## College Physics (4th Edition)

(a) The jetliner is 872.8 km from the starting point. (b) The jetliner could have flown directly to the destination at an angle of $9.9^{\circ}$ south of east. (c) The flight took 2.25 hours. (d) A direct flight would have taken 2.18 hours.
(a) We can find the total distance the jetliner flies east: $600.0~km+(300.0~km)~cos~30.0^{\circ} = 859.8~km$ We can find the total distance the jetliner flies south: $(300.0~km)~sin~30.0^{\circ} = 150.0~km$ We can find the distance from the starting point: $\sqrt{(859.8~km)^2+(150.0~km)^2} = 872.8~km$ The jetliner is 872.8 km from the starting point. (b) We can find the angle $\theta$ south of east: $tan~\theta = \frac{150.0~km}{859.8~km}$ $\theta = tan^{-1}(\frac{150.0~km}{859.8~km})$ $\theta = 9.9^{\circ}$ The jetliner could have flown directly to the destination at an angle of $9.9^{\circ}$ south of east. (c) We can find the time of the flight: $t = \frac{600.0~km+300.0~km}{400.0~km/h} = 2.25~h$ The flight took 2.25 hours. (d) We can find the time of a direct flight: $t = \frac{872.8~km}{400.0~km/h} = 2.18~h$ A direct flight would have taken 2.18 hours.