1 · Thrust on the ice
Air leaves the nozzle at u = 500 m/s. The thrust is the momentum carried away per second:
Ice has no friction, so once the jet stops the car keeps its speed. The fastest straight is: burn first, then coast. Reaching speed v costs
A side jet on the ice gives a sideways velocity that nothing can remove. You either slide off the ice or reach B already skidding.
2 · Holding the curve
To stay on the arc the car needs a net force toward O. Friction can supply up to μN, with N = mg + Ffan; the side jet supplies S:
While the tyres roll this is static friction, and it takes whatever value is needed. So the side thrust must stay inside a window:
Too little and the car slides outward; too much and it is pushed in toward O. Either way the tyres slide, friction becomes kinetic (μN, against the sliding), and the car can no longer be steered round the arc.
3 · The ideal plan
Use the least side thrust, S = mv2/R − μN, for the time t = πR/(2v). The air left after the straight must last the curve:
With the fan on and the textbook value π = 3 this gives , so v = 12 m/s: 48 kg on the straight and 42 kg in the curve, 6.85 s in total. Entering any faster leaves too little air for the curve.
The game draws a true circle (π = 3.14159…), so its ideal plan is slightly slower.
4 · Free drive: steering is friction
The tyres can push the car in any horizontal direction, but never harder than μN in total (the friction circle). Turning the wheels by an angle δ asks the car to follow a circle of radius
with wheelbase L = 4 m. Within the limit the friction is static and the car rolls round the circle. Beyond it the tyres slide, friction becomes kinetic (μN against the slide) and the car runs wide. Brakes use friction too: they take 70 % of μN, so braking hard in a turn leaves little for steering.
On ice μ = 0: steering and brakes have nothing to push against. Only the jets can change the velocity, and the car keeps sliding in a straight line whichever way it points.