A roadway is being design for a speed of 120 km/hr at one horizontal curve. It is know that the super elevation is 8% and the coefficient of side friction is 0.09. Determine the minimum radius of the curve that will provide safe vehicle operation.

The given design speed is $V = 120 \text{ km/hr}$.

$$V = 120 \text{ km/hr} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ hr}}{3600 \text{ s}}$$ $$V = \frac{120000}{3600} \text{ m/s} = 33.333 \text{ m/s}$$

The super elevation is $e = 8\%$.

$$e = \frac{8}{100} = 0.08$$

The formula is $R_{min} = \frac{V^2}{g(e+f)}$, where $g$ is the acceleration due to gravity ($9.81 \text{ m/s}^2$), $e$ is the super elevation, and $f$ is the coefficient of side friction.

$$R_{min} = \frac{(33.333 \text{ m/s})^2}{9.81 \text{ m/s}^2 (0.08 + 0.09)}$$ $$R_{min} = \frac{1111.111}{9.81 \times 0.17}$$ $$R_{min} = \frac{1111.111}{1.6677}$$ $$R_{min} \approx 666.38 \text{ m}$$

The minimum radius of the curve that will provide safe vehicle operation is approximately $666.38 \text{ m}$.