4. A particle starts from the origin at t=0 with a velocity of (161 - 12j) m/s and moves in the xy plane with a constant acceleration of a= (3.01 – 6.0ĵ) m/s². 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.

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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:
   
   \[
   \mathbf{r}(t) = (16t + \frac{3}{2}t^2) \mathbf{i} + (-12t - 3t^2) \mathbf{j}
   \]
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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