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.
Best Answer
I'd go along with Alexander's comment that working in SI units makes life a lot easier. However, assuming you have a good reason for sticking to US units ...
The torque you've calculated is in foot pounds. The easy way to see this is that the 33,000 conversion factor you've used converts horsepower to foot pounds min$^{-1}$, and the rpm is in units of min$^{-1}$. The min$^{-1}$ on the top and bottom of the fraction cancel leaving the units as foot pounds.
In the second equation you've put in the distance as inches, which makes life harder than it need be. If you take the wheel radius to be one foot rather than 12 inches you get the force equal to 682.77 pounds.
For the last step you need to be aware that the "pound" is being used as a unit of force here i.e. it's the force exerted by an object weighing one pound in Earth's gravity. In units of feet per second the acceleration due to gravity is about 32.18 feel/sec$^2$, so the acceleration of an object weighing 2712 pounds will be:
$$a = \frac{682.77}{2712} \times 32.18 = 8.09$$
and that's in units of feet/sec$^2$.
Maybe I'm just used to SI units, but I repeated the calculation using SI units and got the same result a lot quicker! Apart from anything else it makes the distinction between mass and force a lot clearer so you wouldn't have forgotten you need to multiply by the acceleration due to gravity to get the acceleration.
You're quite correct that if power is constant then torque is inversely proportional to rpm. However for most engines the power is roughly proportional to engine speed over a reasonable range, so the torque is approximately constant. If you look at the torque curves for family cars you'll see the torque is roughly constant in the middle of the engine speed range but tails off at high engine speeds. Very highly tuned engines have a torque that peask at higher engine speeds because they're tuned to develop a lot of power at high engine speeds.