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ISBN: 9781305952300
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![### Particle Motion in the XY Plane with Constant Acceleration
**Problem Statement:**
1. A particle starts from the origin at \( t = 0 \) with a velocity of \((16\mathbf{i} - 12\mathbf{j}) \, \text{m/s}\) and moves in the \( xy \)-plane with a constant acceleration of \(\mathbf{a} = (3.0\mathbf{i} - 6.0\mathbf{j}) \, \text{m/s}^2\).
a. What is the position vector at any time \( t \)?
b. Determine the direction (as measured from the positive \( x \)-axis, counterclockwise) of the position at \( t = 2s \).
**Solution:**
a. **Position Vector at Any Time \( t \):**
The position vector \(\mathbf{r}(t)\) of a particle moving with initial velocity \(\mathbf{v_0}\) and constant acceleration \(\mathbf{a}\) can be found using the kinematic equation:
\[
\mathbf{r}(t) = \mathbf{r_0} + \mathbf{v_0} t + \frac{1}{2} \mathbf{a} t^2
\]
Given:
- Initial position \(\mathbf{r_0} = 0\)
- Initial velocity \(\mathbf{v_0} = 16\mathbf{i} - 12\mathbf{j} \, \text{m/s}\)
- Acceleration \(\mathbf{a} = 3.0\mathbf{i} - 6.0\mathbf{j} \, \text{m/s}^2\)
Substituting these values in the equation:
\[
\mathbf{r}(t) = (16\mathbf{i} - 12\mathbf{j})t + \frac{1}{2}(3.0\mathbf{i} - 6.0\mathbf{j})t^2
\]
Simplifying:
\[
\mathbf{r}(t) = (16t + \frac{3}{2}t^2) \mathbf{i} + (-12t - 3t^2) \mathbf{j}
\]](https://content.bartleby.com/qna-images/question/bad9eda7-57c9-475e-b126-0f24ea0e5433/0122785c-2ddb-42d5-b95d-99433bef34e7/36i549_processed.jpeg)
Transcribed Image Text:### Particle Motion in the XY Plane with Constant Acceleration
**Problem Statement:**
1. A particle starts from the origin at \( t = 0 \) with a velocity of \((16\mathbf{i} - 12\mathbf{j}) \, \text{m/s}\) and moves in the \( xy \)-plane with a constant acceleration of \(\mathbf{a} = (3.0\mathbf{i} - 6.0\mathbf{j}) \, \text{m/s}^2\).
a. What is the position vector at any time \( t \)?
b. Determine the direction (as measured from the positive \( x \)-axis, counterclockwise) of the position at \( t = 2s \).
**Solution:**
a. **Position Vector at Any Time \( t \):**
The position vector \(\mathbf{r}(t)\) of a particle moving with initial velocity \(\mathbf{v_0}\) and constant acceleration \(\mathbf{a}\) can be found using the kinematic equation:
\[
\mathbf{r}(t) = \mathbf{r_0} + \mathbf{v_0} t + \frac{1}{2} \mathbf{a} t^2
\]
Given:
- Initial position \(\mathbf{r_0} = 0\)
- Initial velocity \(\mathbf{v_0} = 16\mathbf{i} - 12\mathbf{j} \, \text{m/s}\)
- Acceleration \(\mathbf{a} = 3.0\mathbf{i} - 6.0\mathbf{j} \, \text{m/s}^2\)
Substituting these values in the equation:
\[
\mathbf{r}(t) = (16\mathbf{i} - 12\mathbf{j})t + \frac{1}{2}(3.0\mathbf{i} - 6.0\mathbf{j})t^2
\]
Simplifying:
\[
\mathbf{r}(t) = (16t + \frac{3}{2}t^2) \mathbf{i} + (-12t - 3t^2) \mathbf{j}
\]
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