As humans don't have much in the way of regenerative braking, I've set out the following scenario for two speeds. Please let me know if I've made any errors in the calculations!
The setting: a 75kg person, traveling 100m. Assume 2 seconds are used accelerating to either 1.5m/s or 3m/s.
F=ma, W=Fd, and d=0.5at^2 so:
F_{1} = 75kg * (1.5m/2s^2); F_{1} = 56.25N d_{1} = 0.5 * (1.5m/2s^2) * (2s)^2 d_{1} = 1.5m (He will travel 1.5m in this time.) W_{1} = 56.25N * 1.5m W_{1} = 84.375 J
F_{2} = 75kg * (3m/2s^2); F_{2} = 112.5N d_{2} = 0.5 * (3m/2s^2) * (2s)^2 d_{2} = 3m (He will travel 3m in this time.) W_{2} = 112.5N * 3m W_{2} = 337.5 J
Disregarding friction, we can assume that person then coasts at this speed until near arrival at destination, at which point he digs his heels into the ground to stop.
Now, the equation for net work is: W_{net} = 0.5mv_{final}^2  0.5mv_{initial}^2
Obviously starting, traveling, and stopping would give a net work value of 0, but I think one must consider the energy flow. The initial acceleration has v_{initial} = 0m/s and v_{final} = 1.5 or 3m/s, and represents chemical energy being converted to kinetic energy. The second "acceleration" has v_{initial} = 1.5 or 3m/s and v_{final} = 0m/s, and represents kinetic energy being converted to heat (friction braking). Thus, the caloric cost is mirrored by W_{1} and W_{2} above, i.e. roughly quadruple for the faster speed. So, what am I missing?
Edit: added final W_{1} and W_{2} values, changed "double" to "quadruple".
Edited by requiem on 01/01/2013 23:51:44 MST.
