Answer: Relay Race

The problem was based on a real-life scenario.

First let’s find the position of something that accelerates with constant power, starting at rest.

Its kinetic energy is

$T = Pt$

with $P$ the power, $T$ the kinetic energy, and $t$ the time it’s been accelerating. Kinetic energy is $1/2mv^2$ , so

$\frac{1}{2} mv^2 = Pt$

or solving for $v$

$v = \sqrt{\frac{2Pt}{m}}$ .

Integrating

$x = \frac{2}{3}\sqrt{\frac{2P}{m}}t^{3/2}$ .

Constant power is somewhat less effective than constant acceleration.

The time needed to complete a race, accelerating the whole way under constant power, is

$t = \sqrt[3]{\frac{9mx^2}{8P}}$ .

So the time scales with the minus one-third power of the power output. However, if we add in stopping and turning around, this will get even weaker, because it’s significantly easier to turn around when you aren’t already going fast (although the details depend on the length of the race and the coefficient of friction, and I don’t think it’s very enlightening to work out in full detail.)

This entry was posted on July 21, 2009 at 12:10 am and is filed under problems and solutions. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “Answer: Relay Race”

Paul Murray Says:
July 21, 2009 at 1:16 am
” I don’t think it’s very enlightening to work out in full detail”

On the contrary – have a look at my reply that I just wrote on the post that proposed the problem. I didn’t work out the math, but I think my answer was right.
Mark Eichenlaub Says:
July 21, 2009 at 2:46 am
Hi Paul,

Nice point. I think in a realistic scenario we’ll be stuck somewhere in between time scaling with the zeroth power of power (as you showed to be the limit of high power output) and with the minus one-third.

I guess it’s a bit silly for me to say power output is constant if it’s going to be very high, because if friction is a limiting factor, there’d be no way to apply that much power.

Arcsecond