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I'll go through the math and consider the energy costs. But I see two reasons off hand why that might be the case:
1. From a math/physics standpoint, it's better to spend less time going slowly.
2. People are less efficient runners at higher speeds.
Here's some math:
Ok, well let's say you're running at X+/-Y mph (with or against the wind) with a Z mph tailwind or headwind. Say the force against you is proportional to a*v^2, where a is a constant and v is the relative velocity. The assumption here is that the only thing that matters are the fluid dynamics, that your body's efficiency or running economy isn't terribly important. This makes sense in the limit of small wind velocities. And let's say the track is just out and back (forget about the curves). Then the time it takes you to finish a lap is time = distance / velocity, or
T = d/(X+Y) + d/(X-Y)
T = 2dX/(X^2-Y^2) (1)
and remember that Y has to be less than X
The amount of work done is a*(X+Y-Z)^2*d+a*(X-Y+Z)^2*d = d*a*(2X^2+2Y^2+2Z^2-4XZ-4YZ)
W = 2da*(X^2+Y^2+Z^2-2XZ-2YZ) (2)
I guess I'll try to plug equation 2 into equation 1, first rearrange 2:
W = 2da*(X^2+Y^2+Z^2-2XZ-2YZ) (2)
W/2da-Y^2-Z^2+2Z(X+Y) = X^2
Plug in:
T = 2dX/(W/2da-Y^2-Z^2+2Z(X+Y)-Y^2)
T = 2dX/(W/2da-2*Y^2-Z^2+2Z(X+Y)) (3)
So for a given amount of work, what's the fastest time you can run? What we would want to know is: Should Y be greater than Z, equal to Z, or smaller than Z to minimize T?
We want to maxmize the denominator, which means we want to minimize the quantity
2Y^2+Z^2
this is minimal when Y is less than Z.
So from a fluid dynamics standpoint, you want to put some extra effort into fighting against the wind; you don't want it to slow you down by the entire wind velocity Z. And you also don't want to go faster against the wind than you would have ordinarily (this corresponds to switching the sign of Y, it would make the denominator smaller because of the last term in equation 3).