I'm trying to work out an acceleration curve for a vehicle for a simulation, but despite a bunch of research there are a couple of things I still don't understand.

First we have $P = mav$. Assuming that I know or can set the power for this vehicle, I can rearrange that for acceleration, $a = \frac{P}{mv}$. That doesn't seem to work in practice, though - it's not possible to calculate the acceleration for a given power from standstill, i.e. $v=0$.

Looking around, different sources seem to recommend different things (calculate applied torque, use kinetic energy, include gear ratios, etc, etc), but none that I can find seem to cohesively link them back into the original calculation.

Clearly we can't have infinite acceleration at zero velocity. I'm wondering if friction between the wheel and the running surface is what sets an upper limit on acceleration here, but I can't work out how to factor it into my calculations.

Given a known power, mass, velocity, coefficient of friction - basically any variable can be known other than acceleration - how can I combine what I have into one calculation to work out acceleration at a given velocity?

  • $\begingroup$ "I'm wondering if friction between the wheel and the running surface is what sets an upper limit on acceleration here" Yes it does. The maximum possible acceleration of the vehicle is $$a_{max}=\mu_{s}g$$ Where $\mu_s$ is the coefficient of static friction between the road and tire and $g$=9.8 m/s$^2$ $\endgroup$
    – Bob D
    Commented Mar 25, 2023 at 15:20
  • $\begingroup$ There are a lot of ways to do this at different levels of complexity. But torque is also a relevant limit in cars. One reasonable-ish way to set it up would to have a maximum torque and a maximum power. Calculate the acceleration with either, and choose the lower number (if the torque based acceleration is lower, the engine is torque limited, otherwise its power limited). Figure out how much torque you need to apply to the wheels to get a certain acceleration. $\endgroup$
    – AXensen
    Commented Mar 25, 2023 at 16:03
  • $\begingroup$ Related: physics.stackexchange.com/a/640977/392 $\endgroup$ Commented Mar 26, 2023 at 22:01
  • $\begingroup$ Friction is there, but air resistance is order of magnitude higher than friction for highway speeds and above. $\endgroup$ Commented Mar 26, 2023 at 22:02
  • $\begingroup$ Related - physics.stackexchange.com/a/743639/392 $\endgroup$ Commented Mar 26, 2023 at 22:03

4 Answers 4


Under constant power, the force on the car due to the engine is indeed $$F_{\rm motor} = \frac{P}{v}$$

But you want to account for air resistance

$$ F_{\rm drag} = m \beta\, v^2 $$

where $beta$ is some constant related to the geometry of the car and the density of the air.

And rolling resistance (friction)

$$ F_{\rm roll} = m a_{\rm f} $$

where $a_{\rm f}$ is constant deceleration due to the rolling resistance of the tires and drivetrain.

Combine the above and Newton's 2nd law to get

$$F_{\rm motor} - F_{\rm drag} - F_{\rm roll} = m a $$

which is solved for the acceleration

$$ \boxed{ a = \frac{P}{m v} - a_{\rm f} - \beta v^2 } $$

The above has a direct solution, but is rather complex to write it out here. To find it use the following relationships

$$ \begin{aligned} x - x_1 &= \int _{v_1}^v \frac{ v}{a}\,{\rm d}v \\ t - t_1 &= \int _{v_1}^v \frac{ 1}{a}\,{\rm d}v \\ \end{aligned}$$

where $t_1$ is the initial time, $x_1$ the initial distance and $v_1$ the initial speed.

In a simulation setting, you can use the above as an integrator, given a acceleration is a function of speed only $a = f(v)$ then at each time-frame you have

$$\begin{aligned} \Delta x &= \frac{v}{f(v)} \Delta v \\ \Delta t &= \frac{1}{f(v)} \Delta v \end{aligned}$$

The integrator finds the distance and time needed to go from speed $v$ to speed $v+\Delta v$ when $\Delta v$ is a finite but small change in speed.


Some of the confusion seems to be coming from the equation relating a force to the rate it does work $F = \frac{P}{v}$. Rather than considering $F$ becoming infinite for a fixed power at $v = 0$, it’s more helpful to think of a finite force as applying zero power. This is true the instant the car is at rest. Differentiating the expression for kinetic energy results in $P = mva $. Thus, counterintuitively, KE is not increasing at $v = 0$, even though you are accelerating.

In a real car, of course, the engine will still consume fuel whenever it’s running. It’s just not converted into translational kinetic energy if the car is stationary.


In the first problem you can't just set power as something and do velocity zero because power is not something we can fix,you can fix the force but power depends on the engine(efficiency) also here we only considered the only force which acts on the engine but the motion is a mixture of a number of forces not just one, all of which are variable


The power of the car engine is

$$P=\tau\,\omega_E=\tau\,\frac{n\,\pi}{30}$$ where $~\tau~$ is the engine torque and $~n~$ is the engine speed of rotation $~[1/min]~$

from here you obtain the force that accelerate the car $$~F=a\tau~\quad, n=n_{P}$$ where $~n_P~$ is the speed of rotation in which you obtain $~\tau=\tau_{max}~$

with Newton second low the acceleration of the car


where $~\mu\,M\,g~$ is the friction force and M is the total mass of the car


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