Question

The power that a student generates when walking at a steady pace of vw is the same as when the student is riding a bike at vb = 3vw. The student is going to travel a distance d. The energy the student uses when walking is Ew. The energy the student uses when biking is Eb. The ratio EwEb is

Answer 1

Answer:

1/9

Explanation:

Sorry, I don't know if this is right, but here is what I did. We are ignoring potential energy because we assume that the student is walking and biking on level ground. Power = W/T, W = Mechanical Energy, or just Kinetic for this case. So $P_{w}=\frac{E_{w} }{T}$, and $P_{b} =\frac{E_{b} }{T}$. Ew = $\frac{1}{2} mv_{w}^{2}$, and Eb =$\frac{1}{2} m(3v_{w})^{2}$. Put Ew over Eb. the 1/2's cancel, the m's cancel, and you are left with $\frac{v_{w} ^{2} }{9v_{w}^{2} }$. Finally, this simplifies after cutting out the vw^2's to 1/9.