/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 The velocity of a particle trave... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 20

The velocity of a particle traveling along a straight line is \(v=\left(3 t^{2}-6 t\right) \mathrm{ft} / \mathrm{s},\) where \(t\) is in seconds. If \(s=4 \mathrm{ft}\) when \(t=0,\) determine the position of the particle when \(t=4\) s. What is the total distance traveled during the time interval \(t=0\) to \(t=4 \mathrm{s} ?\) Also, what is the acceleration when \(t=2 \mathrm{s} ?\)

Short Answer

Expert verified
The position of the particle at time t=4s is 0 ft, the total distance travelled from the initial time t=0 to t=4 seconds is also 0 ft (due to no change in position), and the acceleration of the particle when t=2s is 6 ft/s^2.

Step by step solution

01

Finding the position at t=4s

The position, or displacement, from the initial point is found by integrating the velocity equation. Thus, the integration of \(3t^2 - 6t\) leads to \(t^3 - 3t^2 + C\) where \(C\) is the constant of integration. Since we have the condition s=4 when t=0, we substitute these values into the equation to find \(C=4\). Therefore, the equation of the motion becomes \(s = t^3 - 3t^2 + 4\). Now we find the position at time t=4 by substituting this value to the equation, leading to \(s = 4^3 - 3*4^2 + 4 = 0\)
02

Calculation of total distance traveled

The total distance traveled is the absolute sum of distances in all movements, regardless of direction. Before we proceed, we need to find where the particle changes direction by setting the velocity equation (v=\(3t^2-6t\)) to zero. By solving the equation, we find \(t=0\) and \(t=2\) which are the turning points of the particle. When \(t=0\) to \(t=2\), the displacement is \(s(2)-s(0)=0\), and when \(t=2\) to \(t=4\), the displacement is \(s(4)-s(2)=0\). Since we have no change in position during either interval, the total distance is 0.
03

Finding acceleration at t=2s

Acceleration is obtained by differentiating the velocity equation. Taking the first derivative of \(3t^2 - 6t\), we get \(6t-6\). Substituting the value t=2 into the equation, we get acceleration = \(6*2 - 6 = 6 ft/s^2\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Velocity of a Particle
The velocity of a particle is a vector quantity that tells us how fast and in which direction the particle is moving. In this problem, the particle travels along a straight line. Its velocity is given by the function \( v = 3t^2 - 6t \), where \( t \) is time in seconds. By definition, velocity is the derivative of the position function with respect to time. This means that the particle's velocity changes depending on the value of \( t \), which impacts the particle's overall motion.
  • Importance of Velocity: It provides insight into how quickly the particle's position is changing over time.
  • Positive vs Negative Velocity: Positive velocity indicates motion in one direction, while negative velocity indicates motion in the opposite direction.
To find when the particle changes direction, we set the velocity equation to zero: \( 3t^2 - 6t = 0 \). Solving this gives us the critical time points \( t = 0 \) and \( t = 2 \), which are moments when the particle's speed transitions between moving forward and backward.
Thus, understanding velocity is key to analyzing how the particle's position changes over time and predicting its movements.
Acceleration of a Particle
Acceleration represents the rate of change of velocity and is similarly a vector quantity. In this context, acceleration is the derivative of the velocity function. Let's derive that from our velocity equation \( v = 3t^2 - 6t \), where, upon differentiation, we get the acceleration function \( a(t) = 6t - 6 \).
  • Understanding Acceleration: It indicates how fast the velocity of the particle is changing. Positive acceleration means the speed of the particle is increasing, whereas negative acceleration suggests it is slowing down.
  • Instantaneous Acceleration: When \( t = 2 \), substituting into \( a(t) \) gives us the acceleration as \( 6 \times 2 - 6 = 6 \) ft/s虏. This means that at t=2 seconds, the particle accelerates at \( 6 \text{ ft/s}^2 \).
Acceleration helps us understand how the forces exerted on the particle affect its velocity over time, making it essential for a comprehensive analysis of particle motion.
Position of a Particle
Position represents the specific location of a particle at a given time. In this problem, we determine the particle's position by integrating the velocity function \( v = 3t^2 - 6t \). Integration gives \( s(t) = t^3 - 3t^2 + C \), where \( C \) is a constant derived from initial conditions. Given that the initial position is 4 ft when \( t = 0 \), we find \( C = 4 \). Thus, the position function becomes \( s(t) = t^3 - 3t^2 + 4 \).
  • Calculating Position: To find the position at \( t = 4 \) s, substitute \( t \) into the equation: \( s(4) = 4^3 - 3 \times 4^2 + 4 = 0 \).
  • Distance Traveled: Though the computed position at different intervals indicates no net movement, it's crucial to understand that displacement is different from total distance. Displacement considers direction, causing the net change to appear as zero in specific scenarios like this.
Hence, the position function provides the exact location of a particle at various points in time, essential for mapping its journey along its path.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An automobile is traveling on a curve having a radius of \(800 \mathrm{ft}\). If the acceleration of the automobile is \(5 \mathrm{ft} / \mathrm{s}^{2},\) determine the constant speed at which the automobile is traveling.

A man riding upward in a freight elevator accidentally drops a package off the elevator when it is \(100 \mathrm{ft}\) from the ground. If the elevator maintains a constant upward speed of \(4 \mathrm{ft} / \mathrm{s}\), determine how high the elevator is from the ground the instant the package hits the ground. Draw the \(v\) -t curve for the package during the time it is in motion. Assume that the package was released with the same upward speed as the elevator.

The double collar \(C\) is pin connected together such that one collar slides over the fixed rod and the other slides over the rotating rod \(A B .\) If the angular velocity of \(A B\) is given as \(\dot{\theta}=\left(e^{0.5 t^{2}}\right)\) rad \(/ \mathrm{s},\) where \(t\) is in seconds, and the path defined by the fixed rod is \(r=|(0.4 \sin \theta+0.2)| \mathrm{m},\) determine the radial and transverse components of the collar's velocity and acceleration when \(t=1\) s. When \(t=0, \theta=0 .\) Use Simpson's rule with \(n=50\) to determine \(\theta\) at \(t=1\) s.

For a short time the arm of the robot is extending such that \(\quad \dot{r}=1.5 \mathrm{ft} / \mathrm{s}\) when \(r=3 \mathrm{ft}, z=\left(4 t^{2}\right) \mathrm{ft},\) and \(\theta=0.5 t \mathrm{rad},\) where \(t\) is in seconds. Determine the magnitudes of the velocity and acceleration of the grip \(A\) when \(t=3 \mathrm{s}\).

A particle is moving along a circular path having a radius of 6 in. such that its position as a function of time is given by \(\theta=\sin 3 t,\) where \(\theta\) and the argument for the sine are in radians, and \(t\) is in seconds. Determine the magnitude of the acceleration of the particle at \(\theta=30^{\circ} .\) The particle starts from rest at \(\theta=0^{\circ}\).
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Engineering Physics

Read Explanation

Radiation

Read Explanation

Work Energy and Power

Read Explanation

Translational Dynamics

Read Explanation

Wave Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.