Required information A bicycle travels $3.40 \mathrm{~km}$ due east in $0.600 \mathrm{~h}$, then $9.10 \mathrm{~km}$ at $15.0^{\circ}$ east of north in $0.370 \mathrm{~h}$, and finally another $3.40 \mathrm{~km}$ due east in $0.600 \mathrm{~h}$ to reach its destination. The time lost in turning is negligible. Assume that east is in the $+x$-direction and north is in the $+y$-direction. What is the direction of the average velocity for the entire trip? Enter the answer as an angle in degrees north of east. north of east
Solution
Step 1 ：Given that a bicycle travels $3.40 \mathrm{~km}$ due east in $0.600 \mathrm{~h}$, then $9.10 \mathrm{~km}$ at $15.0^{\circ}$ east of north in $0.370 \mathrm{~h}$, and finally another $3.40 \mathrm{~km}$ due east in $0.600 \mathrm{~h}$ to reach its destination. The time lost in turning is negligible. Assume that east is in the $+x$-direction and north is in the $+y$-direction.
Step 2 ：The average velocity is the total displacement divided by the total time. The displacement is a vector quantity, so we need to calculate the x and y components of the displacement for each leg of the trip, and then add them up to get the total displacement.
Step 3 ：The x and y components of the displacement for the first leg of the trip are $x_1 = 3.4$ and $y_1 = 0$ respectively.
Step 4 ：For the second leg of the trip, the x and y components of the displacement are $x_2 = 8.789925019230521$ and $y_2 = 2.3552533104329387$ respectively.
Step 5 ：The x and y components of the displacement for the third leg of the trip are $x_3 = 3.4$ and $y_3 = 0$ respectively.
Step 6 ：The total x and y displacements are $x_{total} = 15.589925019230522$ and $y_{total} = 2.3552533104329387$ respectively.
Step 7 ：The direction of the average velocity is the arctangent of the ratio of the total y displacement to the total x displacement, which is $8.591013066841251$ degrees.
Step 8 ：Thus, the direction of the average velocity for the entire trip is approximately $8.59^{\circ}$ north of east. So, the final answer is \(\boxed{8.59^{\circ}}\).
