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?

Explanation

To solve the problem step by step, we need to analyze the motion of the particle given its initial velocity and acceleration.

### Step 1: Identify the initial conditions
- Initial velocity, \( \mathbf{v_0} = 3.00 \hat{i} \, \text{m/s} \)
- Acceleration, \( \mathbf{a} = -1.00 \hat{i} - 0.50 \hat{j} \, \text{m/s}^2 \)

### Step 2: Determine when the particle reaches its maximum x-coordinate
At the maximum x-coordinate, the x-component of the velocity will be zero (\( v_x = 0 \)). We can use the kinematic equation:
\[ v_x^2 = v_{0x}^2 + 2a_x s_x \]
Substituting the known values:
\[ 0 = (3.00)^2 + 2(-1.00)s_x \]
\[ 0 = 9 - 2s_x \]
Solving for \( s_x \):
\[ 2s_x = 9 \implies s_x = \frac{9}{2} = 4.5 \, \text{m} \]

### Step 3: Calculate the time taken to reach the maximum x-coordinate
Using the equation:
\[ v_x = v_{0x} + a_x t \]
Setting \( v_x = 0 \):
\[ 0 = 3.00 - 1.00 t \]
Solving for \( t \):
\[ t = 3.00 \, \text{s} \]

### Step 4: Find the y-coordinate at \( t = 3 \, \text{s} \)
Using the equation for displacement in the y-direction:
\[ s_y = v_{0y} t + \frac{1}{2} a_y t^2 \]
Since the initial velocity in the y-direction is zero (\( v_{0y} = 0 \)):
\[ s_y = 0 + \frac{1}{2}(-0.50)(3.00)^2 \]
Calculating:
\[ s_y = \frac{1}{2}(-0.50)(9) = -2.25 \, \text{m} \]

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

### Step 6: Calculate the velocity components at \( t = 3 \, \text{s} \)
The x-component of the velocity at maximum x-coordinate is:
\[ v_x = 0 \]
For the y-component:
\[ v_y = v_{0y} + a_y t = 0 - 0.50(3.00) = -1.50 \, \text{m/s} \]

### Final Answers
(a) The velocity when the particle reaches its maximum x-coordinate is:
\[ \mathbf{v} = 0 \hat{i} - 1.50 \hat{j} \, \text{m/s} \]

(b) The position vector when the particle reaches its maximum x-coordinate is:
\[ \mathbf{r} = 4.5 \hat{i} - 2.25 \hat{j} \, \text{m} \]