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?
-
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.
-
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.
Best Answer
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.)