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Is it possible to calculate the velocity of a car accelerating from rest at full capacity given the power of the engine and the mass of the car? I have a method of solving the velocity with respect to time, but I fear it may contain fallacies since I'm somewhat a physics novice. I will try to be as concise as possible with my work but let me know if I need to expand on any of my work.

Given the horsepower of the engine$^1$, P, there is a relation between power, time, and energy: $P= \frac{dE}{dt}$. Thus $E= Pt$.

Next (making sure to use the kW value for P, not the hp value) substitute this value into the kinetic energy equation to get: $Pt = \frac{1}{2} mv^2$. Then solve for the velocity:

$v = \sqrt{\frac{2Pt}{m}}$

This gives the velocity of the car void of any air resistance. To account for air resistance, first take the derivative of the velocity to get the acceleration:

$a(t) = v'(t) = \sqrt{\frac{P}{2mt}}$

The velocity of a free falling object with air resistance with respect to time t can be modeled by this equation$^2$:

$$v = \sqrt{\frac{2mg}{pAC_d}} \tanh\left(t\sqrt{\frac{gpAC_d}{2m}}\right)$$

To modify this to model to suit the situation, simply replace "g" with a(t) since the object is accelerating due to the force of the engine, not gravity:

$$v = \sqrt{\frac{2m\dot{}a(t)}{pAC_d}} \tanh\left(t\sqrt{\frac{a(t)\dot{}pAC_d}{2m}}\right)$$

Is this an accurate method of arriving at the velocity? If not, where did I go wrong and how can I fix it?

  1. In old times, horsepower ratings used to purely measure the power of the engine alone, but over time its been reformed to represent a closer approximation of an engine’s output as actually installed in a car. So luckily, no adjustments will be necessary to compensate for energy lost as heat or any other means.

  2. Excluding "g", Any new variables introduced in this equation are constants that pertain to the drag force equation and are not of any concern in this situation.

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    $\begingroup$ I don't think it worked, the $t\to\infty$ limit for terminal velocity/max speed of your car looks like 0... $\endgroup$
    – innisfree
    Commented Oct 19, 2016 at 0:41
  • $\begingroup$ You can't replace $g$ with $a(t)$. $g$ appears as a constant related to the force of gravity. The actual acceleration changes. $\endgroup$
    – garyp
    Commented Oct 19, 2016 at 2:56
  • $\begingroup$ Can't be all correct? When $t\to 0+$ according to your formula the acceleration $\to\infty$?? $\endgroup$
    – Jyrki Lahtonen
    Commented Jun 7 at 16:20

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Here's a much simpler way to think about this: Air resistance dominates the car's velocity, so the car rapidly reaches the terminal velocity at which the applied force ($F$) and the drag force ($F_D$) balance. The "tanh(...t)" expression is irrelevant at steady state.

The drag force

$F_D = (1/2) \rho C_d A v^2$,

where $\rho$ = the density of air, $C_d$ is the drag coefficient, and $A$ is the frontal area of the car. (To see where this comes from, and for a better explanation, see http://scitation.aip.org/content/aapt/journal/tpt/50/7/10.1119/1.4752039 -- I think the PDF is freely available.)

Power = Force x velocity, so our force balance becomes

$P / v = (1/2) \rho C_d A v^2$,

or

$P = (1/2) \rho C_d A v^3$,

giving a nice relationship between engine power and velocity. (Again, see the paper for more.)

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  • $\begingroup$ How about some $\LaTeX$? try e.g. $$E=mc^2$$, $$E=mc^2$$ $\endgroup$
    – innisfree
    Commented Oct 19, 2016 at 2:57

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