Question

The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the drag force?

Answer 1

Answer: The drag force goes up by a factor of 4

Explanation:

The Drag Force equation is:

$F_{D}=\frac{1}{2}C_{D}\rho A_{D}V^{2}$ (1)

Where:

$F_{D}$ is the Drag Force

$C_{D}$ is the Drag coefficient, which depends on the material

$\rho$ is the density of the fluid where the bicycle is moving (air in this case)

$A_{D}$ is the transversal area of the body or object

$V$ the bicycle's velocity

Now, if we assume $C_{D}$, $\rho$ and $A_{D}$ do not change, we can rewrite (1) as:

$F_{D}=C.V^{2}$ (2)

Where $C$ groups all these coefficients.

So, if we have a new velocity $V_{n}$ , which is the double of the former velocity:

$V_{n}=2V$ (3)

Equation (2) is written as:

$F_{D}=C.V_{n}^{2}=C.(2V)^{2}$

$F_{D}=4CV^{2}$ (4)

Comparing (2) and (4) we can conclude the Drag force is four times greater when the speed is doubled.