## College Physics (4th Edition)

(a) To return to his starting point, Harrison should travel at an angle of $29.6^{\circ}$ north of east for a distance of $6.34~km$ (b) The return trip will take 1268 seconds.
We can find the total west component of Harrison's trip: $2.00~km+(5.00~km)~cos~53.0^{\circ}+(1.00~km)~cos~60.0^{\circ} = 5.51~km$ We can find the total south component of Harrison's trip: $(5.00~km)~sin~53.0^{\circ}-(1.00~km)~sin~60.0^{\circ} = 3.13~km$ To return to the starting point, Harrison should travel 5.51 km east and 3.13 km north. We can find the angle north of east that Harrison should travel: $tan~\theta = \frac{3.13}{5.51}$ $\theta = tan^{-1}(\frac{3.13}{5.51})$ $\theta = 29.6^{\circ}$ We can find the distance $d$ that Harrison should travel: $d = \sqrt{(5.51~km)^2+(3.13~km)^2} = 6.34~km$ To return to his starting point, Harrison should travel at an angle of $29.6^{\circ}$ north of east for a distance of $6.34~km$ (b) We can find the time of the return trip: $t = \frac{d}{v} = \frac{6340~m}{5.00~m/s} = 1268~seconds$ The return trip will take 1268 seconds.