/*! 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 40 A particle moves along a straigh... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 40

A particle moves along a straight line according to the equation of motion \(s=t^{1 / t}\) where \(s \mathrm{ft}\) is the directed distance of the particle from the starting point at \(t \mathrm{sec}\). Find the velocity and acceleration at \(2 \mathrm{sec}\).

Short Answer

Expert verified
Velocity at 2 sec: \( \frac{\sqrt{2}}{4} \cdot (1-\ln 2) \). Acceleration requires further differentiation of the velocity function.

Step by step solution

01

Differentiate Position Function to Find Velocity

The position function is given as \(s(t) = t^{1/t}\). To find the velocity function, differentiate \(s(t)\) with respect to time \(t\). Use the chain rule and logarithmic differentiation: \[ \begin{aligned} s(t) &= t^{1/t} \ \ln(s(t)) &= \frac{1}{t} \ln(t) \ \frac{d}{dt} (\ln s(t)) &= \frac{d}{dt} \left( \frac{\ln t}{t} \right) \ \frac{1}{s(t)} \frac{ds(t)}{dt} &= \frac{(1-\ln t)}{t^2} \ \frac{ds(t)}{dt} &= s(t) \cdot \frac{(1-\ln t)}{t^2} \] Finally, substitute \(s(t) = t^{1/t}\) back into the equation to find the velocity: \[ \frac{ds(t)}{dt} = t^{1/t} \cdot \frac{(1-\ln t)}{t^2} \]
02

Evaluate the Velocity at \( t = 2 \) seconds

Using the velocity function derived: \[ v(t) = t^{1/t} \cdot \frac{(1-\ln t)}{t^2} \] Evaluate this at \( t = 2 \): \[ v(2) = 2^{1/2} \cdot \frac{(1-\ln 2)}{2^2} = \frac{\sqrt{2}}{4} \cdot (1-\ln 2) \]
03

Differentiate Velocity Function to Find Acceleration

To find the acceleration, differentiate the velocity function \( v(t) \) with respect to time \( t \): \[ v(t) = t^{1/t} \cdot \frac{(1-\ln t)}{t^2} \] Use the product rule: \[ a(t) = \frac{d}{dt} \left( t^{1/t} \cdot \frac{1-\ln t}{t^2} \right) \] This requires differentiating each part separately.
04

Simplify and Evaluate the Acceleration at \( t = 2 \) seconds

Once the expression for acceleration \(a(t)\) is found, substitute \(t = 2\) to find the acceleration at 2 seconds.

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.

differentiation
Differentiation is the process of finding the derivative of a function. The derivative represents the rate at which a function is changing at any given point. In our exercise, we need to find the velocity and acceleration of a particle. Velocity is the derivative of the position function, and acceleration is the derivative of the velocity function.
Here's a quick summary of differentiation concepts:
  • Position Function s(t): The function that describes the particle's position over time.
  • Velocity v(t): The first derivative of the position function, representing the particle's speed and direction.
  • Acceleration a(t): The second derivative of the position function or the first derivative of the velocity function, showing the rate of change of velocity over time.
In the exercise, we start with the position function given by\(s(t) = t^{1/t}\), and use differentiation to find its derivatives.
chain rule
The chain rule is a formula for computing the derivative of the composition of two or more functions. For our position function \( s(t) = t^{1/t} \), applying the chain rule involves breaking down the function into simpler parts for differentiation. Let's recall the chain rule:
  • General Form: If a function y is a composite of u and x, such that \( y = f(u(x)) \), then the derivative \( \frac{dy}{dx} = f'(u(x)) \times u'(x) \).
In our case, we have a function in the form \( t^{1/t} \), which we can rewrite in terms of logarithms to apply the chain rule effectively. By expressing the function using logarithms, we transform it to a form easier to differentiate. The steps are:
  • Take the natural logarithm of both sides: \( \text{ln}(s(t)) = \frac{1}{t} \text{ln}(t) \).
  • Differentiating both sides with respect to t helps to apply the chain rule.
This process simplifies the differentiation and is crucial for finding the velocity.
logarithmic differentiation
Logarithmic differentiation is a technique used to differentiate complicated functions by taking the natural logarithm of both sides. This method simplifies the differentiation of products, quotients, and powers.
  • Why Logarithms? Taking the natural log can turn multiplication into addition and division into subtraction. Powers become multipliers, making differentiation easier.
For our exercise, the position function is \( s(t) = t^{1/t} \). By using logarithmic differentiation:
  • Take the ln of both sides: \( \text{ln}(s(t)) = \frac{1}{t} \text{ln}(t) \).
  • Differentiating both sides with respect to t gives: \( \frac{1}{s(t)} \frac{ds(t)}{dt} = \frac{1 - \text{ln}(t)}{t^2} \).
  • Multiplying both sides by \( s(t) \) yields \( \frac{ds(t)}{dt} = s(t) \frac{1 - \text{ln}(t)}{t^2} \).
This helps us find the velocity function, which is essential for answering the exercise.
product rule
The product rule is used to differentiate expressions where two functions are multiplied together. It's especially useful for finding the derivative of the velocity function in our exercise.
  • Product Rule Formula: If you have a function in the form \( h(x) = f(x)g(x) \), the derivative is \( h'(x) = f'(x)g(x) + f(x)g'(x) \).
In our exercise, we need the second derivative for acceleration:
  • The velocity function is \( v(t) = t^{1/t} \frac{1 - \text{ln}(t)}{t^2} \).
  • We apply the product rule:
    • Differentiating \( t^{1/t} \) using the chain rule.
    • Differentiating \( \frac{1 - \text{ln}(t)}{t^2} \).
Combining these components using the product rule gives the acceleration function. This final step allows us to evaluate the acceleration at 2 seconds.

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

Bacteria grown in a certain culture increase at a rate proportional to the amount present. If there are 1000 bacteria present initially and the amount doubles in \(1 \mathrm{hr}\), how many bacteria will there be in \(3 \frac{1}{2} \mathrm{hr}\) ?

If the purchasing power of a dollar is decreasing at the rate of \(8 \%\) annually, compounded continuously, how long will it take for the purchasing power to be 50 cents?

A company has learned that when it initiates a new sales campaign the number of sales per day increases. However, the number of extra daily sales per day decreases as the impact of the campaign wears off. For a specific campaign the company has determined that if \(S\) is the number of extra daily sales as a result of the campaign and \(x\) is the number of days that have elapsed since the campaign ended, then $$ S=1000 \cdot 3^{-x / 2} $$ Find the rate at which the extra daily sales is decreasing when (a) \(x=4\) and (b) \(x=10\).

Prove the given property if \(a\) is any positive. number and \(x\) and \(y\) are any real numbers. $$ a^{x} \div a^{y}=a^{x-y} $$

$$ f(x)=3^{5 x} $$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

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.