/*! 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 19 An object is moving along a coor... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 19

An object is moving along a coordinate line subject to the indicated acceleration a (in centimeters per second per second) with the initial velocity \(v_{0}\) (in centimeters per second) and directed distance \(s_{0}\) (in centimeters). Find both the velocity \(\underline{v}\) and directed distance \(s\) after 2 seconds (see Example 4). $$ a=\sqrt[3]{2 t+1} ; v_{0}=0, s_{0}=10 $$

Short Answer

Expert verified
Velocity at 2s is approximately 3.6 cm/s; distance is approximately 20.2 cm.

Step by step solution

01

Understand Given Values and Find Formula for Velocity

We know the acceleration formula, \( a = \sqrt[3]{2t + 1} \). To find the velocity, integrate the acceleration with respect to time, \( t \). We are given the initial velocity, \( v_0 = 0 \).
02

Integrate Acceleration to Find Velocity

The velocity \( v(t) \) is found by integrating the acceleration function: \[ v(t) = \int \sqrt[3]{2t + 1} \, dt \]Using integration techniques, we can find the antiderivative. Let \( u = 2t + 1 \), then \( du = 2 \, dt \) or \( dt = \frac{1}{2} du \). The integral becomes:\[ v(t) = \frac{1}{2} \int u^{1/3} \, du = \frac{1}{2} \cdot \frac{3}{4} u^{4/3} + C \]\[ v(t) = \frac{3}{8}(2t + 1)^{4/3} + C \]
03

Apply Initial Condition to Solve for Constant C

Use the initial velocity condition \( v(0) = 0 \) to solve for \( C \):\[ v(0) = \frac{3}{8}(2 \cdot 0 + 1)^{4/3} + C = 0 \]\[ \frac{3}{8} \cdot 1 + C = 0 \]\[ C = -\frac{3}{8} \]
04

Write the Complete Velocity Function

Substitute \( C = -\frac{3}{8} \) back into the velocity function:\[ v(t) = \frac{3}{8}(2t + 1)^{4/3} - \frac{3}{8} \]
05

Calculate Velocity at t = 2s

Plug \( t = 2 \) into the velocity function:\[ v(2) = \frac{3}{8}(2 \cdot 2 + 1)^{4/3} - \frac{3}{8} \]\[ v(2) = \frac{3}{8}(5)^{4/3} - \frac{3}{8} \]Evaluating numerically, \( v(2) \approx 3.6 \) cm/s.
06

Integrate Velocity to Find Distance

To find the position \( s(t) \), integrate the velocity function:\[ s(t) = \int \left( \frac{3}{8}(2t + 1)^{4/3} - \frac{3}{8} \right) \, dt \]Breaking it down, we consider the integral\[ s(t) = \frac{3}{8} \int (2t + 1)^{4/3} \, dt - \frac{3}{8} \int \, dt \]
07

Evaluate the Integral for Position

Using substitution and similar steps, integrate to get:\[ s(t) = \frac{9}{40}(2t + 1)^{7/3} - \frac{3}{8}t + D \]Use initial position \( s_0 = 10 \) to find \( D \):\[ s(0) = \frac{9}{40}(1)^{7/3} - 0 + D = 10 \]\[ \frac{9}{40} + D = 10 \]\[ D = 10 - \frac{9}{40} \approx 9.775 \]
08

Write Distance Function and Calculate at t = 2s

Substitute back into position function:\[ s(t) = \frac{9}{40}(2t+1)^{7/3} - \frac{3}{8}t + 9.775 \]Find \( s(2) \):\[ s(2) = \frac{9}{40}(5)^{7/3} - \frac{3}{8} \cdot 2 + 9.775 \]\[ s(2) \approx 20.2 \] cm.

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.

Velocity
Understanding velocity is key to analyzing how an object's speed changes over time. In calculus, velocity
  • represents the rate of change of position
  • is a vector, which means it has both magnitude and direction
In the given exercise, we start by finding the velocity from the acceleration using integration: \[ v(t) = \int \sqrt[3]{2t + 1} \, dt\]When integrating the acceleration, the result is a velocity that tells us how fast the object is moving and in which direction. Since we were given that the initial velocity \( v_0 \) is 0 cm/s, this effect must be considered in our integration to solve for the constant of integration. The velocity at any given time \( t \) can be calculated using the velocity function after the integration.
Acceleration
Acceleration describes how quickly the velocity of an object is changing. In the exercise, acceleration is provided by the equation:\[ a = \sqrt[3]{2t + 1}\]
  • It is the derivative of velocity.
  • Measures the change in velocity per unit time.
The concept of acceleration helps us understand how the speed of an object changes over time. Frequent integration of acceleration provides insights into the nature of motion, allowing you to determine the velocity over time. In this problem, you integrate the acceleration equation to derive an expression for velocity.
Integration
Integration is a fundamental part of calculus used here to transition from acceleration to velocity and from velocity to position.
  • It involves finding the antiderivative or integral of a function.
  • This process helps establish relationships between time \( t \), velocity \( v(t) \), and position \( s(t) \).
In the exercise, integration of the acceleration function produces the velocity function:\[ v(t) = \int \sqrt[3]{2t + 1} \, dt\]To obtain the object's position at a certain time, we integrate the velocity function:\[ s(t) = \int v(t) \, dt\]Each integration stage allows us to understand more about the dynamics of the moving object by calculating changes over time.
Position Function
The position function describes where an object is located on a coordinate line over time. Understanding this concept involves analyzing the following:
  • The initial position and how it changes over time.
  • Integration of the velocity function to describe position.
For the given problem, we find the position function by integrating the velocity function:\[ s(t) = \int \left( \frac{3}{8}(2t + 1)^{4/3} - \frac{3}{8} \right) \, dt\]After integration, don't forget to incorporate the constant of integration based on the initial conditions, \( s_0 = 10 \) cm, to finalize the position function. This results in a detailed description of where the object is at a specific time, such as after 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

(Carbon Dating) All living things contain carbon 12 , which is stable, and carbon 14, which is radioactive. While a plant or animal is alive, the ratio of these two isotopes of carbon remains unchanged, since the carbon 14 is constantly renewed: after death, no more carbon 14 is absorbed. The half-life of carbon 14 is 5730 years. If charred logs of an old fort show only \(70 \%\) of the carbon 14 expected in living matter, when did the fort burn down? Assume that the fort burned soon after it was built of freshly cut logs.

From what height must a ball be dropped in order to strike the ground with a velocity of \(-136\) feet per second?

Use the Mean Value Theorem to show that the graph of a concave up function \(f\) is always above its tangent line; that is, show that $$ f(x)>f(c)+f^{\prime}(c)(x-c), \quad x \neq c $$

Find the value of \(\$ 1000\) at the end of 1 year when the interest is compounded continuously at \(5 \%\). This is called the future value.

First find the general solution (involving a constant \(\mathrm{C}\) ) for the given differential equation. Then find the particular solution that satisfies the indicated condition. $$ \frac{d y}{d x}=(2 x+1)^{4} ; y=6 \text { at } x=0 $$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Geometry

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.