It's exactly how it works and the math is not complicated. Suppose you have a vehicle w/ 3 m^2 frontal area, drag coefficient of 0.3 and it's travelling at 30

It's exactly how it works and the math is not complicated. Suppose you have a vehicle w/ 3 m^2 frontal area, drag coefficient of 0.3 and it's travelling at 30 m/s. Using this calculator https://www.symbolab.com/calculator/physics/drag-equation https://www.symbolab.com/calculator/physics/drag-equation you can find that will require about 500 N of force (use density of 1.225). The power at that operating point is the product of speed and force, or 30 m/s * 500 N = 15 kW. Now let's change the speed to 40 m/s. That bumps the force to about 900 N. Now we would calculate power as 40 m/s * 900 N = 36 kW. Changing these numbers to horsepower we get 20 hp and 48 hp. The force goes with the square, the power goes with cube. There are assumptions here - for example, if you have a headwind this statement is no longer true because the speed in the drag equation is the air speed of the vehicle (goes up with headwind / down with tailwind) but the third speed factor is the ground speed of the vehicle (because that's based on the physics "work is equal to force times distance" and power is the rate of work). The graph of an internal combustion engine's power at wide-open throttle vs. engine speed is irrelevant, unless you are trying to figure out which gear ratios would let the engine make enough power to drive the vehicle under those conditions.