The problem you're describing involves calculating the angle at which a curved road should be banked so that a car can travel safely around it at a given speed without relying on friction. This can be solved using the concept of centripetal force and the banking angle formula.

Formula for the banking angle (θ):

$\tan(\theta) = \frac{v^2}{r g}$

Where:

• $\theta$ is the banking angle.
• $v$ is the speed of the car (29.0 m/s).
• $r$ is the radius of curvature (194 m).
• $g$ is the acceleration due to gravity (9.81 m/s²).

Steps:

1. First, calculate the right-hand side of the equation:

$\frac{v^2}{r g} = \frac{(29.0 \, \text{m/s})^2}{(194 \, \text{m}) (9.81 \, \text{m/s}^2)}$

2. Now substitute the values:

$\frac{v^2}{r g} = \frac{841}{1902.14} = 0.442$

3. Take the arctangent (inverse tangent) to find $\theta$:

$\theta = \tan^{-1}(0.442) \approx 23.9^\circ$

Final answer:

The curve should be banked at approximately 23.9°.

Would you like further details or have any other questions?

Here are some related questions for further exploration:

1. How does the banking angle change if the speed of the car increases?
2. What would happen if friction is taken into account in the calculation?
3. How does the radius of curvature influence the required banking angle?
4. Can the car stay on the road if the road were flat at this speed?
5. What role does the coefficient of friction play in non-banked curves?

Tip: The steeper the banking angle, the less friction is needed to prevent skidding, which is why curves on highways are often banked at sharper angles in high-speed areas.