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A car is traveling with a speed of \(20.0 \mathrm{m} / \mathrm{s}\) along a straight horizontal road. The wheels have a radius of \(0.300 \mathrm{m}\). If the car speeds up with a linear acceleration of \(1.50 \mathrm{m} / \mathrm{s}^{2}\) for \(8.00 \mathrm{s}\), find the angular displacement of each wheel during this period.

Short Answer

Expert verified
The angular displacement is approximately 693.33 radians.

Step by step solution

01

Identify the Initial Conditions

First, identify the given data from the problem. The car's initial linear speed, \(v_0\), is \(20.0 \mathrm{m/s}\). The radius of each wheel, \(r\), is \(0.300 \mathrm{m}\). The linear acceleration, \(a\), is \(1.50 \mathrm{m/s}^2\), and the time interval, \(t\), is \(8.00 \mathrm{s}\).
02

Calculate the Final Linear Speed

Using the formula for linear velocity with constant acceleration, \(v = v_0 + at\), calculate the final speed. Substitute the given values: \[ v = 20.0 + (1.50)(8.00) = 32.0 \mathrm{m/s} \] Thus, the final linear speed is \(32.0 \mathrm{m/s}\).
03

Find the Average Linear Speed

The average linear speed, \(v_{avg}\), over the time period is the average of the initial and final speeds:\[ v_{avg} = \frac{v_0 + v}{2} = \frac{20.0 + 32.0}{2} = 26.0 \mathrm{m/s} \] This is the average speed during acceleration.
04

Determine Linear Displacement

The linear displacement, \(d\), covered by the car in time \(t\) is given by the formula: \[ d = v_{avg} \times t = 26.0 \times 8.00 = 208.0 \mathrm{m} \] The car covers a distance of \(208.0 \mathrm{m}\).
05

Convert Linear Displacement to Angular Displacement

The relationship between linear displacement \(d\) and angular displacement \(\theta\) is given by \(\theta = \frac{d}{r}\), where \(r = 0.300 \mathrm{m}\). Substitute our values in: \[ \theta = \frac{208.0}{0.300} \approx 693.33 \mathrm{rad} \] This is the angular displacement of each wheel.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Acceleration
Linear acceleration is how quickly the velocity of an object increases or decreases over time. In simple terms, it's the rate of change of speed along a straight path. Imagine pressing down on the accelerator pedal of a car; the car begins to speed up, and that's linear acceleration in action.

The formula for linear acceleration is given by:
  • \(a = \frac{\Delta v}{t}\)
Here, \(a\) is the linear acceleration, \(\Delta v\) is the change in velocity, and \(t\) is the time over which the acceleration happens.

Understanding linear acceleration helps in solving problems where you need to calculate how fast an object will be moving after a certain time if it starts from rest, or with a given initial speed. It's essential in a wide variety of practical applications, from car safety testing to roller coaster design.
Linear Velocity
Linear velocity defines how fast an object travels along a straight path in a specific direction. It's a vector quantity, meaning it has both magnitude and direction, like going 20 meters per second eastward.

You can think of linear velocity as the speedometer reading in a car, but with a particular direction. It's a straight-line speed. Linear velocity is crucial in understanding how objects move in the world around us, from vehicles on roads to balls being thrown.

The standard formula for linear velocity when dealing with constant acceleration is:
  • \(v = v_0 + at\)
Where \(v\) is the final velocity, \(v_0\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time elapsed. This formula provides a simple way to calculate how fast an object will be moving after some time of acceleration.
Angular Velocity
Angular velocity refers to how quickly an object rotates or revolves around a point or axis. It's measured in radians per second, providing a sense of how fast something is spinning.

Unlike linear velocity, which measures movement along a straight path, angular velocity is all about circular motion. Think of a revolving door or the hands of a clock; these are moving in circles, showing angular motion.
  • The relationship between linear and angular velocity is given by the formula:
  • \(v = r \omega\)
Here, \(v\) is the linear velocity, \(r\) is the radius of the circle, and \(\omega\) (omega) is the angular velocity.

Understanding angular velocity is key when dealing with objects that rotate, such as wheels, gears, and celestial bodies. It helps to know how fast something is spinning, which is important for both practical and theoretical problems.
Kinematics
Kinematics is a branch of physics that studies motion without considering the forces that cause it. It focuses on the trajectories, velocities, and accelerations of objects.

In the problem of determining the angular displacement of a car's wheels, kinematics provides the tools necessary to relate linear speed and acceleration to angular quantities.

The important equations in kinematics often include:
  • \(v = v_0 + at\)
  • \(s = v_0 t + \frac{1}{2} a t^2\)
  • \(v^2 = v_0^2 + 2as\)
These formulas allow us to calculate how objects move in a straight line over time, whereas angular kinematics might deal with circular paths.

Understanding kinematics is pivotal for anyone hoping to dive deeper into the physics of motion, making it an essential topic for students and researchers alike. Whether analyzing the motion of a falling apple or a speeding car, kinematics provides a foundational understanding of how things move.

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Most popular questions from this chapter

A top is a toy that is made to spin on its pointed end by pulling on a string wrapped around the body of the top. The string has a length of \(64 \mathrm{cm}\) and is wound around the top at a spot where its radius is \(2.0 \mathrm{cm}\). The thickness of the string is negligible. The top is initially at rest. Someone pulls the free end of the string, thereby unwinding it and giving the top an angular acceleration of \(+12 \mathrm{rad} / \mathrm{s}^{2} .\) What is the final angular velocity of the top when the string is completely unwound?

Our sun rotates in a circular orbit about the center of the Milky Way galaxy. The radius of the orbit is \(2.2 \times 10^{20} \mathrm{m},\) and the angular speed of the sun is \(1.1 \times 10^{-15} \mathrm{rad} / \mathrm{s}\). How long (in years) does it take for the sun to make one revolution around the center?

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of \(1.05 \times 10^{4} \mathrm{rad} / \mathrm{s}\) to an angular speed of \(3.14 \times 10^{4} \mathrm{rad} / \mathrm{s} .\) In the process, the bit turns through \(1.88 \times 10^{4} \mathrm{rad}\) Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of \(7.85 \times 10^{4} \mathrm{rad} / \mathrm{s},\) starting from rest?

The earth orbits the sun once a year \(\left(3.16 \times 10^{7} \mathrm{s}\right)\) in a nearly circular orbit of radius \(1.50 \times 10^{11} \mathrm{m} .\) With respect to the sun, determine (a) the angular speed of the earth, (b) the tangential speed of the earth, and (c) the magnitude and direction of the earth's centripetal acceleration.

A bicycle is rolling down a circular portion of a path; this portion of the path has a radius of \(9.00 \mathrm{m}\). As the drawing illustrates, the angular displacement of the bicycle is 0.960 rad. What is the angle (in radians) through which each bicycle wheel (radius \(=0.400 \mathrm{m}\) ) rotates?

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