/*! 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 997 A particle moves in a plane acco... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 39: Problem 997

A particle moves in a plane according to the parametric equations of motion: \(\mathrm{x}=-\mathrm{t}^{2}, \quad \mathrm{y}=\mathrm{t}^{3}\). Find the magnitude and direction of the acceleration when \(\mathrm{t}=(2 / 3)\).

Short Answer

Expert verified
The magnitude of the acceleration vector at t = \(2/3\) is \(|a| = \sqrt{20}\), and its direction can be found by the angle \(\theta = \arctan(-2)\).

Step by step solution

01

Find the velocity vector

To find the velocity vector, we need to differentiate the parametric equations of motion with respect to time: \[ \begin{aligned} \frac{dx}{dt} &= -2t, \\ \frac{dy}{dt} &= 3t^2. \end{aligned} \] So, the velocity vector \(\vec{v}\) as a function of time is: \[ \vec{v}(t) = \begin{bmatrix} -2t \\ 3t^2 \end{bmatrix}. \]
02

Find the acceleration vector

To find the acceleration vector, we need to differentiate the velocity vector with respect to time: \[ \begin{aligned} \frac{d(-2t)}{dt} &= -2, \\ \frac{d(3t^2)}{dt} &= 6t. \end{aligned} \] So, the acceleration vector \(\vec{a}\) as a function of time is: \[ \vec{a}(t) = \begin{bmatrix} -2 \\ 6t \end{bmatrix}. \]
03

Evaluate the acceleration vector at t = 2/3

To find the acceleration vector at the specific time t = 2/3, we just need to plug t = 2/3 into the acceleration vector: \[ \vec{a}\left(\frac{2}{3}\right) = \begin{bmatrix} -2 \\ 6\left(\frac{2}{3}\right) \end{bmatrix} = \begin{bmatrix} -2 \\ 4 \end{bmatrix}. \]
04

Compute the magnitude of the acceleration vector

The magnitude of the acceleration vector, denoted as |a|, can be calculated using the Euclidean norm: \[ |a| = \sqrt{(-2)^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20}. \]
05

Find the direction of the acceleration vector

To find the direction of the acceleration vector, we can compute the angle 胃 between the vector and the positive x-axis using the inverse tangent function: \[ \theta = \arctan{\left(\frac{y}{x}\right)} = \arctan{\frac{4}{-2}} = \arctan(-2). \] Now we have the magnitude \(|a| = \sqrt{20}\) and the direction \(\theta = \arctan(-2)\) of the acceleration vector when t = 2/3.

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 Vector
In the context of motion in a plane, a velocity vector is an essential concept. It tells you how the position of a particle changes over time. To find it, we differentiate the position equations. For the given parametric equations, the velocity vector \(\vec{v}(t)\) comes after differentiating each component with respect to time \(t\):
  • The change in \(x\): \(\frac{dx}{dt} = -2t\)
  • The change in \(y\): \(\frac{dy}{dt} = 3t^2\)
Thus, the velocity vector becomes:
  • \(\vec{v}(t) = \begin{bmatrix} -2t \ 3t^2 \end{bmatrix}\)
This vector provides the instantaneous rate of change of position in each direction. At any time \(t\), just plug in the value, and you'll know how quickly and in what direction the particle is moving.
Acceleration Vector
Like the velocity vector, the acceleration vector describes the change in velocity over time. It tells us how the speed and direction of the particle鈥檚 motion are changing. To find the acceleration vector \(\vec{a}(t)\), we differentiate the velocity components:
  • Acceleration in the \(x\)-direction: \(\frac{d(-2t)}{dt} = -2\)
  • Acceleration in the \(y\)-direction: \(\frac{d(3t^2)}{dt} = 6t\)
Hence, the acceleration vector is:
  • \(\vec{a}(t) = \begin{bmatrix} -2 \ 6t \end{bmatrix}\)
With this vector, you can determine the rate of change of the particle's velocity at any moment \(t\). Evaluate it at any specific time to understand how the velocity is changing at that instance.
Magnitude of Acceleration
The magnitude of acceleration gives an idea of how strong the acceleration is, regardless of its direction. It's a scalar value derived from the acceleration vector by using the Euclidean norm. For our acceleration vector \(\vec{a}(t) = \begin{bmatrix} -2 \ 6t \end{bmatrix}\):
  • Plug in \(t = \frac{2}{3}\): \(\vec{a}\left(\frac{2}{3}\right) = \begin{bmatrix} -2 \ 4 \end{bmatrix}\)
  • The formula used is \(\sqrt{(-2)^2 + 4^2}\)
This results in:
  • \(|a| = \sqrt{20}\)
This value tells us how much the velocity of the particle is changing at that time, showing the dynamic nature of the motion without giving direction.
Direction of Acceleration
While the magnitude tells us the strength of the acceleration, the direction indicates where it's headed. We find this by measuring the angle of the acceleration vector relative to the positive x-axis.

For an acceleration vector like \(\vec{a}\left(\frac{2}{3}\right) = \begin{bmatrix} -2 \ 4 \end{bmatrix}\):
  • Use: \(\theta = \arctan\left(\frac{4}{-2}\right)\)
  • This leads to \(\theta = \arctan(-2)\)
This angle tells us that the acceleration is moving in a direction relative to the x-axis. Understanding both magnitude and direction together paints a full picture of the particle's dynamic motion at \(t = \frac{2}{3}\).

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

A block of weight \(\mathrm{W}\) is to be moved along a flat table by a force inclined at an angle \(\theta\) with the line of motion, where \(0 \leq \theta \leq(1 / 2) \pi\), as shown in the Figure. Assume the motion is resisted by a frictional force which is proportional to the normal force with which the block presses perpendicularly against the surface of the table. Find the angle \(\theta\) for which the propelling force needed to overcome friction will be as small as possible.

A particle travels along the parabola: \(\mathrm{y}=(1 / 3) \mathrm{x}^{2}\), keeping its vertical velocity component constant at the value 8 . Find the magnitude of the resultant velocity when the particle is at the point \([2,(4 / 3)]\)

A particle moves along the right-hand part of the curve: \(4 y^{3}=x^{2}\), with a velocity \(v_{y}=(d y / d t)\), constant at \(2 .\) Find the speed and direction of motion at the point where \(\mathrm{y}=4\).

A vehicle moves along a trajectory having coordinates given by: \(\mathrm{x}=\mathrm{t}^{3}\), and: \(\mathrm{y}=1-\mathrm{t}^{2}\). The acceleration of the vehicle at any point on the trajectory is a vector, having magnitude and direction. Find the acceleration when \(\mathrm{t}=2\).

A particle starts at the origin and travels along the curve: \(\mathrm{y}=\mathrm{x}^{2}\). As it passes through the point \((3,9)\), its velocity is such that the \(\mathrm{x}\) -component, \(\mathrm{V}_{\mathrm{x}}=\mathrm{D}_{\mathrm{t}} \mathrm{x}\), is \(2 \mathrm{ft} / \mathrm{sec}\). Given that \(D_{x} y=\left[\left(D_{t} y\right) /\left(D_{t} x\right)\right]\), find the \(y\) -component: \(V_{y}=D_{t} y\), and the resultant speed \(\left.\mathrm{V}=\sqrt{(} \mathrm{V}^{2} \mathrm{x}+\mathrm{V}^{2}{ }_{y}\right)\). Use the \(\Delta-\) method.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

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.