/*! 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 32 If the velocity \(v\) of a parti... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If the velocity \(v\) of a particle moving along a straight line decreases linearly with its displacement \(s\) from \(20 \mathrm{m} / \mathrm{s}\) to a value approaching zero at \(s=30 \mathrm{m}\) determine the acceleration \(a\) of the particle when \(s=15 \mathrm{m}\) and show that the particle never reaches the 30 -m displacement.

Short Answer

Expert verified
The acceleration at \(s = 15 \mathrm{m}\) is \(-\frac{20}{3} \mathrm{m/s^2}\). The particle never reaches \(s = 30 \mathrm{m}\) as velocity approaches zero.

Step by step solution

01

Understand the velocity-s displacement relationship

The velocity \( v \) decreases linearly from \( 20 \mathrm{m/s} \) to \( 0 \mathrm{m/s} \) as \( s \) changes from \( 0 \mathrm{m} \) to \( 30 \mathrm{m} \). This gives a line equation for velocity vs. displacement.
02

Establish the linear equation for velocity

We can express the velocity \( v \) in terms of displacement \( s \) using the linear equation \( v = m \, s + c \). Given \( v = 20 \mathrm{m/s} \) at \( s=0 \) and \( v = 0 \mathrm{m/s} \) at \( s=30 \mathrm{m} \), solve for \( m \) and \( c \).
03

Solve for the slope \(m\) and intercept \(c\)

The slope \( m = \frac{0 - 20}{30 - 0} = -\frac{2}{3} \), and the y-intercept \(c\) is \(v\) when \(s = 0\), i.e., \(c = 20\). Thus, the velocity equation is \(v = -\frac{2}{3}s + 20\).
04

Calculate acceleration as a derivative of velocity

Acceleration \( a \) is the derivative of velocity with respect to time, but here we'll differentiate \( v(s) \) with respect to \( s \) and multiply by \( \frac{ds}{dt} = v \) (chain rule). Thus, \( a = \frac{dv}{ds} \times v \).
05

Find \(\frac{dv}{ds}\) using velocity equation

Differentiate \( v = -\frac{2}{3}s + 20\) with respect to \( s \). The derivative is \( \frac{dv}{ds} = -\frac{2}{3} \).
06

Calculate acceleration at \(s = 15 \mathrm{m}\)

Use \( a = \frac{dv}{ds} \times v \), where \( \frac{dv}{ds} = -\frac{2}{3} \), and \( v = -\frac{2}{3}(15) + 20 = 10 \mathrm{m/s} \). Then \( a = -\frac{2}{3} \times 10 = -\frac{20}{3} \mathrm{m/s^2} \).
07

Analyze the possibility of reaching \( s = 30 \mathrm{m} \)

As \( s \to 30 \mathrm{m} \), \( v \to 0 \). \( a = \frac{dv}{ds} \times v \) implies \( a \to 0 \) as well, preventing the particle from reaching \( s = 30 \mathrm{m} \) because velocity asymptotically approaches zero but doesn’t reach it.

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.

Velocity-Displacement Relationship
The relationship between velocity and displacement is essential in understanding the motion of a particle along a straight line. In the given exercise, the velocity \( v \) of a particle decreases linearly with its displacement \( s \). This can be visualized as a straight line when plotting velocity against displacement on a graph.
As the particle moves from \( s = 0 \) to \( s = 30 \ \mathrm{m} \), its velocity decreases from \( 20 \ \mathrm{m/s} \) to 0. This linear decrease indicates that for every additional meter the particle travels, its velocity reduces at a constant rate.
  • The linearity indicates a constant change (in this case, a decrease) in velocity per unit of displacement.
  • This change is quantified by the slope of the velocity vs. displacement graph.
Understanding this concept is crucial as it forms the basis for deriving further relations and solving motion-related problems involving linear reduction or increase in speed over displacement.
Differentiation in Mechanics
Differentiation plays a vital role in mechanics as it helps in finding how quantities like velocity and acceleration change with respect to other variables. In the problem at hand, we used differentiation to understand how the velocity \( v \) changes with displacement \( s \).
To determine the acceleration, we needed to find the rate of change of velocity with respect to displacement, \( \frac{dv}{ds} \). Since the velocity is given as a linear equation, differentiating this equation gives us this required rate of change.
This derivative tells us the rate at which velocity is changing for any given displacement, which is directly used to find acceleration.
  • Acceleration is essentially the rate of change of velocity over time. However, in this scenario, it is linked to displacement, and differentiation helps establish this connection.
  • Without understanding differentiation, it would be challenging to determine how the velocity transforms into acceleration in relation to the particle's position.
Grasping differentiation in mechanics is vital for comprehending how physical quantities evolve dynamically in response to changes in other variables.
Linear Equations in Physics
Linear equations are a fundamental tool in physics for describing relationships between different physical quantities. In this case, the linear equation \( v = -\frac{2}{3}s + 20 \) represents the relationship between velocity \( v \) and displacement \( s \).
This equation tells us directly how the velocity changes as a function of displacement:
  • The slope \( -\frac{2}{3} \) indicates the change in velocity for every meter change in displacement.
  • The y-intercept, in this case, is the initial velocity of the particle when \( s = 0 \), which is \( 20 \ \mathrm{m/s} \).
A linear equation like this makes calculating the velocity at any given point straightforward and systematic. It provides a clear picture of the slowing motion of the particle and aids in solving for acceleration and predicting future motion. Linear equations simplify complex physical conditions into manageable mathematical expressions.
Asymptotic Behavior in Motion
Asymptotic behavior refers to how a function behaves as it approaches a certain point. In the context of motion, it helps us understand how certain quantities like velocity and acceleration behave as displacement nears a critical value.
In this exercise, the particle's velocity approaches zero as the displacement reaches \( s = 30 \ \mathrm{m} \). Despite this, the particle never actually reaches a velocity of zero. Instead, the motion slows asymptotically, meaning it comes ever closer to zero without attaining it.
  • This asymptotic behavior is crucial because it hints at a limit to how far the particle can travel given its initial conditions.
  • As the velocity approaches zero, so too does the acceleration, reinforcing that the particle cannot not reach the full \( 30 \ \mathrm{m} \) displacement.
Understanding asymptotic behavior in motion is essential for predicting whether certain positions can ever be reached and how forces might need to be applied differently to overcome such limitations.

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 projectile is ejected into an experimental fluid at time \(t=0 .\) The initial speed is \(v_{0}\) and the angle to the horizontal is \(\theta\). The drag on the projectile results in an acceleration term \(\mathbf{a}_{D}=-k \mathbf{v},\) where \(k\) is a constant and \(\mathbf{v}\) is the velocity of the projectile. Determine the \(x-\) and \(y\) -components of both the velocity and displacement as functions of time. What is the terminal velocity? Include the effects of gravitational acceleration.

A model rocket is launched from rest with a constant upward acceleration of \(3 \mathrm{m} / \mathrm{s}^{2}\) under the action of a small thruster. The thruster shuts off after 8 seconds, and the rocket continues upward until it reaches its apex. At apex, a small chute opens which ensures that the rocket falls at a constant speed of \(0.85 \mathrm{m} / \mathrm{s}\) until it impacts the ground. Determine the maximum height \(h\) attained by the rocket and the total flight time. Neglect aerodynamic drag during ascent, and assume that the mass of the rocket and the acceleration of gravity are both constant.

A batter hits the baseball \(A\) with an initial velocity of \(v_{0}=100 \mathrm{ft} / \mathrm{sec}\) directly toward fielder \(B\) at an angle of \(30^{\circ}\) to the horizontal; the initial position of the ball is \(3 \mathrm{ft}\) above ground level. Fielder \(B\) requires \(\frac{1}{4}\) sec to judge where the ball should be caught and begins moving to that position with constant speed. Because of great experience, fielder \(B\) chooses his running speed so that he arrives at the "catch position" simultaneously with the baseball. The catch position is the field location at which the ball altitude is \(7 \mathrm{ft}\). Determine the velocity of the ball relative to the fielder at the instant the catch is made.

A particle moving in the \(x\) -y plane has a velocity \(\mathbf{v}=7.25 \mathbf{i}+3.48 \mathbf{j} \mathrm{m} / \mathrm{s}\) at a certain instant. If the particle then encounters a constant acceleration \(\mathbf{a}=0.85 \mathbf{j} \mathrm{m} / \mathrm{s}^{2},\) determine the amount of time which must pass before the direction of the tangent to the trajectory of the particle has been altered by \(30^{\circ}\)

A train which is traveling at \(80 \mathrm{mi} / \mathrm{hr}\) applies its brakes as it reaches point \(A\) and slows down with a constant deceleration. Its decreased velocity is observed to be \(60 \mathrm{mi} / \mathrm{hr}\) as it passes a point \(1 / 2 \mathrm{mi}\) beyond \(A\). A car moving at \(50 \mathrm{mi} / \mathrm{hr}\) passes point \(B\) at the same instant that the train reaches point \(A\) In an unwise effort to beat the train to the crossing, the driver "steps on the gas." Calculate the constant acceleration \(a\) that the car must have in order to beat the train to the crossing by 4 seconds and find the velocity \(v\) of the car as it reaches the crossing.

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.