Real-world torque at standstill:
This is not your main area of concern but needs addressing so that you can then properly deal with the main issue.
at t=0 acceleration is infinite.
Indicates (as you are aware :-) ) a basic misunderstanding or misapplication of formulae.
Where one formula produces results that do not match known reality, it can help to do a "sanity check" using other formulae which should give the same answer but by using a different approach. If (when) they fail to do so, looking for the apparent contradictions can help you to identify incorrect assumptions and applications of formulae to reality.
A very well known basic formula is f = m x a. - Force = mass x acceleration.
Rearranging - a = f/m
For acceleration to be infinite mass must be zero or force must be infinite.
In this case the relationship between power and torque and rotational velocity are liable to be misleading you.
In an ideal system torque is inversely proportional to power when both are plotted against rotational velocity.
ie Torque = Power/RPM x k (k adjusts to get units numerically correct).
This suggests that for a given supplied power Torque becomes infinite at zero RPM.
This is true only if there are no losses and if the energy source is able to be properly matched to the load. To transfer power from a rotating energy source ('motor') to a stationary object (wheel, axle, driveshaft, gear, ...) requires an infinite ratio gearbox, 'which we have not got'.
The closest example in 'everyday' automotive reality is electric motor driven vehicles where maximum torque occurs "at rest" as the electric motor can deliver torque without rotation. One limitation to how fast a modern electric sportscar 'comes off the line' is providing a drivetrain that will not shatter under starting torque but which is not excessively oversized and expensive for its task across the rest of its operating range.
In real-world applications the motor (especially a petrol or diesel engine) power decreases at low rpm and is zero at 0 RPM so torque is also zero at 0 engine RPM. This is compensated for by initially "slipping the clutch" in a manual gearbox system and effectively "slipping the torque converter" in an "automatic gearbox" system. At standstill the motor is allowed to operate at nonzero rpm, power and torque are produced and the torque is used to apply force to the drivetrain - either via a slipping clutch or via circulating transmission fluid exerting force but no power or motion at the instantaneous moment of application. Once there is some force available, f=ma can be applied and "we're off".
In electric cars (with Tesla's easily coming to mind) it is positive with some types of electric motor to produce maximum torque at zero motor rpm and "we're off" faster per available engine power than with internal combustion engines that must produce start torque in a lossy manner, as above.
The real problem - air resistance:
If your basic equations for power and energy re correct then it is likely that your comment
... After taking air resistance into account ....
indicates a very significant misunderstanding of the effects and magnitude of energy consumed by air resistance.
For cars of typical size and dimension, air resistance is THE major source of power loss at velocities above about 10 m/s. As speed increases a proper modelling of drag and energy lost to air resistance is utterly essential to calulating acceleration profiles.
A simplistic formula for air drag which is nevertheless extremely useful and which gives results close to those seen in reality is
Force (drag) = 0.5 x d x Cd x A x V^2
Power = 0.5 x d x Cd x A x V^3
d = air density = 1.2 kg/m^3 at sea level
Cd = drag coefficient
Proportion of flat-plate-drag which you can achieve by aerodynamic streamlining.
A = projected frontal area
Area you would see viewing vehicle front on.
V = velocity
In SI units where A = m^2, V = m/2, d = kg/m^2, F = N (or F ~= kg x 10)
Power is in Watts where 1 HP ~= 760 Watts.
At sealevel this reduces to ~=
Power = 0.6 x V^3 watts per square meter for a flat plate, reduced by whatever Cd you can achieve. eg a modern blockish shaped saloon or SUV may achieve Cd = 0.6-0.8 and the very bet sportscars MAY achieve Cds of 0.2-0.3. ie you can reduce power taken by drag by a factor of 3 or 4 compared to driving in a brick shaped vehicle.
As an illustration of how much power is involved.
Assume 2 m^2 frontal area, Cd = 0.5, V = 27 m/s (= 60 mph / 96 kph)
Power = 0.6 x Cd x A x V^3 = 0.6 x 0.5 x 2 x 27^3
= 11,810 Watts
~= 16 HP
At 100 mph that increases by the cube of the velocity so by a factor of (100/60)^3 =
4.6 times to about 74 HP. At 200 mph the power taken to oppose air resistance is 2^3 = 8 times larger again at 590 HP. For vehicles with hundreds of rear wheel HP available the jump from about 75 HP at 100 mph to about 600 HP at 200 mph means top speed will be "somewhere in this range".
A lot (lot lot) more could be said on this topic, but applying the above drag and power formula as a starting point should greatly improve the match of your results to reality.
Torque and power:
There is commonly a very major misunderstanding of the very simple relationship of torque and power.
Torque = Power per RPM.
That's it. Tht IS the formal meaning (or one of the several different but consistent ones. There will be a constant involved depending on the uits used but it's just a "scaling factor". Using foot-pounds and Horsepower
Torque = HP x 5252 / RPM or
HP = Torque x RPM / 5252 or
RPM = HP x 5252 / Torque
Can it be that simple?
Torque is "twisting force" expressed in force x distance.
If you lean on a 1 foot bar with 100 lbf force you produce 100 foot pounds of force. Double the force or the bar length and you double the torque.
If the bar end does not move you are not delivering any power (despite what your muscles and brain may think). Only as you apply rotation do you deliver power - see formulae above.
A useful 'trick':
An "almost right" formula is Watts = kg.m.RPM.
This almost works because various constants almost cancel. It's about 2% out but very useful for in-the-head estimations.
The point where I am stuck is figuring out what the rpms will be at any given point for a maximum accelerating car... from there I can find the force and then the acceleration.
If you have an automatic transmission you can ~~~ assume that the combination of the correct gear plus torque converter action*
THEN you can assume RPM is variable around engine optimum at all times.
Optimum is (probably/usually) such that mean torque is maximised in that gear.
*A fluid drive torque converter all by itself is effectively an automatically variable gearbox across a limited range, using fluid flow to 'match' driving and driven surfaces.
eg if the
"pump" (engine) rotates faster than the driven surface then the system
is designed so that fluid efficiently transfers energy to the slower
moving driven "plate".
If you have a manual gearbox and a competent driver then you can assume a driver will enter a gear and exit a gear at such revs that Rmax ~= redline and Rmin is the Redline /gear_ratio between this gear and the next lower. For VERY narrow power bands, redline may be above optimum speed for engine so driver will exit at such revs that overall revs in that gear produce maximum torque. If the above does not make sense then draw diagrams with rpm on and power curves and gear ratios nd play until it becomes clear (assuming no typos above :-).
Very simplistic example: Assume a gearbox with 1.5:1 ratios between all gear changes (1 to 2, 2 to 3, 3 to 4, ...) - which is very high in most cases, and an engine that keeps on building power all the way to redline with power increasing fater than RPM and redline at 9,000 RPM.
The optimum way to use this engine is to change up AT redline and to try to start it from rest such that when the clutch fully engages the engine is doing 6000 rpm (assuming your clutch can stand that). I'll leave out the 1st gear startup as that is somewhat arguable. So:
Accelerate in 1st gear to 9000 RPM.
2nd gear is now running at 9000/1.5 = 6000 RPM.
Accelerate in 2nd gear to 9000 RPM.
3rd gear is now running at 9000/1.5 = 6000 RPM.
Accelerate in 3rd gear to 9000 RPM.
4th gear is now running at 9000/1.5 = 6000 RPM.
I had a TY250 trials bike (still have but now sleeping in my basement waiting engine repair from long ago death).
The ratio 1st:2nd is an immense 2:1
The ratio 2nd:3rd is an immense 1.5:1
4:3 and 5:4 are somewhat more sane.
Most fun and (just) possibly optimum way to accelerate was to pull wheelies in 1st 2nd 3rd and 4th successively staring with a near loop in 1st and successively lower ones as the bike desired and changing up as the front wheel kissed the ground after each wheelie. Hitting 5th usually did not produce a wheelie and you then chugged way at about 50 kph :-).