Final answer:
The maximum speed a car can negotiate a turn on a wet road without sliding out of control is approximately 8.52 m/s.
Step-by-step explanation:
To find the maximum speed at which a car can negotiate a turn on a wet road without sliding out of control, we need to consider the coefficient of static friction and the radius of the turn. The formula we can use is v = sqrt(μs * g * r), where v is the maximum speed, μs is the coefficient of static friction, g is the acceleration due to gravity (approximately 9.8 m/s^2), and r is the radius of the turn.
Plugging in the given values, we get v = sqrt(0.205 * 9.8 * 22.5) = 8.52 m/s. Therefore, the maximum speed at which the car can negotiate the turn without sliding out of control is approximately 8.52 m/s.
The maximum speed at which a car can negotiate a turn on a wet road with a coefficient of static friction of 0.205, without sliding out of control, can be calculated using the formula for centripetal force, which is Fc = mv2/r, where m is the mass of the car, v is the velocity, and r is the radius of the turn. Since the force of static friction is the force providing the centripetal force needed to make the turn, and is equal to the coefficient of static friction multiplied by the normal force (μsN), and given that the normal force is equal to the weight of the car (mg), we set up the equation μsmg = mv2/r. Mass (m) cancels out from both sides, leading to μsg = v2/r, and upon rearranging for v, we get v = √(μsgr). Plugging in the numbers, v = √(0.205*9.8*22.5), we can calculate the maximum speed