Is my answer right?

A coin with a diameter of 2.4cm is dropped on edge onto a horizontal surface. The coin starts out with an initial angular speed of 18 rad/s and rolls in a straight line without slipping. If the rotation slows with an angular accleration of magnitude 1.9 rad/s^2, how far does the coin roll before coming to a rest?

Using the equation:

$\displaystyle \omega^{2}=\omega^{2}_{0}+2\alpha \Delta \theta$ where $\displaystyle \omega$ and $\displaystyle \omega_{0}$ are the final and initial angular speeds respectively; $\displaystyle \alpha$ is the angular acceleration; and $\displaystyle \Delta \theta$ is the angular displacement.

$\displaystyle 0=(18)^{2}+2(-1.9)\Delta \theta$

So $\displaystyle \Delta \theta = 85.26 \text{rad}$.

Now $\displaystyle \Delta \theta = \dfrac{\Delta s}{r}$ where $\displaystyle \Delta s$ is the change in arc length and r is the radius.

$\displaystyle 85.26=\dfrac{\Delta s}{0.012}$

So $\displaystyle \Delta s=1.023 \text{m}$. Is this right?