## College Physics (4th Edition)

(a) The average speed is $112~km/h$ (b) The average velocity is $97.3~km/h$
(a) We can find the distance traveled directly east: $(96~km/h)(1.0~h) = 96~km$ We can find the distance traveled east of north: $(128~km/h)(1.0~h) = 128~km$ We can find the average speed for the trip: $average~speed = \frac{distance}{time}$ $average~speed = \frac{224~km}{2.0~h}$ $average~speed = 112~km/h$ The average speed is $112~km/h$ (b) We can find the east component of the displacement: $96~km+(128~km)~sin~30.0^{\circ} = 160~km$ We can find the north component of the displacement: $(128~km)~cos~30.0^{\circ} = 110.9~km$ We can find the magnitude of the displacement: $\sqrt{(160~km)^2+(110.9~km)^2} = 194.68~km$ We can find the average velocity $v_{av}$: $v_{av} = \frac{displacement}{time}$ $v_{av} = \frac{194.68~km}{2.0~h}$ $v_{av} = 97.3~km/h$ The average velocity is $97.3~km/h$