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I need to estimate a drag race quarter mile time given the car's weight, bhp and preferably the drive (FWD, RWD, 4WD). I know $v(t) = ds/dt$ and $a(t) = dv/dt = d^2s/dt^2$, but how can I get the function $s(t)$ to calculate $v(t)$ and $a(t)$? I thought maybe Power to Weight ratio would help for solving the acceleration but I have no idea what to do with it and maybe the drive would just be a constant which is then subtracted from the calculated time. Can you help me?

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migrated from math.stackexchange.com Aug 16 '12 at 13:51

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The fundamental difficulty in trying to apply physics 101 concepts here (if you care about getting the right answer) is that internal combustion engines are complicated and do not offer their maximal power or torque over the whole operating range. Electric cars and those equipped with CVTs can come closer to the results of these naive calculations. –  dmckee Aug 16 '12 at 17:46
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3 Answers

As an alternative approach, one can use calculus to calculate $v(t)$ from $a(t)$ and then $s(t)$ from $v(t)$. In particular, $v(t)=v(0)+\int a(t)dt$ and so $s(t)=s(0)+\int v(t)dt$. One can model acceleration over time using the power-to-weight ratio (though other factors, such as gear ratio, will also come into play). You may either assume the weight is constant, or account for change of weight over time due to burning fuel. You can also take wind resistance into account, or ignore it. Ideally, for simplicity, assume zero wind resistance and a constant weight, but be aware that this will introduce some (not much) error.

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How to get a(t)? And isn't v(0) and s(0) equals 0? –  Cobold Jun 30 '12 at 20:01
    
Correct about the initial values. As for how to get $a(t)$, I'm afraid it has been too long, and I no longer recall, but I did want to make the point about the alternate approach, anyway. I'm sure you can find out some methods of modelingthe the acceleration from the physics stackexchange. –  Cameron Buie Jun 30 '12 at 20:45
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You can get the acceleration from Newton's second law $F=ma$ with the net driving force on the car. Then as @CameronBuie indicated, you integrate twice to find the expression for $s(t)$ from which you can solve for $t$.

If you assume the force is constant, then it proceeds like this: $$ a = F/m \\ v = Ft/m \\ s(t) = \frac{Ft^2}{2m} \\ t_D = \sqrt{2Dm/F} $$ where the integration constants are zero since $s(0)=v(0)=0$ and we can solve for the quarter-mile time $t_D$ by putting $D=0.25\mathrm{miles}$.

Here's an example. Assume a 1-ton car delivers 1000 ft-lbs of torque to the 12in-radius wheels the whole time and there is no drag, then approximately we have $$ F= 1000\mathrm{lb_f}\simeq 4448N \\ t_D = \sqrt{2\cdot 402m \cdot 1000kg / 4448N}\simeq 13.4s $$

To determine the force from the horsepower and drive type is partly a physics problem and partly an automotive engineering question, since it seems to depend a least on RPMs, gear ratios and transmission loss. You could try asking at physics.stackexchange.com or Google for some pointers.

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-1. Sorry the force is definitely not constant. You can maybe assume the power output of the engine is constant ($a=\frac{P}{m v}-{\rm drag}$ after it is in 2nd gear. –  ja72 Aug 17 '12 at 12:54
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Power is energy/time, so (you'll have to work out the units) bhp$=\frac d{dt} \frac {mv^2}2=mav$ Note that this says acceleration is infinite at startup ($v=0$) but you have a traction limit. So $a=\max(traction limit, \frac {bhp}{mv})$ While you are traction limited, $s=\frac 12 at^2$. When you are power limited, $\frac {dv}{dt}=\frac{bhp}{mv}$, so $\frac {v^2}2=\frac {bhp}m t+C$ where you evaluate $C$ to match the end of the traction limit phase.

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