Consider again the problem of a car traveling along a banked turn. Sometimes roads have a "reversed" banking angle. That is, the road is tilted "away" from the center of curvature of the road. If the coefficient of static friction between the tires and the road is ìs = 0.4, the radius of curvature is 22 m, and the banking angle is 12°, what is the maximum speed at which a car can safely navigate such a turn?

To find the maximum speed at which a car can safely navigate a turn with a reversed banking angle, we can use the equation for the maximum speed on a banked turn.

The equation for the maximum speed on a banked turn with a reversed banking angle is:

v = sqrt( g * r * tan(θ + arctan(μs)) )

Where:
v is the maximum speed
g is the acceleration due to gravity (approximately 9.8 m/s^2)
r is the radius of curvature
θ is the banking angle
μs is the coefficient of static friction

Plugging in the given values:
g = 9.8 m/s^2
r = 22 m
θ = -12° (reversed banking angle)
μs = 0.4

Converting the banking angle from degrees to radians:
θ = -12° * (π/180) ≈ -0.20944 radians

Now we can calculate the maximum speed:

v = sqrt( 9.8 * 22 * tan(-0.20944 + arctan(0.4)) )

Using the inverse tangent function:
v = sqrt( 9.8 * 22 * tan(-0.20944 + 0.38051) )

Calculating the tangent:
v = sqrt( 9.8 * 22 * tan(0.17107) )

v ≈ sqrt( 9.8 * 22 * 0.17379 )

v ≈ sqrt( 36.55612 )

v ≈ 6.05 m/s

Therefore, the maximum speed at which a car can safely navigate a turn with a reversed banking angle is approximately 6.05 m/s.

To find the maximum speed at which a car can safely navigate a turn with a "reversed" banking angle, we need to consider the forces acting on the car.

In this case, the forces involved are the gravitational force (mg), the normal force (N), and the frictional force (f). The normal force can be broken down into two components: one perpendicular to the surface (N⊥) and one parallel to the surface (N∥).

First, let's find the components of the normal force. The perpendicular component (N⊥) counteracts the gravitational force, so we have:

N⊥ = mg

The parallel component (N∥) provides the necessary centripetal force to keep the car moving in a circle. It can be found using trigonometry:

N∥ = N * sin(θ)

where θ is the angle of the banking.

Now, we need to find the maximum frictional force that can be exerted by the tires. This can be found using the coefficient of static friction (μs):

f(max) = μs * N∥

Finally, we can equate the maximum frictional force to the centripetal force:

f(max) = (mv²) / r

where m is the mass of the car, v is the velocity, and r is the radius of curvature.

Now, we can solve for the maximum velocity (v):

μs * N * sin(θ) = (m * v²) / r

Rearranging the equation, we get:

v² = r * μs * N * sin(θ) / m

Plugging in the given values, with the mass of the car (m) canceled out, we have:

v² = 22 m * 0.4 * N * sin(12°)

To proceed further, we need to find the value of the normal force (N). In this case, the normal force can be found by balancing the forces in the vertical direction:

N = mg - N⊥

Substituting the value of N⊥ from earlier and plugging in the value of mass (m) and acceleration due to gravity (g), we have:

N = m * g - m * g

Simplifying, we get:

N = 0

This means that there is no normal force acting on the car in the vertical direction. In other words, the car is experiencing weightlessness on the banked curve.

Therefore, the maximum speed at which a car can safely navigate such a turn with a reversed banking angle is infinite.