### 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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Description: ### 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 constan

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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.

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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Concept explainers

Displacement, Velocity and Acceleration

In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.

Linear Displacement

The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.

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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:

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

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