With the CVT operating boundary conditions as you've stated, and assuming also that in both cars the CVT and rest of the drivetrain have exactly equal efficiency, which is defined as the power out of the driveshaft as a fraction of the power being produced at the crankshaft, then the answer is simple and it is FACT: They will accelerate at the SAME rate. Argument closed. This is a good question because on Thursday I am writing an exam in Automotive Engineering 1: Longitudinal Dynamics that will cover the exact same subject.

Edit: reading your question again, you've made the answer even more obvious and simple: you said the CVT has no losses, and you quoted 200** WHEEL** HP for both cases.

The acceleration you feel in your car is equal to the excess accelerative force divided by the mass of the car, the term "excess" referring to what's left over after overcoming rolling resistance, aerodynamic drag, etc.

This excess force that propels the cars, between the road surface and the tire contact patch, is equal to the torque at the wheels divided by the dynamic radius of the driving wheels.

The torque at the wheels is equal to the power available at the wheels divided by the angular speed of the wheels.

If the CVT adjusts the ratio to maintain the same engine RPM as a function wheel rotation rate, then power (which is constant between the two cars) divided by angular speed at the wheels (also constant between the two cars), is also exactly the same.

The only things we are left to assume is that the acceleration occurs with no tire slip and that there isn't a time dependency from when the light turns green to when the conditions described in your question is reached.

This should clearly illustrate that at the end of the day, it's the *FORCE *(analogous to torque) at the **WHEELS **that matter, and if you follow the path back to through the drivetrain and ultimately to the engine, it can be be shown that it is a function of *POWER (the arithmetic product of torque ***AND ROTATIONAL SPEED)** rather than torque alone. It is a ridiculous notion that engine torque independent of RPM accelerates the vehicle.

IF the question were reworded and you set the gear ratios to be the same between the two cars and you stay in the same gear during the acceleration run, then you change everything, and then YES, engine TORQUE alone will determine the acceleration because now you've set the relationship of the rotational speeds of the wheels relative to the engine between the two cars to be constant, and then, the engine with the greater torque wins the race.

Aside: Practically speaking this race scenario can never happen from a standstill, becase the power at the wheels is exactly zero at a standstill, and yes, there is a finite time dependency of the engine and drivetrain reaching the conditions you described in the question. If you had such an ideal CVT that can lock the engine RPM regardless of wheel speed, and if you did the race from a rolling start from, say, 30 MPH, then the analysis I stated would be true.

However, for interest's sake let's look at another scenario where we have a conventional single speed, fixed ratio transmission, two cars identical in every way except on one, the engine has a characteristic of developing constant POWER throughout the RPM range, and the other developing constant torque. Both cars have the same peak HP value.

At any initial point where both cars are travelling at the same speed, the demanded power to overcome rolling resistance and drag is equal. Now both cars take-off to race each other. The car whose engine develops constant power will outaccelerate the one that develops constant torque.

The reason I bring up this analogy is that the engine developing constant power can be very roughly analogous to a Diesel engine (TDI, if you look at the RPM range between 3000-4000 RPM or so), while the one with constant torque again analogous to a gasser (e.g. 1.8T with an RPM range of perfectly constant maximum torque). Road tests support the fact the the in-gear acceleration of the TDI is greater than that of the 1.8T when both engines have roughly the same peak HP.