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. `-2.0 ms^(-1)`
B. `-1.0 ms^(-1)`
C. `-1.5 ms^(-1)`
D. `-1.0 ms^(-1)`

Explanation

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

### Given Data:
- Initial velocity: \( \mathbf{v} = 3.00 \hat{i} \) m/s
- Acceleration: \( \mathbf{a} = -1.00 \hat{i} - 0.500 \hat{j} \) m/s²

### Step 1: Determine the time taken to reach maximum x-coordinate
At the maximum x-coordinate, the velocity in the x-direction becomes zero. We can use the first equation of motion:
\[ v_x = u_x + a_x t \]
Where:
- \( v_x = 0 \) (final velocity in x-direction)
- \( u_x = 3 \) m/s (initial velocity in x-direction)
- \( a_x = -1 \) m/s² (acceleration in x-direction)

Substituting the values:
\[ 0 = 3 - 1 \cdot t \]
Solving for \( t \):
\[ t = 3 \text{ seconds} \]

### Step 2: Calculate the final velocity in the y-direction
Using the equation for the final velocity in the y-direction:
\[ v_y = u_y + a_y t \]
Where:
- \( u_y = 0 \) m/s (initial velocity in y-direction)
- \( a_y = -0.5 \) m/s² (acceleration in y-direction)

Substituting the values:
\[ v_y = 0 - 0.5 \cdot 3 = -1.5 \text{ m/s} \]

### Step 3: Write the final velocity vector
The final velocity vector at the maximum x-coordinate is:
\[ \mathbf{v}_{\text{final}} = 0 \hat{i} - 1.5 \hat{j} = -1.5 \hat{j} \text{ m/s} \]

### Step 4: Calculate the position vector at maximum x-coordinate
Using the second equation of motion to find the x-coordinate:
\[ x = u_x t + \frac{1}{2} a_x t^2 \]
Substituting the values:
\[ x = 3 \cdot 3 + \frac{1}{2} \cdot (-1) \cdot (3^2) \]
\[ x = 9 - 4.5 = 4.5 \text{ m} \]

Now, calculate the y-coordinate:
\[ y = u_y t + \frac{1}{2} a_y t^2 \]
Substituting the values:
\[ y = 0 \cdot 3 + \frac{1}{2} \cdot (-0.5) \cdot (3^2) \]
\[ y = 0 - \frac{1}{2} \cdot 0.5 \cdot 9 = -2.25 \text{ m} \]

### Step 5: Write the position vector
The position vector at the maximum x-coordinate is:
\[ \mathbf{r} = 4.5 \hat{i} - 2.25 \hat{j} \text{ m} \]

### Final Answers:
(a) The final velocity is \( \mathbf{v}_{\text{final}} = -1.5 \hat{j} \) m/s.
(b) The position vector is \( \mathbf{r} = 4.5 \hat{i} - 2.25 \hat{j} \) m.