From the origin, a particle starts at t = 0 s with a velocity `vecv=7.0hatim//s` and moves in the xy plane with a constant acceleration of `veca=(-9.0hati+3.0hatj)m//s^(2)`. At the time the particle reaches the maximum x coordinate, what is its (a) velocity and (b) position vector?

Explanation

To solve the problem step by step, we will analyze the motion of the particle in both the x and y directions separately.

### Given Data:
- Initial velocity: \(\vec{v}_0 = 7.0 \hat{i} \, \text{m/s}\)
- Acceleration: \(\vec{a} = -9.0 \hat{i} + 3.0 \hat{j} \, \text{m/s}^2\)

### Step 1: Determine the time at which the particle reaches the maximum x-coordinate.
At the maximum x-coordinate, the final velocity in the x-direction (\(v_{fx}\)) will be 0. We can use the equation of motion:
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