When I started to write this I was making it more difficult than it ought to be and came up with the following which, as I have done it, I'm going to include as it might be of interest.
As the car is going down the road the battery must supply energy to:
1)Push the car against a force (drag) that is proportional to the square of the airspeed i.e. to (s + w)^2 where s is the ground speed and w the wind speed component in the direction of travel. In time ∆t the car travel s*∆t and as energy is force times distance this energy is Cd*(s^3 + 2*s^2*w + s*w^2)∆t. Cd is a drag coefficient which is a function of (s + w) but lets assume it is constant here to make things simple.
2)Change the kinetic energy of the car by m*a*s*∆t where m is the mass of the car and a is the acceleration which is negative when you take your foot off the accelerator.
3)Change the potential energy of the vehicle by m*g*G*s*∆t in which g is the acceleration due to gravity and G is the grade as a fraction (G - 0.02 for a 2% grade).
4)Run the cabin and battery heaters, the coolant pump, the windshield wipers, the window motors, the seat motors the radio, and charge your cell phone. If the power demanded by all those things is W then the energy taken from the battery in time ∆t is W*∆t.
Dividing through by ∆t we get the total power demand on the battery, ∆E/∆t
P = ∆E/∆t = Cd*s^3 +2*Cd*s^2*w + Cd*s*w^2 + m*a*s + m*g*G*s + W
We don't want, for the display, ∆E/∆t. We want ∆E/∆x where x is the distance traveled. That's easily obtained from
∆E/∆x = (∆E/∆t)/(∆x/∆t) where clearly ∆x/∆t = s, the ground speed of the vehicle. So
∆E/∆x = P*∆t/∆x = Cd*s^2 + 2*Cd*s*w + Cd*w^2 + m*a + m*g*G + W/s = Wh/mi
This makes it clear that your main strategy for extending range is slow down and wait for headwinds to die down if they are forecast.
It also shows that if you stop for any reason that the instantaneous Wh/mi becomes infinite if there is any electric consumption other than that of the traction motors which, as discussed earlier, is not a problem for the hypothecated display calculation because ∆x is never 0 when a computation is done.