Chapter 16, Problem 41RPP

Fuel used to counter air resistance The resistive drag force that the air exerts on an automobile is

$F_{\text{Air on Car}}=(1/2)CA\rho {\upsilon }^2$ where C is a drag coefficient that depends on now streamlined the car is. $\rho$ is the density of the air. A is me cross-sectional area of the car along its line of motion, and v is the car's speed relative to the air. Consider the effect of air resistance on fuel consumption in a typical car during a 160-km (100- ml) trip while traveling at 22 m/s (50 ml /h). The resistive force exerted by the air on the car is about 200 N. Thermal energy produced due to this resistive force during the trip is

$\Delta U_{\text{thermal}}=F_{\text{Air on Car}}d=(200\text{ N)(1}\text{.6}\times {\text{10}}^5\text{m)=3}\text{.2}\times {\text{10}}^7\text{ J}$

One liter of gas releases about $3.2\times {10}^7$

J of energy. So the gasoline energy that is equivalent to the resistive thermal energy produced during this trip is

$(3.2\times {10}^7\text{ J)/(3}\text{.25}\times {\text{10}}^7\text{J/L)=0}\text{.98 L=0}\text{.26 gallons}$

In summary, you use about one-quarter gallon of gasoline to counter air resistance during the 160-km trip. Unfortunately, the car is only about 13% efficient, where

$\text{Efficiency=}\frac{\text{useful energy output}}{\text{energy input}}=0.13$

Consequently, your useful output of 0.98 liters requires an energy input of (0.98 L)/0.13=7.5 L of gas to overcome this air resistance. This is about half of the gasoline a typical car consumes when driving at a steady speed for 160 km. The other half is needed to overcome rolling resistance.

The amount of fuel used to counter air resistance should do what?

a. Increase in proportion to the speed squared

b. Increase in proportion to the speed

c. Be the same independent of the speed

d. Decrease in proportion to the inverse of the speed

e. Decrease in proportion to the inverse of the speed squared