Related to this answer to a question about the speed of an accelerating car, I'm looking for a formula for the speed of a car decelerating in neutral (with the transmission disengaged, i.e. no torque, either positive or negative, from the engine.) Although similar to the modeling made in that answer, the constant $A$ should be negative since there is no torque from the engine. Assuming $c_r$ is the constant related to the rolling resistance force and $c_{drag}$ that due to aerodynamic drag, we can write

$$ \frac{dv}{dt} = -c_r - c_{drag} v^2 $$

This can be rewritten as

$$ \frac{dv}{-c_r - c_{drag} v^2} = dt $$

Solving this without taking integration constants into account, I found

$$ v = \sqrt{\frac{c_r}{c_{drag}}} \tan \left( -\sqrt{c_r c_{cdrag}} t \right) $$

Although there is a problem with this formula in that $c_r = 0$ when $v = 0$, which is not captured by the formula, in my case I don't really consider it an issue as I'm interested in the deceleration curve between two different speeds, both greater than zero.

The real issue for me is that I haven't been able to properly fit an integration constant to this, so that I start with a certain desired speed $v_0$ at $t = 0$. My first thought was to add $v_0$ as an integration constant after solving the left-hand side of the second equation above. This gives:

$$ v = \sqrt{\frac{c_r}{c_{drag}}} \tan\left(-\sqrt{c_r c_{drag}} (t - v_0)\right), $$

but when I replaced some reasonable values for $c_r$ and $c_{drag}$ (which actually worked fine in the acceleration case), and taking $v_0 = 30$, the initial speed was over 350 m/s, which makes no sense.

So how do I correctly add integration constants to the solution above, so that I get a correct answer for the desired $v_0$?

  • $\begingroup$ One thing you should note above $(t - v_0)$ is nonsensical because the units are incapable with addition (that is of it refers to a velocity and not a time). $\endgroup$ – Triatticus Mar 16 '20 at 6:52

While writing the question, it dawned on me how to correctly proceed. Consider again the second equation:

$$ \frac{dv}{-c_r - c_{drag} v^2} = dt. $$

Solving the integral, we get

$$ - \frac{\tan^{-1} \left( \sqrt{\frac{c_{drag}}{c_r}} v \right)}{\sqrt{c_r c_{drag}}} + c = t, $$

where without loss of generality a single integration constant was considered.

The point I was missing is that I know $v(t = 0) = v_0$. Therefore, I need to find the value of $c$ in this equation that ensures $v = v_0$ when $t = 0$. Replacing these values into the equation, I get the following:

$$ c = \frac{\tan^{-1} \left( \sqrt{\frac{c_{drag}}{c_r}} v_0 \right)}{\sqrt{c_r c_{drag}}}. $$

So, the actual equation for deceleration, assuming $v(t = 0) = v_0$, should be:

$$ v = \sqrt{\frac{c_r}{c_{drag}}} \tan\left(-\sqrt{c_r c_{drag}} \left( t - \left( \frac{\tan^{-1} \left( \sqrt{\frac{c_{drag}}{c_r}} v_0 \right)}{\sqrt{c_r c_{drag}}} \right) \right) \right). $$

This is now giving reasonable values, except, as stated in the question, when $v$ reaches zero, where friction should subside. However, this is to be expected as the vanishing friction is not modeled.


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