How Much Does Increasing Engine Power Affect Top Speed Due to Air Drag?

[Linus Pauling]

1. A certain car has an engine that provides a maximum power P_0. Suppose that the maximum speed of the car, v_0, is limited by a drag force proportional to the square of the speed (as in the previous part). The car engine is now modified, so that the new power P_1 is 10 percent greater than the original power (P_1=110\%P_0)

Assume the following:

* The top speed is limited by air drag.
* The magnitude of the force of air drag at these speeds is proportional to the square of the speed.

By what percentage, (v_1-v_0)/v_0, is the top speed of the car increased?



2. F_drag = C_d*Av^2
P = -C_d*Av^3



3. P_1 is proportional to v_1^2
Thus, 1.1P_o is proportional to (1 + alpha)^n * v_o^2

Divide by P_o proportional to v_o^n to obtain:

alpha = 1.1^(1/n) - 1

My answer was 4.8% and was wrong. Again, I want the percentage increase (two sig figs) in speed: (v_1 - v_o)/v_o

 

[Delphi51]

Linus, I find the notation confusing. Could we just say F = kv^2 so
P = F*v = k*v^3.
You could then write another equation with 1.1P and some other letter for the faster velocity. It should be easy to eliminate the P between the two equations and come up with an equation relating the two v's. I get considerably less than 4.8% that way.

 

[Linus Pauling]

I'm sorry but I don't understand how you're eliminating P, and furthermore how you are getting to a raw percentage.

If I take P = kv_o^3
and
1.1P = kv_1^3

substitute for P in the 1.1P equation:

1.1(kv_o^3) = kv_1^3
1.1v_o^3 = v_1^3

the new speed v_1 would be cuberoot(1.1v_o^3)

?

 

[Delphi51]

P = kv^3 and 1.1P = kV^3
Put the first in the second:
1.1*kv^3 = kV^3
1.1*v^3 = V^3
V = cube root of 1.1 * v
V = 1.032*v
That is a 3.2% increase in speed.

 

[Linus Pauling]

Ah, I didn't realize I could take the cube root of 1.1 and v separately. thank you.

 

[Delphi51]

Ah, yes. And I didn't make it very clear, not having a cube root symbol handy. Most welcome!