I am currently trying to understand a certain relation between power $P$, velocity $v$ and acceleration $a$.

We are looking at a car driving along a horizontal road and at static friction, and how static friction translates to movement with the cars tires. We are given the maximum static friction coefficient $\mu$ which is $0.6$.

The question I was given is, "at which velocity $v$ will the maximum possible acceleration $a$ be limited by the power of the engine $P$?"

I know that we can write $P(t) = F(t)v(t)$ and therefore $P(t) = ma(t)v(t)$ and with that $v(t) = \cfrac{P(t)}{ma(t)}$

I don't quite understand the relation of the given quantities however. It says, that $v_1 = \frac{P}{ma_{max}}$ is the velocity at which the acceleration is limited by the static friction, where $a_{max} = \mu \cdot g$. And then, that above $v_1$ the acceleration will be limited by the power, and that only then we can bring the entirety of the cars power onto the street without the wheels spinning through.

My questions now:

  1. What exactly happens up to $v_1$ ?
  2. Why can we suddenly neglect the static friction once we have hit this threshold velocity of $v_1$? Why can we only bring the full power of the engine onto the street there?

It all makes somewhat sense to me, in an intuitive way, having driven a car many times already, but I am just so damn blind in translating this into these forumlas. I am sure that the answers are probably right in front of my eyes and that I just can't really interpret them yet due to a lack of experience and knowledge and intuition about physics. I would be really happy if someone could elaborate on this in a clear way.

  • $\begingroup$ "at which velocity 𝑣 will the maximum possible acceleration 𝑎 limited by the power of the engine 𝑃 ?" This isn't even a sentence. There's no main verb! $\endgroup$ Jan 2 at 12:15
  • 1
    $\begingroup$ @PhilipWood I think the OP means "be limited". $\endgroup$
    – Bob D
    Jan 2 at 14:28
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    $\begingroup$ Yes indeed. I was just upset that the op had been set a question that didn't make sense. I was being over-sensitive. $\endgroup$ Jan 2 at 14:32
  • $\begingroup$ "at which velocity $v$ will the maximum possible acceleration $a$ limited by the power of the engine $P$?" is not a valid sentence. Please fix. $\endgroup$ Jan 2 at 23:09
  • $\begingroup$ "what exactly happens up to v1" in practice is that the car is accelerating at more or less constant rate, but also is trying to get off-course. At v1 the driver feels a sudden "kick from the back" and the directional stability happens all of a sudden. The forward acceleration starts going down and down. $\endgroup$
    – fraxinus
    Jan 3 at 9:32

3 Answers 3


A car's acceleration (ignoring air resistance) is limited by two separate quantities. You can plot acceleration vs. speed and describe these limits as curves. The actual acceleration $a$, must be below those two curves at all times.


One is the maximum power the engine can put out $P$, shown above in the dashed blue curve, and the second is the available traction of the tires with the surface $\mu_s$.

  • Power is defined as $P = F\,v$ and with $F= m a$, you have the limiting power curve as $$a = \frac{P}{m v} \tag{1}$$ for all speeds. This is a hyperbola, with acceleration going down as speed increases.

  • Traction is defined by the available coefficient of friction of the tire with the surface $\mu_s$. This limits the torque on the wheels since the tractive force is $F \leq \mu_s m g$ where $m g$ is the weight of the car. Again with $F = m a$, you have the limiting traction curve as $$a = \mu_s g \tag{2}$$ for all speeds. This is a flat curve, as it does not depends on the speed of the car.

Caution: All of the above are extreme oversimplifications of the actual physics of cars.

Now consider the situation where the engine power and gearing are such that in 1st gear the car's acceleration would exceed available traction $a > \mu_s g$


What would happen in this case, the acceleration curve would be capped (chopped off) at the traction limit as the tires would start to spin, instead of providing the acceleration provided by the engine power.

This phenomenon cannot happen once the speed $v$ as above the intersection of the two limits curves. To find where this speed $v_1$ is, equate (1) and (2) to get

$$ \frac{P}{m v_1} = \mu_s g $$ and solve for $v_1$.

As a simplification of the car acceleration model, you then have two branches that switch at $v_1$

$$ a = \begin{cases} \mu_s g & v \leq v_1 \\ \frac{P}{m v} & v > v_1 \end{cases} $$

If you want to include air resistance, you have $\tfrac{P}{v} - F_{\rm drag} = m a$ for the power limit curve, and with $F_{\rm drag} = \beta v^2$ as a generalized air resistance curve, the power limit curve is now

$$ a = \frac{P - \beta v^3}{m v} \tag{3} $$

But now the calculation for $v_1$ involves solving a cubic equation which is not easy.

I think the solution is

$$ v_1 = \left( \sqrt{ \tfrac{P^2}{4 \beta^2} + \tfrac{\mu_s^3 m^3 g^3}{27 \beta^3} } + \tfrac{P}{2\beta} \right)^{1/3} - \left( \sqrt{\tfrac{P^2}{4 \beta^2} + \tfrac{\mu_s^3 m^3 g^3}{27 \beta^3}} - \tfrac{P}{2\beta}\right)^{1/3} \tag{4} $$

The solutions of limiting acceleration at all speeds involve calculus and the relationships for time and distance as a function of speed.

$$ \begin{aligned} \Delta t &= \int \limits_{\Delta v} \tfrac{1}{a} \,{\rm d}v & \Delta x & = \int \limits_{\Delta v} \tfrac{v}{a}\,{\rm d}v \end{aligned} \tag{5}$$

  • $\begingroup$ I realized that my acceleration curves aren't accurate in shape, but the idea of clipping off the excess acceleration above traction remains. $\endgroup$ Jan 3 at 15:58
  1. What exactly happens up to $v_1$ ?

Once the maximum possible static friction force of $\mu_{s}mg$ is reached, the drive wheel(s) lose traction (spin} and static friction becomes kinetic (skidding) friction, which is generally less than static friction. Since static friction is responsible for the maximum acceleration of the car, once the maximum possible static friction force is reached the acceleration can no longer increase. That occurs at $v_1$ per your equation.

  1. Why can we suddenly neglect the static friction once we have hit this threshold velocity of $v_1$?

Because the maximum possible static friction force is reached and friction becomes kinetic (sliding)..

Why can we only bring the full power of the engine onto the street there?

Because once the maximum possible static friction force is exceeded, part of the power will be unproductive (not contribute to additional acceleration) as it is dissipated as kinetic friction heating.

Hope this helps.

  • $\begingroup$ It’s not true that kinetic friction contributes no acceleration. It does contribute substantially less acceleration than static friction, and the skidding friction is independent of the wheel speed. To see, consider the case of hard braking, where the wheels lock to the car and skid on the ground. Locked wheels are less effective at stopping the car (thus anti-lock brake systems on modern cars), and the vehicle is effectively unsteerable if the front wheels get locked, but a car with locked brakes will still eventually stop. $\endgroup$
    – rob
    Jan 3 at 3:49
  • $\begingroup$ Kinetic friction is much more complex and in a lot of cases is assumed to be both dependent on the wheel speed and more than the static friction $\endgroup$
    – fraxinus
    Jan 3 at 9:20
  • $\begingroup$ @rob so I suppose we can only say static friction determines the maximum possible acceleration. Which still leaves the question on how to determine the acceleration due to kinetic friction. See my edits. $\endgroup$
    – Bob D
    Jan 3 at 9:28

I think the static friction is irrelevant to this scenario. Power does not limit vehicle velocity when wheels are spinning. Power limits velocity when the all power is required to overcome aerodynamic drag of the vehicle.

Aerodynamic drag force is given by $F_D = \frac{\rho}{2}C_D A v^2$, where $C_D$ is drag coefficient, A is the frontal area of the vehicle, and $\rho$ is air density. Power required to overcome drag is, therefore, $$P_D = F_D v = \frac{\rho}{2}C_D A v^3$$ Restated, the maximum achievable velocity for a given power (neglecting rolling resistance and other power losses) is $$v= \left(\frac{2 P}{\rho C_D A}\right)^\frac{1}{3}$$

In reality, rolling resistance (from tires, bearings, etc.) and other losses (air conditioning, power steering, etc.) will result in a lower maximum velocity, since some engine power will be consumed by these losses.

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    $\begingroup$ The question is not about power limiting velocity. It is about power limiting acceleration, and asks for the conditions (velocity particularly) where the power limit becomes the binding constraint (instead of traction). The coefficient of static friction is quite important to finding where the traction limit and power limit curves intersect. $\endgroup$
    – Ben Voigt
    Jan 2 at 20:16
  • $\begingroup$ So when Pd(v)=Pengine, engine power does limit acceleration to zero at velocity v. To resplit the hair. $\endgroup$
    – AndyW
    Jan 6 at 16:54

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