/*! 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 36 A position function of a particl... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 36

A position function of a particle moving along a coordinate line is provided. (a) Evaluate \(s\) and \(v\) when \(a=0\) Evaluate \(s\) and \(a\) when \(v=0\). $$s=t^{3}-6 t^{2}+1$$

Short Answer

Expert verified
When \(a=0\), \(s=-15\) and \(v=-12\); when \(v=0\), \(s=1\), \(a=-12\) at \(t=0\), and \(s=-31\), \(a=12\) at \(t=4\).

Step by step solution

01

Understand the Position Function

The position function is given as \(s(t) = t^3 - 6t^2 + 1\). This function describes the position \(s\) of a particle at any time \(t\). To find velocity and acceleration, we need to differentiate this function.
02

Find the Velocity Function

Velocity \(v(t)\) is the first derivative of the position function \(s(t)\). Thus, differentiate \(s(t)\) to find \(v(t)\): \[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 1) \]Calculate: \[ v(t) = 3t^2 - 12t \]
03

Find the Acceleration Function

Acceleration \(a(t)\) is the derivative of the velocity function \(v(t)\). Differentiate \(v(t)\) to obtain \(a(t)\):\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t) \]Calculate:\[ a(t) = 6t - 12 \]
04

Evaluate for \(a = 0\)

Set \(a(t) = 0\) and solve for \(t\):\[ 6t - 12 = 0 \]\[ t = 2 \]With \(t = 2\), find \(s(2)\) and \(v(2)\):For \(s(2)\):\[ s(2) = 2^3 - 6 \, (2)^2 + 1 = 8 - 24 + 1 = -15 \]For \(v(2)\):\[ v(2) = 3(2)^2 - 12 \, (2) = 12 - 24 = -12 \]
05

Evaluate for \(v = 0\)

Set \(v(t) = 0\) and solve for \(t\):\[ 3t^2 - 12t = 0 \]\[ 3t(t - 4) = 0 \]\[ t = 0 \quad \text{or} \quad t = 4 \]Evaluate for both \(t=0\) and \(t=4\):For \(t = 0\):\(s(0) = 0^3 - 6(0)^2 + 1 = 1\)\(a(0) = 6(0) - 12 = -12\)For \(t = 4\):\(s(4) = 4^3 - 6(4)^2 + 1 = 64 - 96 + 1 = -31\)\(a(4) = 6(4) - 12 = 24 - 12 = 12\)

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.

Position Function
The position function is a mathematical representation of how the position of a particle changes over time. In our example, the function is given as \(s(t) = t^3 - 6t^2 + 1\), where \(s(t)\) represents the position of the particle at any time \(t\). This function helps us understand exactly where the particle is along a coordinate line at any point in time.
To explore further characteristics of the particle's motion, such as velocity and acceleration, we need to perform differentiation. Differentiation effectively gives us the rate of change of a quantity. In this context, it helps us find how the position is changing with time.
  • Position at any time \(t\) can be identified directly using \(s(t)\).
  • To find the velocity, which is the derivative of the position function, we differentiate \(s(t)\).
Velocity Function
Velocity is the first derivative of the position function with respect to time. It tells us how fast the particle's position is changing. Essentially, velocity indicates the speed and direction of the particle's motion. For the given position function \(s(t) = t^3 - 6t^2 + 1\), its velocity function \(v(t)\) is derived from differentiating \(s(t)\), resulting in:
\[ v(t) = \frac{ds}{dt} = 3t^2 - 12t \]
Key points about velocity:
  • Velocity is zero where the position is neither increasing nor decreasing.
  • It becomes negative when the particle moves in a reverse direction along the coordinate line.
  • Evaluating \(v(t) = 0\) helps find points where the particle might change direction or stop momentarily.
In this problem, at \(t = 2\), when the acceleration is zero, we find that the velocity is \(-12\), indicating the particle is moving backward.
Acceleration Function
Acceleration is the second derivative of the position function, or equivalently, the derivative of the velocity function. In simple terms, acceleration tells us how the velocity of the particle is changing over time. A positive acceleration means the velocity is increasing, while a negative acceleration indicates a decrease in velocity. For the velocity function \(v(t) = 3t^2 - 12t\), the acceleration function \(a(t)\) is derived as follows:
\[ a(t) = \frac{dv}{dt} = 6t - 12 \]
Important aspects of acceleration include:
  • Determining when the acceleration is zero, \(a(t) = 0\), helps find when the velocity stops changing, which can be significant for identifying turning points.
  • Positive acceleration may indicate the particle is speeding up in the direction of motion.
  • Negative acceleration, often termed deceleration, refers to the particle slowing down.
In our exercise scenario, solving \(a(t) = 0\) gives \(t = 2\), showing a key point where the particle's change in velocity temporarily stops.

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

Analyze the trigonometric function \(f\) over the specified interval, stating where \(f\) is increasing, decreasing, concave up, and concave down, and stating the \(x\) -coordinates of all inflection points. Confirm that your results are consistent with the graph of \(f\) generated with a graphing utility. $$f(x)=2 x+\cot x ;(0, \pi)$$

The concentration \(C(t)\) of a drug in the bloodstream \(t\) hours after it has been injected is commonly modeled by an equation of the form $$C(t)=\frac{K\left(e^{-b t}-e^{-a t}\right)}{a-b}$$ where \(K > 0\) and \(a > b > 0\) (a) At what time does the maximum concentration occur? (b) Let \(K=1\) for simplicity, and use a graphing utility to check your result in part (a) by graphing \(C(t)\) for various values of \(a\) and \(b\)

Use a graphing utility to make a conjecture about the relative extrema of \(f,\) and then check your conjecture using either the first or second derivative test. $$f(x)=10 \ln x-x$$

Functions similar to $$f(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}$$ arise in a wide variety of statistical problems. (a) Use the first derivative test to show that \(f\) has a relative maximum at \(x=0,\) and confirm this by using a graphing utility to graph \(f\) (b) Sketch the graph of $$f(x)=\frac{1}{\sqrt{2 \pi}} e^{-(x-\mu)^{2} / 2}$$ where \(\mu\) is a constant, and label the coordinates of the relative extrema.

Use a CAS to graph \(f^{\prime}\) and \(f^{\prime \prime},\) and then use those graphs to estimate the \(x\) -coordinates of the relative extrema of f. Check that your estimates are consistent with the graph of \(f\). $$f(x)=\frac{\tan ^{-1}\left(x^{2}-x\right)}{x^{2}+4}$$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

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.