Answer :
Answer: The drag force increases by a factor of 4
Explanation:
The Drag Force equation is:
[tex]F_{D}=\frac{1}{2}C_{D}\rho A_{D}V^{2}[/tex] (1)
Where:
[tex]F_{D}[/tex] is the Drag Force
[tex]C_{D}[/tex] is the Drag coefficient, which depends on the material
[tex]\rho[/tex] is the density of the fluid where the bicycle is moving (air in this case)
[tex]A_{D}[/tex] is the transversal area of the body or object
[tex]V[/tex] the bicycle's velocity
Now, if we assume [tex]C_{D}[/tex], [tex]\rho[/tex] and [tex]A_{D}[/tex] are constant (do not change) we can rewrite (1) as:
[tex]F_{D}=C.V^{2}[/tex] (2)
Where [tex]C[/tex] groups all these coefficients.
So, if we have a new velocity [tex]V_{n}[/tex] , which is the double of the former velocity:
[tex]V_{n}=2V[/tex] (3)
Equation (2) is written as:
[tex]F_{D}=C.(V_{n})^{2}=C.(2V)^{2}[/tex]
[tex]F_{D}=4CV^{2}[/tex] (4)
Comparing (2) and (4) we can conclude the Drag force is four times greater when the speed is doubled.