A particle leaves the origin with an initial velodty `v= (3.00 hati) m//s` and a constant acceleration `a= (-1.00 hati-0.500 hatj) m//s^2.` When the particle reaches its maximum x coordinate, what are
(a) its velocity and (b) its position vector?

Options

A. `v=-2hati`
B. `v=(-1.5hatj)m//s`
C. `r=(4.5hati-2.25hatj)m`
D. `r=(3hati-2hatj)m`

Answer

To solve the problem step by step, we need to analyze the motion of the particle under the given conditions.

### Given:
- Initial velocity, \( \mathbf{v_0} = 3.00 \, \hat{i} \, \text{m/s} \)
- Acceleration, \( \mathbf{a} = -1.00 \, \hat{i} - 0.50 \, \hat{j} \, \text{m/s}^2 \)

### (a) Finding the velocity when the particle reaches its maximum x-coordinate:

1. **Determine the time taken to reach maximum x-coordinate**:
- The particle will stop moving in the x-direction when its velocity in the x-direction becomes zero.
- Using the equation of motion:
\[ v_x = v_{0x} + a_x t \]
- Setting \( v_x = 0 \) (the final velocity in the x-direction):
\[ 0 = 3.00 - 1.00 t \]
- Rearranging gives:
\[ t = \frac{3.00}{1.00} = 3.00 \, \text{s} \]

2. **Calculate the velocity in the y-direction at this time**:
- Using the equation of motion for the y-direction:
\[ v_y = v_{0y} + a_y t \]
- Here, \( v_{0y} = 0 \) (initial velocity in the y-direction):
\[ v_y = 0 - 0.50 \times 3.00 = -1.50 \, \text{m/s} \]

3. **Combine the velocities to form the velocity vector**:
- The velocity vector when the particle reaches its maximum x-coordinate is:
\[ \mathbf{v} = 0 \, \hat{i} - 1.50 \, \hat{j} = -1.50 \, \hat{j} \, \text{m/s} \]

### (b) Finding the position vector when the particle reaches its maximum x-coordinate:

1. **Calculate the position in the x-direction**:
- Using the equation of motion:
\[ x = v_{0x} t + \frac{1}{2} a_x t^2 \]
- Substituting the known values:
\[ x = 3.00 \times 3.00 + \frac{1}{2} \times (-1.00) \times (3.00)^2 \]
- Calculating:
\[ x = 9.00 - 4.50 = 4.50 \, \text{m} \]

2. **Calculate the position in the y-direction**:
- Using the equation of motion:
\[ y = v_{0y} t + \frac{1}{2} a_y t^2 \]
- Substituting the known values:
\[ y = 0 + \frac{1}{2} \times (-0.50) \times (3.00)^2 \]
- Calculating:
\[ y = 0 - 2.25 = -2.25 \, \text{m} \]

3. **Combine the positions to form the position vector**:
- The position vector when the particle reaches its maximum x-coordinate is:
\[ \mathbf{r} = 4.50 \, \hat{i} - 2.25 \, \hat{j} \, \text{m} \]

### Final Answers:
(a) The velocity when the particle reaches its maximum x-coordinate is:
\[ \mathbf{v} = -1.50 \, \hat{j} \, \text{m/s} \]

(b) The position vector when the particle reaches its maximum x-coordinate is:
\[ \mathbf{r} = 4.50 \, \hat{i} - 2.25 \, \hat{j} \, \text{m} \]