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This problem involves kinematics in two dimensions, with the particle's motion given by initial velocity and constant acceleration vectors.

### Given:
- Initial velocity: \(\vec{v_0} = (1.95 \hat{i}) \, \text{m/s}\)  
  (This indicates the initial velocity is 1.95 m/s in the \(x\)-direction and 0 in the \(y\)-direction.)
- Acceleration: \(\vec{a} = (-4.47 \hat{i} - 4.46 \hat{j}) \, \text{m/s}^2\)
  (This indicates acceleration in both \(x\)- and \(y\)-directions.)

### To find:
- (a) The velocity when the particle reaches its maximum \(x\)-coordinate.
- (b) The position vector when the particle reaches its maximum \(x\)-coordinate.

### Step-by-step solution:

#### Part (a): Velocity at Maximum \(x\)-coordinate

At the maximum \(x\)-coordinate, the velocity in the \(x\)-direction is 0. This is because the particle stops moving forward in the \(x\)-direction before reversing direction due to the negative acceleration.

For the velocity in the \(y\)-direction, we use the first equation of motion for velocity:
\[
v_y = v_{0y} + a_y t
\]
Since the initial velocity in the \(y\)-direction is zero (\(v_{0y} = 0\)):
\[
v_y = 0 + (-4.46)t = -4.46t
\]

The time \(t\) when the particle reaches the maximum \(x\)-coordinate can be found using the equation for motion in the \(x\)-direction. At this point, the \(x\)-component of velocity becomes 0:
\[
v_x = v_{0x} + a_x t
\]
Setting \(v_x = 0\) (since the particle stops moving in the \(x\)-direction):
\[
0 = 1.95 + (-4.47)t
\]
Solving for \(t\):
\[
t = \frac{1.95}{4.47} \approx 0.436 \, \text{seconds}
\]

Now substitute this \(t\) value into the equation for \(v_y\):
\[
v_y = -4.46 \times 0.436 \approx -1.945 \, \text{m/s}
\]
Thus, the velocity at the maximum \(x\)-coordinate is \(\vec{v} = (0 \hat{i} - 1.945 \hat{j}) \, \text{m/s}\).

#### Part (b): Position Vector at Maximum \(x\)-coordinate

We now find the position vector when the particle reaches the maximum \(x\)-coordinate. The position vector is given by:
\[
\vec{r} = \vec{r_0} + \vec{v_0} t + \frac{1}{2} \vec{a} t^2
\]
Since the particle starts at the origin (\(\vec{r_0} = 0\)):
\[
r_x = v_{0x} t + \frac{1}{2} a_x t^2
\]
\[
r_y = v_{0y} t + \frac{1}{2} a_y t^2
\]

For the \(x\)-component:
\[
r_x = (1.95)(0.436) + \frac{1}{2}(-4.47)(0.436)^2
\]
Calculating each term:
\[
r_x = 0.8502 + \frac{1}{2}(-4.47)(0.190096) \approx 0.8502 - 0.4248 = 0.4254 \, \text{m}
\]

For the \(y\)-component:
\[
r_y = \frac{1}{2}(-4.46)(0.436)^2
\]
\[
r_y = \frac{1}{2}(-4.46)(0.190096) = -0.4248 \, \text{m}
\]

Thus, the position vector is \(\vec{r} = (0.425 \hat{i} - 0.425 \hat{j}) \, \text{m}\).

### Final Answers:
- (a) The velocity at the maximum \(x\)-coordinate is \( \vec{v} = (0 \hat{i} - 1.945 \hat{j}) \, \text{m/s}\).
- (b) The position vector at the maximum \(x\)-coordinate is \(\vec{r} = (0.425 \hat{i} - 0.425 \hat{j}) \, \text{m}\).

Would you like any further details or have questions on the steps?

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Here are some related questions to deepen understanding:
1. What does the sign of the acceleration vector indicate about the particle's motion?
2. How does the particle's initial velocity affect its maximum \(x\)-coordinate?
3. What would happen to the motion if the acceleration in the \(x\)-direction were zero?
4. How can we interpret the position vector geometrically at the maximum \(x\)-coordinate?
5. What additional forces might affect the particle in a real-world scenario?

**Tip:** When calculating kinematic quantities, always check whether you need to treat each component (like \(x\) and \(y\)) separately.

A particle leaves the origin with an initial velocity of \(\vec{v_0} = (1.95 \hat{i})\) m/s and a constant acceleration \(\vec{a} = (-4.47 \hat{i} - 4.46 \hat{j})\) m/s². When the particle reaches its maximum x-coordinate, what are (a) its velocity, (b) its position vector?

Find the velocity and position vector of a particle with initial velocity of (1.95 i) m/s and constant acceleration (-4.47 i - 4.46 j) m/s² when it reaches its maximum x-coordinate. Detailed step-by-step kinematic equations are used to solve this problem.

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