Part D I am having trouble on d) During the time interval 0

Name: Part D I am having trouble on d) During the time interval 0
Uploaded: 2024-03-20T07:07:37.000Z
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Description: Part D I am having trouble on d) During the time interval 0

Given: A particle moves along the x-axis with the velocity v(t)=t6-13t4+1210t3+3 At the time t = 0…

Answered: d) During the time interval 0

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Part D I am having trouble on

Expert Solution

Step 1

Given:

A particle moves along the x-axis with the velocity $v (t) = \frac{t^{6} - 13 t^{4} + 12}{10 t^{3} + 3}$

At the time t = 0 the particle's initial position is x= 7.

Step 2

Let us assume the position of the particle is $x (t)$ at time t.

Implies that, $x' (t) = v (t)$

$\begin{array}{rcl} \int_{x_{0}}^{x} d x (t) & = & \int_{0}^{4} v (t) d t \\ x - x_{0} & = & \int_{0}^{4} \frac{t^{6} - 13 t^{4} + 12}{10 t^{3} + 3} d t \end{array}$

Use a graphing utility to calculate the integral of the right side.

$x - x_{0} = - 0.288$

It is known that the initial position $x_{0} = 7$

Therefore,

$\begin{array}{rcl} x - 7 & = & - 0.288 \\ x & = & 7 - 0.288 \\ x & = & 6.712 \end{array}$

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