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Find the position vector for a particle with acceleration, initial velocity, and initial position given below. ā(t) = (3t, 6 sin(t), cos(5t)) v(0) = ( − 3, — 1,0) - 7(0) = (0, – 5,0) - r(t) =

Find the position vector for a particle with acceleration, initial velocity, and initial position given below. ā(t) = (3t, 6 sin(t), cos(5t)) v(0) = ( − 3, — 1,0) - 7(0) = (0, – 5,0) - r(t) =

Calculus: Early Transcendentals
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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### Problem Statement: Find the position vector for a particle with acceleration, initial velocity, and initial position given below. ### Given Data: **Acceleration:** \[ \vec{a}(t) = \langle 3t, 6 \sin(t), \cos(5t) \rangle \] **Initial Velocity:** \[ \vec{v}(0) = \langle -3, -1, 0 \rangle \] **Initial Position:** \[ \vec{r}(0) = \langle 0, -5, 0 \rangle \] ### Required Solution: **Position Vector:** \[ \vec{r}(t) = \left\{\begin{array}{c} \ \text{[enter x-component here]}\ , \\ \ \text{[enter y-component here]}\ , \\ \ \text{[enter z-component here]}\ \end{array}\right\} \] ### Explanation: You need to determine the position vector \(\vec{r}(t)\) by integrating the given acceleration vector \(\vec{a}(t)\) to find the velocity vector \(\vec{v}(t)\), and then integrating \(\vec{v}(t)\) to find the position vector \(\vec{r}(t)\).
Transcribed Image Text:### Problem Statement: Find the position vector for a particle with acceleration, initial velocity, and initial position given below. ### Given Data: **Acceleration:** \[ \vec{a}(t) = \langle 3t, 6 \sin(t), \cos(5t) \rangle \] **Initial Velocity:** \[ \vec{v}(0) = \langle -3, -1, 0 \rangle \] **Initial Position:** \[ \vec{r}(0) = \langle 0, -5, 0 \rangle \] ### Required Solution: **Position Vector:** \[ \vec{r}(t) = \left\{\begin{array}{c} \ \text{[enter x-component here]}\ , \\ \ \text{[enter y-component here]}\ , \\ \ \text{[enter z-component here]}\ \end{array}\right\} \] ### Explanation: You need to determine the position vector \(\vec{r}(t)\) by integrating the given acceleration vector \(\vec{a}(t)\) to find the velocity vector \(\vec{v}(t)\), and then integrating \(\vec{v}(t)\) to find the position vector \(\vec{r}(t)\).
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1/2-1/25cos(5t) is wring could you please check it again? thank you!

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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