[jakobnev]

Originally Posted by briank

Now I would say the car in scenario (1) above is probably getting almost 50% higher mpg than (2) above because it's covering 1.5 times the distance with the same effective air drag.

The force acting on the car is the same, but since speed is 50% grater in one example you need 50% more power! (P=V*F)

But the energy is the same, it's simply force times distance: W=F*D

So it's all down to what power lets the engine reach the better position on it's BSFC-chart.