91Ó°ÊÓ

Uniform acceleration is a key concept in physics where an object increases its velocity at a constant rate over a period of time. This means that the speed of the object changes by the same amount every second. In the context of the exercise, the car starts from rest, which means its initial velocity is zero. It then reaches a speed of 22.0 m/s in 9.00 seconds. To find the acceleration, we use the formula: \[ a = \frac{\Delta v}{\Delta t} \]Where:

• \( \Delta v \) is the change in speed (final velocity - initial velocity)
• \( \Delta t \) is the time interval

For this problem:

• Initial velocity \( v_{initial} = 0 \) m/s
• Final velocity \( v_{final} = 22.0 \) m/s
• Time \( t = 9.0 \) s

Let's calculate the acceleration by substituting: \[ a = \frac{22.0 \, \text{m/s} - 0 \, \text{m/s}}{9.0 \, \text{s}} = \frac{22.0}{9.0} \, \text{m/s}^2 = 2.44 \, \text{m/s}^2 \]This uniform acceleration means the car's speed increases by 2.44 m/s every second over the 9-second interval.

The tire circumference is the distance around the outer edge of the tire. It is crucial for calculating the number of revolutions a tire makes, as it links the linear distance traveled by the car to the rotations of the tires.Circumference is determined by the tire's diameter. The formula for circumference \( C \) of a circle is:\[ C = 2\pi r \]Here, the diameter of the tire is given as 58.0 cm, so to find the radius \( r \), we use:\[ r = \frac{\text{diameter}}{2} = \frac{58.0 \, \text{cm}}{2} = 29.0 \, \text{cm} = 0.29 \, \text{m} \]Thus, the circumference is calculated as:\[ C = 2\pi \times 0.29 \, \text{m} = 1.82 \, \text{m} \]This means that for each complete revolution, the tire travels 1.82 meters. By dividing the total distance traveled by this circumference, we find out how many times the tire revolves during the movement.

Angular speed refers to how quickly an object rotates or spins around an axis. In this exercise, we are interested in the angular speed of the tires of the car in revolutions per second.Initially, we know the total distance covered by the car, which has helped us determine the number of revolutions:A complete revolution corresponds to a change of angle of \( 2\pi \) radians. The problem already calculates the change in angle \( \Delta \theta \) for 54 revolutions:\[ \Delta \theta = 54 \times 2\pi = 108\pi \, \text{rad} \]The final angular speed \( \omega \) is calculated as:\[ \omega = \frac{\Delta \theta}{\Delta t} = \frac{108\pi \, \text{rad}}{9 \, \text{s}} = 12\pi \, \text{rad/s} \]To find \( \omega \) in revolutions per second (rev/s), convert from radians per second by using the fact that one complete revolution is \( 2\pi \) radians:\[ \omega = \frac{12\pi \, \text{rad/s}}{2\pi \, \text{rad/rev}} = 6 \, \text{rev/s} \]Therefore, the tires rotate at a final angular speed of 6 revolutions per second by the end of the acceleration period.