**Step 1: Introduction/Overview of the Question** The problem concerns a banked highway curve designed for a specific speed (65 km/h) and radius (205 m). On a rainy day, cars are moving at a lower speed (40 km/h), and we must find the minimum coefficient of friction (\(\mu\)) required between tires and road so that vehicles can safely make the turn without sliding. The solution involves using the physics of banked curves, centripetal force, friction, and trigonometry. --- **Step 2: Calculation of the Banking Angle (\(\theta\))** First, find the angle at which the curve is banked for the design speed, using the formula: \[ \tan\theta = \frac{v^2}{rg} \] Where: - \(v\) = design speed = 65 km/h = \(65 \times \frac{1000}{3600} = 18.06\) m/s - \(r\) = radius = 205 m - \(g\) = 9.8 m/s² \[ \tan\theta = \frac{(18.06)^2}{205 \times 9.8} = \frac{326.16}{2009} = 0.1624 \] \[ \theta = \arctan(0.1624) \approx 9.22^\circ \] --- **Step 3: Calculation of Minimum Coefficient of Friction (\(\mu\)) for 40 km/h** Now, calculate the minimum coefficient of friction required at 40 km/h: - \(v' = 40\) km/h = \(40 \times \frac{1000}{3600} = 11.11\) m/s The formula for the minimum coefficient of friction for a banked curve is: \[ \mu = \frac{v'^2/(rg) - \tan\theta}{1 + (v'^2/(rg))\tan\theta} \] \[ v'^2/(rg) = \frac{(11.11)^2}{205 \times 9.8} = \frac{123.46}{2009} = 0.0615 \] \[ \mu = \frac{0.0615 - 0.1624}{1 + (0.0615)(0.1624)} \] \[ \mu = \frac{-0.1009}{1.00998} = -0.0999 \] Since friction cannot be negative, **the minimum friction required is the magnitude (positive value), since friction must act up the bank to prevent sliding down**: \[ \mu = 0.10 \] --- **Final Answer:** \[ \boxed{0.10} \] **Units:** (no units, coefficient of friction is dimensionless)