The aerodynamic drag on a sports car depends on its shape. The car has a drag coefficient of 0.1 with the windows and the roof closed. With the windows and the roof open, the drag coefficient becomes 0.8. The car travels at 44 km/h with the windows and roof closed. For the same amount of power needed to overcome the aerodynamic drag, the speed of the car with the windows and roof open (round off to two decimal places), is ________km/h (the density of air and frontal area may be assumed to be constant).

The correct answer is 22.

Step-by-step Derivation:

The aerodynamic drag force ($F_d$) acting on a car is defined by the formula:

$F_d=\frac{1}{2}{C}_d\rho A{V}^2$

where:
• $C_d$ is the drag coefficient,
• $\rho$ is the density of air,
• $A$ is the frontal area of the car, and
• $V$ is the velocity (speed) of the car.

The power ($P$) required to overcome this aerodynamic drag is the product of the drag force and the speed of the car:

$P=F_d\cdot V$

Substituting the expression for $F_d$ into the power equation gives:

$P=\frac{1}{2}{C}_d\rho A{V}^3$

According to the problem description, the power needed to overcome drag remains the same in both cases (windows and roof closed vs. open), and the density of air ($\rho$) and frontal area ($A$) are assumed to be constant. Therefore, we can set up the equality:

$P_{\text{closed}}=P_{\text{open}}$

${\left(\frac{1}{2}{C}_d\rho A{V}^3\right)}_{\text{closed}}={\left(\frac{1}{2}{C}_d\rho A{V}^3\right)}_{\text{open}}$

Since $\frac{1}{2}$, $\rho$, and $A$ are constant, they cancel out from both sides, leaving:

$C_{d,\text{closed}}\cdot V_{\text{closed}}^3=C_{d,\text{open}}\cdot V_{\text{open}}^3$

Given values:
• $C_{d,\text{closed}}=0.1$
• $C_{d,\text{open}}=0.8$
• $V_{\text{closed}}=44\text{km/h}$

Substituting these values into the equation:

$0.1\cdot {44}^3=0.8\cdot V_{\text{open}}^3$

Solving for $V_{\text{open}}$:

$V_{\text{open}}^3=\frac{0.1}{0.8}\cdot {44}^3$

$V_{\text{open}}^3=\frac{1}{8}\cdot {44}^3$

Taking the cube root of both sides:

$V_{\text{open}}=\sqrt[3]{\frac{1}{8}\cdot {44}^3}$

$V_{\text{open}}=\frac{1}{2}\cdot 44=22\text{km/h}$