It's a little bit hard to follow your reasoning. Let me try to give the method I would use - simple balance of energy.
If the car reaches a velocity $v$ after time $t$, the power of the engine will have been used mostly for five components:
- Kinetic energy of car: $\frac12 m v^2$
- Rotational kinetic energy of tires: $4 \times \frac12 I_t \omega_t^2$
- Rotational kinetic energy of engine: $\frac12 I_e \omega_e^2$
- Rolling friction: $mg\mu d$
- Air drag - becomes significant as speed increases.
The rotational component of the tires is not insignificant: a uniform cylinder of mass $m$ and radius $r$, rolling with velocity $v$ has angular velocity $\omega = \frac{v}{r}$ and rotational kinetic energy
$$\begin{align}KE &= \frac12 I \omega^2\\
&=\frac12 (\frac12 m r^2 ) \left(\frac{v}{r}\right)^2\\
&=\frac14 m v^2\end{align}$$
In other words, the total kinetic energy of the tires (linear plus rotational) is 50% greater than just the linear component (for a uniform tire). In other words, it's like carrying the mass of two additional tires in the car.
Similarly, the engine, running at 6000 rpm = 100 revs/sec, is storing a significant amount of energy as well - if you think about it, when you rev an engine it takes a bit of time to get up to speed, which tells you that maybe half a second's worth of engine power is spent just bringing the engine up to speed.
The other components you would have to estimate from the numbers you have - it's a bit hard to understand the distance your car moved, or how long it was moving for.
The air drag is a function of the square of the velocity, so it will be most significant near the top speed of the car. You can write the instantaneous force of drag as
$$F=\frac12 c_d \rho A v^2$$
where $c_d$ is typically around 0.3 and $C_d A$ around 0.55 m2. With $\rho = 1.3 kg/m^3$, the force is approximately
$$F = 0.27\times \left(\frac{v}{3.6}\right)^2$$
with $v$ in km/h - meaning that at 100 km/h, the force is for a streamlined car is a little over 200 N, and the power needed to overcome that drag ($P = Fv$) is around 6 kW, or 7.5 hp.
For rolling resistance, car tires have a coefficient around 0.03, so the force (for your 1143 kg car) is
$$F_{rolling}=mgc=336 N$$
and the power used is
$$P = Fv = 9.3 kW ~ 12 hp$$
For a speed of 100 km/h, and a tire mass of 30 kg / tire, I compute kinetic energy of
$$KE = \frac12(m_{car} + 2m_{tire})v^2 = 464 kJ$$
Forgetting for a moment the power curve of the engine, if the car accelerates to 100 km/h in 10 seconds, the distance covered will be
$$d = \frac12 a t^2 ~ 138 m$$
At this point, the energy spent rolling was
$$E_{roll} = 336\times 138 = 46.7 kJ$$
Integrating the air drag over 10 seconds:
$$E_{drag} = \int F(v) v dt \\
= \int 0.27 \left(v(t)\right)^3 dt \\
= \int 0.27 \left(at\right)^3 dt \\
= \frac{0.27\times 2.8^3}{4}t^4\\
~14.8 kJ$$
The air drag will be less (since the drag increases with square of velocity, and is less than rolling friction even at 100 km/h), so the major component will be the kinetic energy of the car.
Just from the above, the car that takes 10 seconds to accelerate has spent 464+47+15 = 526 kJ, or 52.6 kW - meaning an average power output of around 71 hp.
As you can see, the time to accelerate really matters if you want to use acceleration to get the power. It scales roughly inversely - but there's quite a bit to making this even vaguely accurate. And the "power output" of an engine will be a strong function of rpm, so while it may peak at 6800 rpm, that's not where it spent a lot of the accelerating time...
I hope this gives you food for thought - good simulation takes time.
The torque and power curves are of course related (power is related to torque times RPM) but the maximum torque and maximum power are achieved at different RPMs and tell us different things about the car, therefore both are specified.
Best is of course to look at a diagram like the one below, showing the relationships of the BMW 335i twin-turbo engine as an example:
The maximum power is usually achieved near the highest rated RPM for the engine, but since you normally don't drive the car near the max RPM, this figure by itself is less useful for the non-racing driver than looking at the torque curve I think. The maximum torque is usually achieved at a lower RPM so just getting that parameter (and the RPM it relates to) could be more useful actually.
You might be able to estimate that a car with a higher maximum power might have a higher torque at relevant RPMs of course so it's still useful for quick comparisons.
In a turbo engine (like the 335i engine above) the maximum torque will be very quickly achieved at low RPMs and it will then stay flat through a long range; this is where you typically will be driving the car between gearshifts and this is what will give you the "oomph".
A non-turbo engine will have a more gradually increasing torque curve.
But just to be clear, both parameters are related and similar engines with different maximum powers also probably have some similar relationship between their maximum torques (and oomphs).
Best Answer
The gear ratios are gearing down, i.e. they reduce the angular velocity at the wheels and increase the torque. The correct formula is:
$$ T_{wheels} = T_{engine} G $$
Plus don't forget that there is a differential gear as well as the gearbox, and this reduces the angular velocity and increases the torque as well. The value of $G$ in your equation is the gearbox ratio multiplied by the differential ratio.
Response to comment:
I don't know what the differential gearing is in your car, but differentials are generally around 4:1 so let's take this value. In that case the torque at the wheels in first gear is:
$$T_{wheels} = 500 \times 2.97 \times 4 = 5940Nm $$
I would guess the tyre radius is about 0.3m, so the force at the contact between the tyre and ground is:
$$ F = \frac{T_{wheels}}{r} = \frac{5940}{0.3} = 19800N $$
So the acceleration is:
$$ a = \frac{F}{m} = \frac{19800}{2500} = 7.92ms^{-2} $$
That seems pretty good to me. 0.8g!