Aerodynamic drag

Power

Under the assumption that the fluid is not moving relative to the currently used reference system, the power required to overcome the aerodynamic drag is given by: $${\displaystyle P_{D}=\mathbf {F} _{D}\cdot \mathbf {v} ={\tfrac {1}{2}}\rho v^{3}AC_{D}}$$ The power needed to push an object through a fluid increases as the cube of the velocity increases. For example, a car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome aerodynamic drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speeds, the drag/force quadruples per the formula. Exerting 4 times the force over a fixed distance produces 4 times as much work. At twice the speed, the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, 4 times the work done in half the time requires 8 times the power.