A particle starts from the origin at `t=Os` with a velocity of `10.0 hatj m//s` and moves in the `xy`-plane with a constant acceleration of `(8hati+2hatj)m//s^(-2)`. What time is the `x`-coordinate of the particle `16m`?

Explanation

To solve the problem, we need to find the time at which the x-coordinate of the particle is 16 meters. The particle starts at the origin with an initial velocity and is subject to constant acceleration. Let's break down the solution step by step.

### Step 1: Identify the given information
- Initial position: \( x_0 = 0 \, m \) (origin)
- Initial velocity in the x-direction: \( u_x = 0 \, m/s \) (since the initial velocity is only in the y-direction)
- Initial velocity in the y-direction: \( u_y = 10 \, \hat{j} \, m/s \)
- Acceleration in the x-direction: \( a_x = 8 \, \hat{i} \, m/s^2 \)
- Acceleration in the y-direction: \( a_y = 2 \, \hat{j} \, m/s^2 \)
...