I want to find the car speed "v" that minimizes the energy lost to air friction and climbing the hill.

Assume there is no rolling resistance so that the force against the car from friction is all air resistance

F1 = cv^2

where c is the drag coefficient that involves the surface area of the car, air density, and air viscosity. "v" is the speed. The v^2 is why it's more efficient to go at slower speeds despite reaching your destination faster (unless you are going uphill). It comes from having to accelerate air out of the way and is because the kinetic energy increase of the air "before" or "as" it gets turned to heat energy is some constant times the kinetic energy 1/2 mv^2.

So the energy lost to air friction is:

E1 =d F1 = dcv^2 where d is the distance travelled.

The net work energy the engine must provide to climb the hill by itself is:

E2 = mgh where h=d sin(angle)

The total net energy provided by the engine (after engine inefficiency losses) is:

E1+E2 = dcv^2 + mgd sin(A).

Let d = t/v where t is time that d was travelled:

E1+E2 = tcv + mg(t/v) sin(A)

I want to find the v that minimizes E1+E2, so take the derivative with respect to v and set it to zero:

0 = tc - mgt/v^2 sin (A)

Solving, I get

v = SQRT(mg sin(A) / c)

is the most efficient speed up hill.

I can make this in more practical terms, especially since I don't know c. I mean I can measure c in a way to make sure it cancels. Let constant V be the speed at which the car wants to roll downhill which is when the air resistance equals the force downward.

F1 = F2 = cV^2 = mg sin(A)

gives

c = mg sin(A) / V^2

plugging into the v equations above:

**v = V**

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