The dominant form of drag experienced by vehicles (bikes, cars, planes, etc.) at operating speeds is called form drag. It increases quadratically with velocity (essentially because the amount of air you run into increases with v and so does the amount of force you must exert on each small volume of air). Thus Fdrag=CdAv2, where A is the cross-sectional area of the vehicle and Cd is called the coefficient of drag.Consider a vehicle moving with constant velocity v . Find the power dissipated by form drag.
A certain car has an engine that provides a maximum power P0. Suppose that the maximum speed of the car, v0, is limited by a drag force proportional to the square of the speed (as in the previous part). The car engine is now modified, so that the new power P1 is 10 percent greater than the original power (P1=110%P0).
Assume the following:
The top speed is limited by air drag.
The magnitude of the force of air drag at these speeds is proportional to the square of the speed.
By what percentage, (v1−v0)/v0, is the top speed of the car increased?

The power is defined as the change of work by the unit of time, as the work is the product of the force by the distance, we can rewrite the power, where the bold ones indicate vectors and the point is the scalar product.

P = F .v

In the current case, the force always opposes the speed, so the product is negative, which indicates a dissipative power.

Let's write the dissipation force and calculate

Fdrag = Cd A v²

P = Cd A v² v

P = cd A v³

this is the power dissipated

Second part

We write the expressions for the dissipated power for Po and P1

Po = Cd A vo³

P1 = 110 Po

P1 = Cd A v1³

We replace

110 Po = Cd A v13

110 (Cd A Vo³) = Cd A v1³

simplify

110 Vo3 = v13

v1 = ∛110 vo

We look for the one that increases the speed

(v1-vo) / vo = (∛110 vo-vo) / vo

(v1-vo) / vo = ∛110 -1

(v1-vo) / vo = 3.79

The speed increase percentage is 3.79