The position of a particle after t seconds is given by r(t) = (t², 4t, 4 lnt) for t≥ 1. (a) Calculate the position, velocity, and acceleration of the particle when t = 2. (b) Calculate the total distance traveled by the particle from t = 1 to t = e². (c) Is the object speeding up or slowing down at t = e² seconds?
The position of a particle after t seconds is given by r(t) = (t², 4t, 4 lnt) for t≥ 1. (a) Calculate the position, velocity, and acceleration of the particle when t = 2. (b) Calculate the total distance traveled by the particle from t = 1 to t = e². (c) Is the object speeding up or slowing down at t = e² seconds?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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can you help me with these three problems and can you do these three parts because I don't know how to do it those parts. Can you please show how to do these step by step so I can follow along
![Title: Particle Motion Analysis
The position of a particle after \(t\) seconds is given by \( \mathbf{r}(t) = \langle t^2, 4t, 4 \ln t \rangle \) for \( t \geq 1 \).
---
### (a) Calculation of Position, Velocity, and Acceleration
**Objective:** Calculate the position, velocity, and acceleration of the particle when \( t = 2 \).
1. **Position** at \(t = 2\):
\[
\mathbf{r}(2) = \langle (2)^2, 4(2), 4 \ln(2) \rangle = \langle 4, 8, 4 \ln 2 \rangle
\]
2. **Velocity**:
Velocity \(\mathbf{v}(t)\) is the derivative of the position function \(\mathbf{r}(t)\):
\[
\mathbf{v}(t) = \frac{d}{dt}\mathbf{r}(t) = \frac{d}{dt} \langle t^2, 4t, 4 \ln t \rangle = \langle 2t, 4, \frac{4}{t} \rangle
\]
Therefore, at \(t = 2\):
\[
\mathbf{v}(2) = \langle 2(2), 4, \frac{4}{2} \rangle = \langle 4, 4, 2 \rangle
\]
3. **Acceleration**:
Acceleration \(\mathbf{a}(t)\) is the derivative of the velocity function \(\mathbf{v}(t)\):
\[
\mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t) = \frac{d}{dt} \langle 2t, 4, \frac{4}{t} \rangle = \langle 2, 0, -\frac{4}{t^2} \rangle
\]
Therefore, at \(t = 2\):
\[
\mathbf{a}(2) = \langle 2, 0, -\frac{4}{(2)^2} \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe5147bf2-ddd5-4450-99d9-d8f97b210b53%2Fd7fe96f7-39f4-4873-af2b-c28e80227e87%2Fdkmifxe_processed.png&w=3840&q=75)
Transcribed Image Text:Title: Particle Motion Analysis
The position of a particle after \(t\) seconds is given by \( \mathbf{r}(t) = \langle t^2, 4t, 4 \ln t \rangle \) for \( t \geq 1 \).
---
### (a) Calculation of Position, Velocity, and Acceleration
**Objective:** Calculate the position, velocity, and acceleration of the particle when \( t = 2 \).
1. **Position** at \(t = 2\):
\[
\mathbf{r}(2) = \langle (2)^2, 4(2), 4 \ln(2) \rangle = \langle 4, 8, 4 \ln 2 \rangle
\]
2. **Velocity**:
Velocity \(\mathbf{v}(t)\) is the derivative of the position function \(\mathbf{r}(t)\):
\[
\mathbf{v}(t) = \frac{d}{dt}\mathbf{r}(t) = \frac{d}{dt} \langle t^2, 4t, 4 \ln t \rangle = \langle 2t, 4, \frac{4}{t} \rangle
\]
Therefore, at \(t = 2\):
\[
\mathbf{v}(2) = \langle 2(2), 4, \frac{4}{2} \rangle = \langle 4, 4, 2 \rangle
\]
3. **Acceleration**:
Acceleration \(\mathbf{a}(t)\) is the derivative of the velocity function \(\mathbf{v}(t)\):
\[
\mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t) = \frac{d}{dt} \langle 2t, 4, \frac{4}{t} \rangle = \langle 2, 0, -\frac{4}{t^2} \rangle
\]
Therefore, at \(t = 2\):
\[
\mathbf{a}(2) = \langle 2, 0, -\frac{4}{(2)^2} \
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