The following post doesn't really add anything that hasn't been said before, but I hope it might illustrate what others have said, cement everything in people's minds and hopefully make understanding the subject a little easier.
Lets say we have two engines, with the torque curves as shown below. These two engines essentially have the same power curve.
Engine one has a peak flywheel torque of 63.5lbft, and engine two has a peak torque of 127lbft.
Assuming that the gearing can be optimised for each engine which one can accelerate the car quicker at a speed of 60mph? (N.B. this IS a trick question!)
Lets say the driven wheels have a Diameter of 0.6metres. That means at 60mph the rear wheels need to turn at approx 850rpm.
Given that the wheels of both cars are the same size this means that the car which can produce the highest torque at the road wheels will be able to produce the most motive force at the tyre contact patch, and since f=mA it will have the higher acceleration (assuming mass, inertia and drag are all equal).
First we will optimise the gearing of each engine such that the
peak torque can be transmitted to the road at 60mph.
Engine 1 creates it peak torque at 9000rpm. In order to match this engine speed to the road wheel speed of 850rpm we need a gear ratio of 10.59:1. That means the flywheel torque is increased by 10.59 times: 63.5 x 10.59 = 672lbft.
Engine 2 creates its peak torque at 4500rpm. In order to match this engine speed to the road wheel speed of 850rpm we need a ratio of 5.29:1. That means the flywheel torque is increased by 5.29 times; 127 x 5.29 = 672lbt!
So the engines have exactly the same capability to accelerate the car. The clue was in the statement at the very beginning; "Both engines have essentially the same power curve".
Hopefully that has shown that it is difficult to compare one engine to another on flywheel torque, and easy to compare them on flywheel power (higher power = higher acceleration potential).
Now lets look at how much torque each engine can generate at the road wheels, at 60mph, if we gear the engines such that the engine is at a higher RPM than the peak torque figure (an RPM where both engines generate more power)
Lets take Engine 1 as an example. At 15,000 rpm it can generate 52lbft (considerably less than its peak value of 63.5lbft). To match 15,000 rpm to the 850 rpm required to turn the wheels at a speed equivalent to 60mph it needs a ratio of 17.65:1. This means the flywheel torque is multiplied by 17.65; 52 x 17.65 = 917.8lbft!!!! A lot more than the 672lbft generated earlier.
Since the two engines generate the same power, engine 2 also will be able to generate 917.8lbft at the driven wheels, I'll leave that check to anyone keen enough to prove it for themselves (the numbers are 104lbft and 7500rpm).
I hope that illustrates why it is easier to compare the output of an engine in terms of its power output rather than its torque output, even though either can be used to determine the acceleration, and both will give the same result.
A word of warning, as others have said; What we've done is pick two points on the torque/power curve. In reality the acceleration is determined by the driven wheel torque/power that the car can create at all engine speeds which are used during acceleration. For a typical 5 or 6 speed gearbox equipped car this means the top 30 to 40% of the rev-range. So what we need to do is compare the top 30 to 40% of the power curve -the one with the higher power over that rev range will be able to accelerate the car faster (mass, inertia and drag being equal), assuming gearing can be optimised in each case.
