Maximum velocity for RWD vehicle on 6% grade without gear ratio

[eqm]

Homework Statement

Determine the maximum velocity attainable by a vehicle with the following conditions:

• RWD
• 6% grade
• Weight = 20 kN
• CG is 1.25 m behind front axle and 0.5 m above ground level.
• Wheel base is 2.8 m.
• Effective rolling radius of wheel = 30 cm
• Coefficient of aerodynamic drag = 0.45 with frontal area 2.3m²
• ρ = 1.25 kg/m³
  
• Engine develops peak torque at 45 kW and 4000 rpm
• The rotating inertia of the gearbox is and engine is 0.454 kgm²
• The rotating inertia of each wheel with driveline is 1.76 kgm²
• coefficient of friction between road and tire μ = 0.8

Homework Equations

W_r = (W l₁cosθ + R_ah + W h sinθ)/L

Max Tractive Effort = μ W_r

R_a = 1/2 ρ V² A C_D
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The Attempt at a Solution

To solve this problem, I was going to first determine if the wheel torque is limited by the vehicle motor or by road adhesion, then find the maximum velocity using the limiting torque. I can get motor torque but am stuck at this point as I am not given a gear ratio to be able to convert motor torque to wheel torque.

T_M = 30 P n_rpm / π = 107.43 N/m

I know how to solve for everything else, its really just the the conversion to wheel torque that I'm stuck at (unless I'm approaching this problem entirely wrong).

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[jack action]

Instead of converting engine power to engine torque, convert all resistances $R$ to power by multiplying them by the car velocity. So instead of:

$F_{wheels} - \sum{R} = ma$

Since wheel power is equal to engine power $P_{eng}$, you can use:

$P_{eng} - \left(\sum{R}\right)v = mav$