# Formula for deceleration of a car with transmission disengaged

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 $$v_0$$?

• 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). – 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.