/*! 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 24 The acceleration of a particle i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 24

The acceleration of a particle is defined by the relation \(a=-k \sqrt{v}\) where \(k\) is a constant. Knowing that \(x=0\) and \(v=81 \mathrm{m} / \mathrm{s}\) at \(t=0\) and that \(v=36 \mathrm{m} / \mathrm{s}\) when \(x=18 \mathrm{m}\), determine \((a)\) the velocity of the particle when \(x=20 \mathrm{m},(b)\) the time required for the particle to come to rest.

Short Answer

Expert verified
The velocity at \( x = 20 \) m is 27 m/s, and it takes 9 seconds to stop.

Step by step solution

01

Understand the Relation

The acceleration is defined by the relation \( a = -k \sqrt{v} \). We need to relate acceleration \( a \), velocity \( v \), and position \( x \). Integrating with respect to position will allow us to find a relationship between \( v \) and \( x \).
02

Express Acceleration as a Function of Velocity and Position

We use the fact that \( a = \frac{dv}{dt} \) and \( a = v \frac{dv}{dx} \), so \( v \frac{dv}{dx} = -k \sqrt{v} \). By separating variables, we can write \( v^{1/2} dv = -k dx \).
03

Integrate the Separated Equation

Integrate both sides: \( \int v^{1/2} dv = -k \int dx \). The left side integrates to \( \frac{2}{3}v^{3/2} \) and the right side integrates to \( -kx + C \). This yields: \( \frac{2}{3}v^{3/2} = -kx + C \).
04

Determine the Constant C

By substituting the initial conditions \( x = 0 \), \( v = 81 \), we can solve for \( C \). So, \( \frac{2}{3} \cdot 81^{3/2} = C \). Calculating this gives \( C = 486 \).
05

Solve for Velocity at a Given Position

We now use the equation \( \frac{2}{3}v^{3/2} = -kx + C \) to find the velocity when \( x = 20 \). From step 4, \( C = 486 \). With the condition \( x = 18 \), \( v = 36 \), find \( k \) first. \( \frac{2}{3} imes 36^{3/2} = -18k + 486 \), solve to find \( k = 9 \). Then, for \( x = 20 \), solve \( \frac{2}{3}v^{3/2} = -9 \times 20 + 486 \) to find \( v \).
06

Calculate Final Result for Velocity

With \( C = 486 \), \( k = 9 \), for \( x = 20 \), \( -180 + 486 = \frac{2}{3}v^{3/2} \) gives \( \frac{2}{3}v^{3/2} = 306 \). Solving for \( v \), we get \( v = 27 \text{ m/s} \).
07

Determine Time Required for Particle to Stop

To find the time when \( v = 0 \), use \( v = 81 - kt \), where \( a = \frac{dv}{dt} = -k \sqrt{v} \) and initial velocity is \( 81 \). At rest, \( v = 0 \), thus \( 0 = 81 - kt' \). Using the value of \( k = 9 \), \( t' = \frac{81}{9} = 9 \text{ s} \).

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.

Understanding Acceleration
Acceleration tells us how quickly the velocity of a particle changes over time. It is a rate of change, calculated using calculus. In this exercise, acceleration is given by the formula \( a = -k \sqrt{v} \), where \( k \) is a constant and \( v \) is velocity. This means the acceleration depends on the current velocity of the particle. The negative sign implies that the acceleration decreases the particle's speed as it moves, meaning the particle is slowing down.
  • *Acceleration* can be positive or negative, indicating speeding up or slowing down, respectively.
  • The magnitude of acceleration affects how quickly these changes occur.
  • In calculus terms, acceleration is the derivative of velocity with respect to time.
By understanding these points, we can derive relationships between acceleration and other motion attributes, like velocity and position.
Exploring Velocity
Velocity is a vector quantity that denotes the rate at which an object changes its position. In simpler terms, it tells you how fast something is moving and in which direction. The velocity of the particle changes as dictated by its acceleration. Initially, the particle is given a velocity \( v = 81 \text{ m/s} \). As it moves, this velocity alters due to the acceleration \( a = -k \sqrt{v} \).
  • *Velocity* is often confused with speed, but unlike speed, it includes direction.
  • High initial velocity results in significant initial movement, which decreases over time if negatively accelerated.
  • The velocity can be derived from integrating acceleration with respect to time.
Understanding how velocity and acceleration interplay helps predict the motion of particles under various conditions.
Insight into Kinematics
Kinematics focuses on describing the motion of objects without delving into the causes of this motion. It uses concepts like displacement, velocity, and acceleration to describe how a particle moves along a path. In this problem, we are given specific kinematic conditions: the particle starts at rest, with its position \( x=0 \) and velocity \( v=81 \text{ m/s} \) at time \( t=0 \). By applying kinematic equations, we can solve for velocity at any position, as well as the time required for a particle to come to rest.
  • By relating acceleration, velocity, and displacement, kinematics provides a comprehensive picture of motion.
  • Kinematic equations are fundamental for solving problems involving motion, such as finding the velocity at a specific point.
  • These equations do not account for the forces causing the motion, focusing purely on the trajectory.
With the given relations, we can calculate how long the particle takes to stop and its velocity at specific positions like \( x = 20 \).
Application of Calculus
Calculus is essential when dealing with kinematics, as it allows us to find precise relationships between various motion-related concepts. Integration and differentiation are powerful tools used here to derive velocity from acceleration and displacement from velocity.
  • *Differentiation* enables us to find instantaneous rates of change, like acceleration from velocity.
  • *Integration* is used to determine accumulative quantities, like velocity from acceleration.
  • Understanding the limits for these operations is crucial for their correct application in physics problems.
In the solution, integrating the separated equation \( v^{1/2} dv = -k dx \) helps find a relationship between velocity and position. By substituting initial conditions, the constants can be calculated to derive specific outcomes for the problem. Calculus therefore bridges the gap between theoretical motion equations and practical applications for predicting movement.

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

The motion of a vibrating particle is defined by the position vector \(\mathbf{r}=10\left(1-e^{-3 t}\right) \mathbf{i}+\left(4 e^{-2 t} \sin 15 t\right) \mathbf{j}\), where \(\mathbf{r}\) and \(t\) are expressed in millimeters and seconds, respectively. Determine the velocity and acceleration when \((a) t=0,(b) t=0.5 \mathrm{s}\).

An airplane begins its take-off run at \(A\) with zero velocity and a constant acceleration \(a\). Knowing that it becomes airborne 30 s later at \(B\) and that the distance \(A B\) is \(900 \mathrm{m}\), determine ( \(a\) ) the acceleration \(a\) (b) the take-off velocity \(v_{B}\).

In a water-tank test involving the launching of a small model boat, the model's initial horizontal velocity is \(6 \mathrm{m} / \mathrm{s}\) and its horizontal acceleration varies linearly from \(-12 \mathrm{m} / \mathrm{s}^{2}\) at \(t=0\) to \(-2 \mathrm{m} / \mathrm{s}^{2}\) at \(t=t_{1}\) and then remains equal to \(-2 \mathrm{m} / \mathrm{s}^{2}\) until \(t=1.4 \mathrm{s}\). Knowing that \(v=1.8 \mathrm{m} / \mathrm{s}\) when \(t=t_{1}\), determine \((a)\) the value of \(t_{1},(b)\) the velocity and the position of the model at \(t=1.4 \mathrm{s}\).

Conveyor belt \(A,\) which forms a \(20^{\circ}\) angle with the horizontal, moves at a constant speed of \(4 \mathrm{ft} / \mathrm{s}\) and is used to load an airplane. Knowing that a worker tosses duffel bag \(B\) with an initial velocity of \(2.5 \mathrm{ft} / \mathrm{s}\) at an angle of \(30^{\circ}\) with the horizontal, determine the velocity of the bag relative to the belt as it lands on the belt.

The acceleration due to gravity of a particle falling toward the earth is \(a=-g R^{2} / r^{2},\) where \(r\) is the distance from the center of the earth to the particle, \(R\) is the radius of the earth, and \(g\) is the acceleration due to gravity at the surface of the earth. If \(R=3960 \mathrm{mi}\), calculate the escape velocity, that is, the minimum velocity with which a particle must be projected vertically upward from the surface of the earth if it is not to return to the earth. (Hint: \(v=0 \text { for } r=\infty .)\)
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Dynamics

Read Explanation

Space Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Translational Dynamics

Read Explanation

Force

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.