Question

A car accelerates uniformly from rest and reaches a speed of \(22.0 \mathrm{~m} / \mathrm{s}\) in 9.00 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?

Short answer

Question: During a car's motion, a tire of diameter 58.0 cm starts from rest and accelerates uniformly, reaching a final speed of 22.0 m/s in 9.00 seconds. Calculate a) the number of revolutions that the tire made during the motion, and b) the final angular speed of the tire in revolutions per second. Answer: a) The tire makes approximately 54.3 revolutions. b) The final angular speed of the tire is approximately 6.03 revolutions per second.

Step-by-step solution

1. Calculate Distance Traveled

Using the kinematic equation for acceleration, we will find the distance traveled by the car. We know that the car starts from rest (initial velocity \(v_0 = 0\)) and accelerates uniformly, reaching a final velocity of \(22.0 \mathrm{~m/s}\) in \(9.00\,\mathrm{s}\). Let's call the acceleration \(a\) and the distance traveled \(d\). The kinematic equation we will use is: \(v^2 = v_0^2 + 2ad\). Since the car starts from rest, \(v_0 = 0\). We can rearrange the equation to solve for \(d\) as follows: \(d = \frac{v^2}{2a}\).

2. Find the Acceleration

To find the acceleration, we can use the equation: \(v = v_0 + at\). Since the car starts from rest, \(v_0 = 0\), so \(a = \frac{v}{t}\). Plugging in the given values, we get \(a = \frac{22.0 \mathrm{~m/s}}{9.00\,\mathrm{s}} = 2.\overline {44} \mathrm{~m/s^2}\).

3. Calculate Distance Traveled

Now we can use our equation for \(d\) and plug in the calculated acceleration, \(a = 2.\overline {44} \mathrm{~m/s^2}\), and the final velocity, \(v = 22.0 \mathrm{~m/s}\): \(d = \frac{(22.0 \mathrm{~m/s})^2}{2(2.\overline {44} \mathrm{~m/s^2})} \approx 99.0\,\mathrm{m}\).

4. Find the Circumference of the Tire

Given the diameter of the tire is \(58.0\,\mathrm{cm}\), we can find the circumference. Recall that the circumference of a circle is given by \(C = \pi D\), where \(D\) is the diameter. Converting the diameter from centimeters to meters, we have \(D = 0.580\,\mathrm{m}\). So, \(C = \pi(0.580\,\mathrm{m}) \approx 1.822\,\mathrm{m}\).

5. Calculate the Number of Revolutions

Now that we have the total distance traveled (\(99.0\,\mathrm{m}\)) and the circumference of the tire (\(1.822\,\mathrm{m}\)), we can find the number of revolutions. The number of revolutions, \(n\), can be found by dividing the total distance traveled by the circumference of the tire: \(n = \frac{d}{C} = \frac{99.0\,\mathrm{m}}{1.822\,\mathrm{m}} \approx 54.3\). The tire makes approximately 54.3 revolutions during the car's motion.

6. Calculate the Final Angular Speed

To find the final angular speed, we need to divide the number of revolutions (\(54.3\)) by the time it took to make those revolutions (\(9.00\,\mathrm{s}\)). Final angular speed, \(\omega_\mathrm{f}\), can be calculated as \(\omega_\mathrm{f} = \frac{n}{t} = \frac{54.3}{9.00\,\mathrm{s}} = 6.03\,\mathrm{rev/s}\). The final angular speed of the tire is approximately 6.03 revolutions per second. Answer: a) The tire makes approximately 54.3 revolutions. b) The final angular speed of the tire is approximately 6.03 revolutions per second.

Key concepts

Uniform Acceleration

Understanding uniform acceleration helps us to analyze motion scenarios where an object changes its velocity at a constant rate. When acceleration, denoted as \(a\), is uniform, it means that the object will speed up or slow down steadily over time.

For instance, in our exercise, the car accelerates uniformly from a standstill, meaning that the velocity increases by equal amounts in equal time intervals until it reaches \(22.0 \text{ m/s}\) in 9 seconds. This concept is critical as it defines the motion's nature and helps us predict future positions and velocities of the object using kinematic equations. Incorporating this into exercises and solutions, especially for students unfamiliar with physics, could encompass graphics illustrating the concept of uniform acceleration, or provide real-life examples, such as a car's cruise control maintaining a steady increase in speed.

Kinematic Equations

Kinematic equations allow us to solve problems involving objects in motion with uniform acceleration. These equations relate acceleration, distance, time, and velocity in various combinations to provide solutions to motion-related problems.

In the provided solution, kinematic equations are used to calculate the distance traveled and the acceleration of the car. For educational clarity, each equation can be introduced with a brief explanation of when and why it is used. To enhance understanding:

• Emphasize the conditions under which each equation is applicable, such as starting from rest or constant acceleration.
• Use visual aids to show the relationship between acceleration, velocity, and time graphically.
• Provide step-by-step examples using the equations to solve different types of problems, including ones with graphical representations of motion.

By doing this, students will gain a solid foundation for applying these equations to a variety of motion scenarios.

Revolutions of a Tire

The concept of tire revolutions is a practical application of kinematic principles and directly relates to angular motion. One revolution of a tire is a complete turn of the wheel, causing the point of contact to return to its starting position. This concept can be confusing for students, and hence, visual aids showing the tire rotations and a reference point can be incredibly helpful.

In the exercise, the number of revolutions is obtained by dividing the distance the car travels by the circumference of the tire. Illustrating how distance traveled translates into circular motion and consequently into revolutions can make the concept clearer. Furthermore, discussing the link between linear and angular speed could provide deeper insights:

• Explain that linear speed is along the path, whereas angular speed is how fast the angle changes.
• Discuss the relationship between the circumference (related to linear distance), revolutions (rotational cycles), and angular speed.

Using these explanations can improve understanding of how vertical motion is converted into circular motion for objects like tires.