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
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 39RE
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**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 \(\mathbf{r}(t)\):

\[
\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt}
\]
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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