[cojaro]

With some Physics, you could find out the drag yourself! =)

KE(i) + PE(i) = KE(f) + PE(f) + E(lost)
The sum of the initial kinetic and potential energies is equal to the sum of the final kinetic and potential energies and energy lost, which in the case of cars, is the rolling resistance and drag

We get:

½mv(i)² + mgh(i) = ½mv(f)² + mgh(f) + E(r.r.) + E(drag)
m = mass
v = velocity
g = gravitational acceleration
h = height
And we can change this from an energy problem to a work problem!
↓
½mΔv² + mgΔh = F(r.r.)d + F(C)d
d = distance traveled
d can be found by d = ½(v(i) + v(f))Δt
↓
½mΔv² + mgΔh = C(r.r.)Nd + ½ρv(f)²C(d)Ad
ρ = air density
A = cross-sectional area (largest)
C(d) = drag coefficient
C(r.r.) = coefficient of rolling friction of the tire(CRF)
↓
½mΔv² + mgΔh = d(C(r.r.)mg + ½ρv(f)²C(d)A)
ρ ≈ 1.2kg/m³
m = 1103.14kg
mg = 1103.14kg * 9.81m/s² = 10821.80N
↓
551.57Δv² + 10821.80Δh = C(r.r.)10821.80d + 0.6v(f)²C(d)Ad
↓
[551.57Δv² + 10821.80Δh - C(r.r.)10821.80d] \ [0.6v(f)²Ad] = C(d)

Ta-da! I ♥ Physics. And I hope I got all of that right!

What you'll need is the rolling resistance rating of your tires, and the cross-sectional area of the fit. Everything else can be found! =)