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edited Nov 24, 2017 at 15:17

Considering the complexity of the answers proposed and @Spirine answer which I think is wrong because confuses wheel torque with engine torque I want to give my own answer to the question. I think my approach is very straightforward and correct.

In this approachanswer I willfully ignore talking about the transmission in order to keep everything very straightforward. We can ignore the transmission because in practice any automatic or manual transmission can be approximated to a perfect torque converter which always use the engine at the maximum power to obtain the maximum torque at the wheel and thus the maximum acceleration in the vehicle speed range. The key concept here is that engine torque does not directly translate into wheel torque, the faster you go the longer the gear ratio the lower the torque at the wheel.

This let us draw some conclusions:This let us draw some conclusions:

But both are lower so it is a lot faster :)

Addendum, The ideal transmission approximation:

The ideal transmission can be thought as a lossless CVT (Continuous Variable Transmission) with no gear ratio limit. In practice a gearbox have a limited number of discrete ratios and this means that the engine will have to adapt the RPM to the current car speed and the active gear. The power loss here happens mostly in first gear when the car is so slow that it cannot be at peak power RPM, but at this stage, more than by power, cars are limited by tires grip, especially in the case of high powered cars; so it is a non-problem.

The ideal transmission perfectly convert power of the engine into torque at the wheel so that at 0 speed torque is infinite. As speed increases torque at the wheel decreases following the ideal equation: $$ τ_w = {P \over v} $$

In practice power varies as function of rpm and rpm varies as function of the gear and v, this makes everything unnecessarily complex and car-specific. This approximation, let us use very simple math to draw the above conclusions.

Addendum: Drive-train, tire and other ignored losses As speed increases drag, unlike other losses, increases with the square of the speed. Other losses are usually very small compared to drag at high speed. At low speed power is in excess compared to the grip tires offer so other losses do not really matter.

In this approach I willfully ignore talking about the transmission in order to keep everything very straightforward. We can ignore the transmission because in practice any automatic or manual transmission can be approximated to a perfect torque converter which always use the engine at the maximum power to obtain the maximum torque at the wheel and thus the maximum acceleration in the vehicle speed range.

This let us draw some conclusions:

But both are lower so it is a lot faster :)

In this answer I willfully ignore talking about the transmission in order to keep everything very straightforward. We can ignore the transmission because in practice any automatic or manual transmission can be approximated to a perfect torque converter which always use the engine at the maximum power to obtain the maximum torque at the wheel and thus the maximum acceleration in the vehicle speed range. The key concept here is that engine torque does not directly translate into wheel torque, the faster you go the longer the gear ratio the lower the torque at the wheel.

This let us draw some conclusions:

But both are lower so it is a lot faster :)

Addendum, The ideal transmission approximation:

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created Nov 24, 2017 at 12:51

fededevi

Consider the formula for kinetic energy $[E_k]$ : $$E_k = {1 \over 2}mv^2$$ Which can be solved for $[v]$ to obtain: $$ v = \sqrt{ {2 E_k} \over m }$$

This give us the speed as a function of kinetic energy and mass. To increase kinetic energy we need to be able to do work, engines produce work over time at a rate given by its power. The work produced over time is transferred to the car as kinetic energy by our ideal transmission and drive-train. By deriving over time we can get acceleration as function of power, velocity and mass:

$$ a = { P \over v m }$$

From this equation is straightforward to see that torque of the engine does not matter at all when an ideal transmission is used. Acceleration depends only on power/weight ratio.

If we want to calculate the maximum speed of the car we can use the same equation and add the (negative) acceleration provided by drag, let's ignore drive-train and ground-friction tire losses since those are low at high speed compared to drag. Usually air resistance force is approximated to $f = kv^2$ which can be integrated in the equation:

$$ a = { P \over v m } - {kv^2 \over m}$$ $$ k = \text drag coeff.$$

Maximum speed is reached when engine acceleration equals the drag deceleration, or simply we put $a = 0$ which give us the equation:

$$ { P \over v_m m } = {kv_m^2 \over m}$$

solved for max velocity $v_m$:

$$v_m = \sqrt[3] {P \over k}$$

This let us draw some conclusions:

Maximum speed is a function of power and drag coefficient.
Mass does not affect maximum speed.
Torque of the engine does not affect maximum speed.
Torque of the engine does not affect acceleration.
The Lamborghini would be faster than the truck even at the same weight and drag coefficient.

But both are lower so it is a lot faster :)