Question

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 = C dAv2, 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.

Answer 1

To solve this problem it is necessary to resort to the concepts related to energy, power, speed and force.

By definition we know that Power is the Amount of Energy used over a period of time, that is

$P = \frac{E}{t}$

Energy in turn is the amount of force used during a given distance, that is

$E = F*d$

Replacing at the first equation we have that

$P = \frac{F*d}{t}$

For the kinematic equations of movement description we have that velocity is equivalent to

$v = \frac{d}{t}$

Then,

$P = F*v$

Here the force is equal to the drag force, then

$P = \vec{F_{D}}*\vec{v}$

Since $\vec{F_{D}} \angle \vec{v}$  are opposite therefore cos180=-1

$P = -F_D*v$

$P = -C_d Av^2 v$

Where,

$C_d =$ Drag coefficient

A = Area

V = Velocity

Finally we have the expression:

$P = -C_d A v^3$