A car accelerates uniformly from rest and reaches a speed of \(22.0 \mathrm{~m} / \mathrm{s}\) in \(9.00 \mathrm{~s}\). The diameter of a tire on this car is \(58.0 \mathrm{~cm}\). a) Find the number of revolutions the tire makes during the car's motion, assuming that no slipping occurs. b) What is the final angular speed of a tire in revolutions per second?

Angular speed is the rate at which an object rotates or spins around an axis. In the context of a tire on a moving vehicle, it refers to how fast the tire is spinning. The angular speed is typically measured in radians per second (rad/s) or revolutions per second (rev/s).

To convert to a more intuitive unit like rev/s, we use the fact that one full revolution is equal to an angle of 2π radians. Therefore, if you find the angular speed in rad/s, you would divide by 2π to obtain the angular speed in rev/s. This conversion is crucial when comparing the rotational speed of tires to the vehicle's linear speed (measured in m/s or km/h).

In the exercise where a final angular speed is asked, the student must understand the conversion between radians and revolutions to interpret the tire's performance accurately.

Linear acceleration refers to the change in linear velocity of an object per unit time, moving along a straight path. This is directly related to the force applied according to Newton's second law of motion and is usually measured in meters per second squared (m/s²).

In the exercise provided, to determine the linear acceleration of the car, the formula \(v = u + at\) was used, where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time taken. Knowing the linear acceleration allows us to understand how quickly the car is speeding up and provides the information needed to calculate other values like tire revolutions and angular speed.

An understanding of acceleration is fundamental since it directly affects the angular acceleration of the tire, leading to a greater angular speed over time, given a constant linear acceleration.

The term 'tire revolutions' refers to the number of full turns a tire makes over a certain period or distance. It's an essential measure for understanding how far a vehicle has travelled since each turn of the tire corresponds to a specific distance – the tire's circumference. It's a useful conversion factor in mechanics and kinematics.

To calculate the tire revolutions from the angular distance in radians, as seen in the exercise, you divide the angular distance by \(2\text{π}\) since there are \(2\text{π}\) radians in one revolution. This calculation provides the total number of complete rotations a wheel makes, helping understand vehicle speed and distance covered, as long as there's no slipping between the tire and the road.

Angular distance measures how much an object has rotated about a point or axis. It is usually represented in degrees or radians. For a wheel, it indicates how far it has turned over a particular linear distance.

In kinematic problems like the exercise, the linear distance covered by the car (calculated from linear acceleration and time) can be used along with the radius of the tire to find the angular distance the tire has covered. The formula \(\theta (in \text{ radians}) = \frac{s}{r}\) converges the linear progress of the car into the rotational progress of its tires given that the tire's radius remains constant. The knowledge of angular distance is pivotal for connecting linear and rotational kinematics in objects like wheels.