Required information A bicycle travels $7.10 \mathrm{~km}$ due east in $0.270 \mathrm{~h}$, then $12.00 \mathrm{~km}$ at $15.0^{\circ}$ east of north in $0.800 \mathrm{~h}$, and finally another $710 \mathrm{~km}$ due east in $0.270 \mathrm{~h}$ 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 magnitude of the average velocity for the entire trip? $\mathrm{km} / \mathrm{h}$

Step 1 ：The displacement for the first leg of the trip is \(7.10 \mathrm{~km}\) due east, which is \(7.10 \mathrm{~km}\) in the \(+x\) direction.

Step 2 ：The displacement for the second leg of the trip is \(12.00 \mathrm{~km}\) at \(15.0^{\circ}\) east of north. This can be broken down into components in the \(+x\) and \(+y\) directions. The \(+x\) component is \(12.00 \mathrm{~km} \cos(15.0^{\circ})\) and the \(+y\) component is \(12.00 \mathrm{~km} \sin(15.0^{\circ})\).

Step 3 ：The displacement for the third leg of the trip is \(710 \mathrm{~km}\) due east, which is \(710 \mathrm{~km}\) in the \(+x\) direction.

Step 4 ：The total time is \(0.270 \mathrm{~h} + 0.800 \mathrm{~h} + 0.270 \mathrm{~h}\).

Step 5 ：The total displacement in the \(+x\) direction is \(7.10 \mathrm{~km} + 12.00 \mathrm{~km} \cos(15.0^{\circ}) + 710 \mathrm{~km}\) and in the \(+y\) direction is \(12.00 \mathrm{~km} \sin(15.0^{\circ})\).

Step 6 ：The magnitude of the total displacement is the square root of the sum of the squares of the \(+x\) and \(+y\) displacements.

Step 7 ：The average velocity is the total displacement divided by the total time.

Step 8 ：Final Answer: The magnitude of the average velocity for the entire trip is \(\boxed{543.80 \, \mathrm{km/h}}\).