Final answer:
The fastest speed a car could navigate a 30° banked turn with a 400 m radius without friction is 39 m/s, calculated using principles of centripetal force and the banking angle.
Step-by-step explanation:
A student has asked the question: What is the fastest a car could go around a corner banked at 30° above the horizontal on a race track with a radius of 400 m with no assistance from friction? To solve this, we use physics principles related to circular motion and banking angles. The centripetal force required to turn the car is provided entirely by the horizontal component of the normal force when friction is absent.
Using the formula for centripetal force, \( F_c = \frac{mv^2}{r} \) where m is the mass of the car, v is the velocity, and r is the radius of the turn, we can equate this to the horizontal component of the normal force acting on the car, which is \( N \cdot sin(\theta) \) where \( \theta \) is the banking angle. Given that \( \theta = 30° \) and \( r = 400 m \), and considering that gravity, \( g \), is the only force acting vertically, we have the following relationship: \( mg = N \cdot cos(\theta) \). Combining these equations and solving for v gives us \( v = \sqrt{rg \cdot tan(\theta)} \).
Plugging in the values, \( g = 9.8 m/s^2 \), \( r = 400 m \), and \( \theta = 30° \), we calculate the velocity v. This simplifies to \( v = \sqrt{400 m \cdot 9.8 m/s^2 \cdot tan(30°)} \), which calculates to approximately 39 m/s. Therefore, 39 m/s is the fastest a car could go around this corner with no assistance from friction.