/*! 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 23 A particle is moving along a str... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A particle is moving along a straight line such that its acceleration is defined as \(a=(-2 v) \mathrm{m} / \mathrm{s}^{2}\), where \(v\) is in meters per second. If \(v=20 \mathrm{~m} / \mathrm{s}\) when \(s=0\) and \(t=0\), determine the particle's position, velocity, and acceleration as functions of time.

Short Answer

Expert verified
The particle's acceleration as a function of time is \(a(t) = -40e^{-2t}\ m/s²\), its velocity as a function of time is \(v(t) = 20e^{-2t}\ m/s\), and its position as a function of time is \(s(t) = -10e^{-2t} + 10\ m\).

Step by step solution

01

Solve for Velocity

Firstly, we have to solve the differential equation for the velocity. The equation for acceleration is given as \(a=-2v\). We can rewrite it to express it as a function of velocity. This gives us: \(\frac{dv}{dt} = -2v\). To solve this differential equation, we can separate the variables on either side, which gives us: \(\frac{dv}{v} = -2dt\). Integrate the both sides: \(\int \frac{dv}{v} = \int -2 \, dt\). The integral of \(\frac{1}{v}\) is \(ln|v|\) and the integral of \(dt\) is simply \(t\). Therefore: \(ln|v| = -2t + ln|C|\). Solve for \(v\) by taking the exponent of both sides: \(v(t) = Ce^{-2t}\).
02

Find the Initial Condition for Velocity

The problem provides that initially when \(t = 0\) , \(v = 20\ m/s\). Substitute \(t = 0\) and \(v = 20\) into the equation \(v(t) = Ce^{-2t}\) to get the constant of integration \(C\). Therefore: \(C = 20\ m/s\). Substituting the value of \(C\) in the velocity equation we get: \(v(t) = 20e^{-2t}\ m/s\).
03

Solve for Position

To get the position equation we will have to integrate the velocity equation again. The equation for velocity is \(v(t) = 20e^{-2t}\ m/s\). Integrate it to get the position function. The integral of \(e^{-2t}\) will be \(\frac{-1}{2}e^{-2t}\). Therefore, the position function will be: \(s(t) = -10e^{-2t} + D\).
04

Find the Initial Condition for Position

The problem states that initially when \(t = 0\) , \(s = 0\). Substitute \(t = 0\) and \(s = 0\) into the equation \(s(t) = -10e^{-2t} + D\) to get the constant of integration \(D\). Therefore: \(D = 10\ m\). Substituting the value of \(D\) in the position equation we get: \(s(t) = -10e^{-2t} + 10\ m\).
05

Write out the final Result

Finally, we can write out the final result. The acceleration function is \(a(t) = -2v(t) = -2(20e^{-2t}) = -40e^{-2t}\ m/s²\). The velocity function is \(v(t) = 20e^{-2t}\ m/s\). The position function is \(s(t) = -10e^{-2t} + 10\ m\).

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 91Ó°ÊÓ!

Key Concepts

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

Acceleration-Velocity Relationship
Understanding the relationship between acceleration and velocity is fundamental in analyzing particle motion. In physics, acceleration is defined as the rate of change of velocity over time. Mathematically, this relationship is denoted as:
\[ a = \frac{dv}{dt} \]
In the given problem, the acceleration is expressed in terms of the velocity itself:\(a = -2v\). This negative sign indicates that the acceleration is in the opposite direction of the velocity, implying a deceleration, or a slowing down of the particle as it moves. Integration is then used to find velocity as a function of time from this differential equation, which is necessary for tracking how the particle's speed changes as time progresses.
Differential Equations in Physics
Differential equations play a critical role in physics as they describe how physical quantities change over time or space. Here's a glance at how they are used:
  • Differential equations can represent physical phenomena, such as motion, heat conduction, or wave propagation.
  • In the context of kinematics, they often relate various motion attributes like position, velocity, and acceleration.
  • Solving differential equations may require integration, particularly for obtaining a function from its derivative, as seen with velocity and position in the exercise.
These equations can be simple or complex, depending on the forces involved in a system. The challenge in physics is often not just to set up the correct equation but to solve it for the concerned variables.
Kinematic Equations
Kinematic equations are the toolbox for solving motion problems in classical mechanics. These equations provide the necessary relations between displacement (s), velocity (v), acceleration (a), and time (t).

Common Kinematic Equations include:

  • \( v = u + at \)
  • \( s = ut + \frac{1}{2}at^2 \)
  • \( v^2 = u^2 + 2as \)
  • \( s = vt - \frac{1}{2}at^2 \) when starting from rest
While these equations are powerful, they apply only to constant acceleration scenarios. Our textbook problem involves a variable acceleration, hence it isn't straightforward to apply these standard equations; instead, we need to solve a differential equation.
Integration in Kinematics
Integration is a mathematical tool that enables us to find velocity from acceleration and position from velocity. It's essentially the reverse process of differentiation:
  • Acceleration to Velocity: The integral of acceleration with respect to time gives the velocity function.
  • Velocity to Position: Similarly, the integral of velocity over time yields the position equation.
In the textbook example, after finding the velocity function, we integrate it once more to determine the position function. This stepwise integration is key to determining the complete state of motion for the particle at any given moment. It's crucial to note that each integration may introduce an integration constant, which can be determined using initial conditions provided in the problem statement.

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 cord is attached to the pin at \(C\) and passes over the two pulleys at \(A\) and \(D .\) The pulley at \(A\) is attached to the smooth collar that travels along the vertical rod. Determine the velocity and acceleration of the end of the cord at \(B\) if at the instant \(s_{A}=4 \mathrm{ft}\) the collar is moving upward at \(5 \mathrm{ft} / \mathrm{s}\), which is decreasing at \(2 \mathrm{ft} / \mathrm{s}^{2}\).

The velocity of a particle is given by \(v=\left\\{16 t^{2} \mathbf{i}+\right.\) \(\left.4 t^{3} \mathbf{j}+(5 t+2) \mathbf{k}\right\\} \mathrm{m} / \mathrm{s}\), where \(t\) is in seconds. If the particle is at the origin when \(t=0\), determine the magnitude of the particle's acceleration when \(t=2 \mathrm{~s} .\) Also, what is the \(x, y, z\) coordinate position of the particle at this instant?

A boat is traveling along a circular path having a radius of \(20 \mathrm{~m} .\) Determine the magnitude of the boat's acceleration when the speed is \(v=5 \mathrm{~m} / \mathrm{s}\) and the rate of increase in the speed is \(\dot{v}=2 \mathrm{~m} / \mathrm{s}^{2}\)

The car travels along a road which for a short distance is defined by \(r=(200 / \theta) \mathrm{ft}\), where \(\theta\) is in radians. If it maintains a constant speed of \(v=35 \mathrm{ft} / \mathrm{s}\), determine the radial and transverse components of its velocity when \(\theta=\pi / 3\) rad.

Tests reveal that a normal driver takes about \(0.75 \mathrm{~s}\) before he or she can react to a situation to avoid a collision. It takes about \(3 \mathrm{~s}\) for a driver having \(0.1 \%\) alcohol in his system to do the same. If such drivers are traveling on a straight road at \(30 \mathrm{mph}\) (44 \(\mathrm{ft} / \mathrm{s}\) ) and their cars can decelerate at \(2 \mathrm{ft} / \mathrm{s}^{2}\), determine the shortest stopping distance \(d\) for each from the moment they see the pedestrians. Moral: If you must drink, please don't drive!

See all solutions

Recommended explanations on Physics Textbooks

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.