~speed = 70.0~mi/h$ (b) $average~velocity = 58.9~mi/h$
(a) We can find the distance driven in the west direction. $d = v~t = (55.0~mi/h)(1.20~h) = 66.0~mi$ He must drive the remaining 56.0 miles in 48.0 minutes. We can find the average speed: $average~speed = \frac{distance}{times}$ $average~speed = \frac{56.0~mi}{48.0~min}$ $average~speed = (\frac{7.0~mi}{6.0~min})(\frac{60~min}{1~h})$ $average~speed = 70.0~mi/h$ (b) We can find the total displacement west: $66.0~mi+(56.0~mi)~cos~30.0^{\circ} = 114.5~mi$ We can find the displacement south: $(56.0~mi)~sin~30.0^{\circ} = 28.0~mi$ We can find the magnitude of the displacement: $\sqrt{(114.5~mi)^2+(28.0~mi)^2} = 117.87~mi$ We can find the average velocity: $average~velocity = \frac{displacement}{time}$ $average~velocity = \frac{117.87~mi}{2.00~h}$ $average~velocity = 58.9~mi/h$