Let $m$ be the mass of the car. The forces acting on the car are gravity ($mg$), the normal force ($N$), and static friction ($f_s$). The net force provides the centripetal acceleration $a_c = v^2/R$.

First, we determine the banking angle $\theta$ from the design speed $v_d$. At this speed, no friction is needed ($f_s = 0$). The horizontal component of the normal force provides the centripetal force. Vertical force balance:

$$N \cos\theta = mg$$

Horizontal force balance:

$$N \sin\theta = \frac{mv_d^2}{R}$$

Dividing the second equation by the first gives the bank angle:

$$\tan\theta = \frac{v_d^2}{gR}$$

Next, consider the car moving at speed $v_a < v_d$. The required centripetal force is less than the horizontal component of the normal force at that speed. Thus, the car tends to slide down the incline. The static friction force $f_s$ must point up the incline to prevent this. For the minimum coefficient of friction, $f_s$ is at its maximum value, $f_s = \mu_s N$.

We resolve forces into horizontal and vertical components: Horizontal: $N \sin\theta - f_s \cos\theta = \frac{mv_a^2}{R}$ Vertical: $N \cos\theta + f_s \sin\theta = mg$

Substitute $f_s = \mu_s N$:

$$N(\sin\theta - \mu_s \cos\theta) = \frac{mv_a^2}{R}$$ $$N(\cos\theta + \mu_s \sin\theta) = mg$$

Divide the first new equation by the second to eliminate $N$ and $m$:

$$\frac{\sin\theta - \mu_s \cos\theta}{\cos\theta + \mu_s \sin\theta} = \frac{v_a^2}{gR}$$

Divide the numerator and denominator of the left side by $\cos\theta$:

$$\frac{\tan\theta - \mu_s}{1 + \mu_s \tan\theta} = \frac{v_a^2}{gR}$$

Now, we solve for $\mu_s$:

$$\mu_s = \frac{\tan\theta - \frac{v_a^2}{gR}}{1 + \frac{v_a^2}{gR}\tan\theta}$$

Finally, substitute the expression for $\tan\theta = v_d^2/(gR)$:

$$\mu_s = \frac{\frac{v_d^2}{gR} - \frac{v_a^2}{gR}}{1 + \frac{v_a^2}{gR} \frac{v_d^2}{gR}} = \frac{\frac{1}{gR}(v_d^2 - v_a^2)}{1 + \frac{v_a^2 v_d^2}{g^2 R^2}}$$ $$\mu_s = \frac{gR(v_d^2 - v_a^2)}{g^2 R^2 + v_a^2 v_d^2}$$