Find r(t) and v(t) if a(t) = tk, v(0) = 15i, r(0) = 14j.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.3: Algebraic Expressions
Problem 32E
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Question
![**Problem Statement:**
Find \(\mathbf{r}(t)\) and \(\mathbf{v}(t)\) if \(\mathbf{a}(t) = t \mathbf{k}\), \(\mathbf{v}(0) = 15 \mathbf{i}\), \(\mathbf{r}(0) = 14 \mathbf{j}\).
(Use symbolic notation and fractions where needed.)
**Solution Steps:**
Given:
\(\mathbf{a}(t) = t \mathbf{k}\)
\(\mathbf{v}(0) = 15 \mathbf{i}\)
\(\mathbf{r}(0) = 14 \mathbf{j}\)
1. **Determining Velocity \(\mathbf{v}(t)\):**
We know that acceleration \(\mathbf{a}(t)\) is the derivative of velocity \(\mathbf{v}(t)\):
\[
\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt}
\]
Given \(\mathbf{a}(t) = t \mathbf{k}\), we integrate with respect to \(t\):
\[
\mathbf{v}(t) = \int t \mathbf{k} \, dt = \left( \frac{t^2}{2} + C_1 \right) \mathbf{k}
\]
We use the given initial condition \(\mathbf{v}(0) = 15 \mathbf{i}\) to find the constant of integration \(C_1\):
\[
\mathbf{v}(0) = 15 \mathbf{i} \implies \left( \frac{0^2}{2} + C_1 \right) \mathbf{k} = 15 \mathbf{i} \implies C_1 \mathbf{k} = 15 \mathbf{i}
\]
Thus, we must add the \(15 \mathbf{i}\) component to our velocity function:
\[
\mathbf{v}(t) = 15 \mathbf{i} + \frac{t^2}{2} \mathbf{k}
\]
2. **Determining Position \(\mathbf{r}(t)\):**
We know that velocity \(\mathbf{v}(t)\) is the derivative of position \(\mathbf{r}(t)\):
\[
\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdb0e1959-c10c-4eae-bad7-9452177a0286%2F042a8cc1-002d-43ca-9005-30c3d76e8c8c%2Flemupln_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find \(\mathbf{r}(t)\) and \(\mathbf{v}(t)\) if \(\mathbf{a}(t) = t \mathbf{k}\), \(\mathbf{v}(0) = 15 \mathbf{i}\), \(\mathbf{r}(0) = 14 \mathbf{j}\).
(Use symbolic notation and fractions where needed.)
**Solution Steps:**
Given:
\(\mathbf{a}(t) = t \mathbf{k}\)
\(\mathbf{v}(0) = 15 \mathbf{i}\)
\(\mathbf{r}(0) = 14 \mathbf{j}\)
1. **Determining Velocity \(\mathbf{v}(t)\):**
We know that acceleration \(\mathbf{a}(t)\) is the derivative of velocity \(\mathbf{v}(t)\):
\[
\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt}
\]
Given \(\mathbf{a}(t) = t \mathbf{k}\), we integrate with respect to \(t\):
\[
\mathbf{v}(t) = \int t \mathbf{k} \, dt = \left( \frac{t^2}{2} + C_1 \right) \mathbf{k}
\]
We use the given initial condition \(\mathbf{v}(0) = 15 \mathbf{i}\) to find the constant of integration \(C_1\):
\[
\mathbf{v}(0) = 15 \mathbf{i} \implies \left( \frac{0^2}{2} + C_1 \right) \mathbf{k} = 15 \mathbf{i} \implies C_1 \mathbf{k} = 15 \mathbf{i}
\]
Thus, we must add the \(15 \mathbf{i}\) component to our velocity function:
\[
\mathbf{v}(t) = 15 \mathbf{i} + \frac{t^2}{2} \mathbf{k}
\]
2. **Determining Position \(\mathbf{r}(t)\):**
We know that velocity \(\mathbf{v}(t)\) is the derivative of position \(\mathbf{r}(t)\):
\[
\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt}
\]
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