To solve for the radius of the curve, we can use the principles of circular motion and the condition of the road being banked. The key here is that the car will stay on the road without any friction, so the normal force provides the necessary centripetal force for the circular motion.

Formula for the radius:

For a banked curve with no friction, the following equation governs the motion of the car:

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

Where:

• $\theta$ is the angle of the bank (30.5°),
• $v$ is the speed of the car (18.0 m/s),
• $r$ is the radius of the curve (which we need to solve for),
• $g$ is the acceleration due to gravity (approximately 9.8 m/s²).

Rearranging the equation to solve for $r$:

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

Now, plug in the known values:

• $v = 18.0 \, \text{m/s}$,
• $g = 9.8 \, \text{m/s}^2$,
• $\theta = 30.5^\circ$.

Calculation:

$r = \frac{(18.0)^2}{9.8 \times \tan(30.5^\circ)}$

First, calculate $\tan(30.5^\circ)$: $\tan(30.5^\circ) \approx 0.587$

Now substitute the values: $r = \frac{324}{9.8 \times 0.587}$ $r = \frac{324}{5.75}$ $r \approx 56.5 \, \text{m}$

Final Answer:

The radius of the curve is approximately 56.5 meters.

Would you like further details on any of the steps or have other questions?

Related Questions:

1. How does the friction between the tires and road affect the car’s motion on a banked curve?
2. What would happen if the car were traveling faster than the required speed?
3. How does the angle of banking affect the required speed for a car to stay on the curve?
4. What other factors could affect the stability of the car on a banked curve?
5. How would the radius change if the speed of the car were doubled?

Tip:

When dealing with banked curves, understanding the relationship between speed, radius, and banking angle is crucial. The steeper the bank angle, the less friction is required to keep the car on the road.