Regardless of the path taken, at the end of the trip Mrs. Patil will be 8.0 miles away from home. If the path taken involves a trip lasts 0.25 hours, then the average velocity over the trip is
[tex]\dfrac{8.0\,\mathrm{mi}}{0.25\,\mathrm h}=32\,\dfrac{\mathrm{mi}}{\mathrm h}[/tex]
