Question

At high speed, the main retarding force on an automobile is air drag. We haven’t specifically studied air drag yet, because it’s quite simple. The drag force is very well approximated as

Fdrag = γv^2

where γ depends on the shape and size of the car.

Consider a car with γ = 1.9N/ (m/s)^2 and an engine that develops a maximum power s of 75 kW. (Remember, this means that the engine does 75 kJ of work per second.)

(a) What is the maximum speed of this car, in miles (or km) per hour?

(b) Suppose that, rather than driving at its maximum speed, the driver travels only at 100 km/hr. What power is required from the engine to go this fast?

(c) It’s said that one of the most important things a driver can do to save fuel is to slow down. Suppose a given trip requires 10 gallons of gas at 70 miles per hour. How much gas will it require at 60 miles per hour?
Note: If you wind up doing a great many unit conversions for this problem, you’re making it far harder than it should be. The easiest approach here is to figure out the energy required for a trip as a function of distance and speed, and then see how it depends on the speed.

Answer 1

Here,

Fdarg = 1.9 * v^2

a) let the velocity of the car is v

as Power = Fdrag * velocity

75 *10^3 = 1.9 * v^2 * v

v = 34.05 m/s

v = 122.6 km/hr

the speed of the car is 122.6 km/hr

b)

for maximum speed , v = 100 km/hr

v = 27.8 m/s

for the power

Power = Fdrag * v

Power = 1.9 * v^2 * v

Power = 1.9 * (27.8)^3

Power = 40756 W = 40.8 kW

the power of the engine needed is 40.8 kW

c)

for the energy required , E = power * time

E = Y * v^3 * distance/v

E = Y * v^2/distance

hence , the energy needed for trip is proporational to square of speed

Now, let the gas required is E2

E2/E1 = v2^2/v1^2

E2/10 = 60^2/70^2

E2 = 7.34 gallon

the gas required at 60 miles per hour is 7.34 gallon