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`?

Options

A. `t=2s`
B. `t=4s`
C. `t=3s`
D. `t=1s`

Explanation

To solve the problem, we need to determine the time at which the x-coordinate of the particle reaches 16 meters. We will use the equations of motion to find this time.

### Step-by-step Solution:

1. **Identify the Given Information:**
- Initial velocity in the y-direction, \( u_y = 10.0 \, \hat{j} \, \text{m/s} \)
- Initial velocity in the x-direction, \( u_x = 0 \, \hat{i} \, \text{m/s} \)
- Acceleration in the x-direction, \( a_x = 8 \, \hat{i} \, \text{m/s}^2 \)
- Acceleration in the y-direction, \( a_y = 2 \, \hat{j} \, \text{m/s}^2 \)
- We need to find the time \( t \) when the x-coordinate \( x = 16 \, \text{m} \).

2. **Use the Equation of Motion for the x-coordinate:**
The equation for the displacement in the x-direction is given by:
\[ x = u_x t + \frac{1}{2} a_x t^2 \]
Substituting the known values:
\[ 16 = 0 \cdot t + \frac{1}{2} \cdot 8 \cdot t^2 \]
This simplifies to:
\[ 16 = 4 t^2 \]

3. **Solve for \( t^2 \):**
Rearranging the equation gives:
\[ t^2 = \frac{16}{4} = 4 \]

4. **Find \( t \):**
Taking the square root of both sides:
\[ t = \sqrt{4} = 2 \, \text{s} \]
Since time cannot be negative, we take the positive root.

5. **Conclusion:**
The time at which the x-coordinate of the particle is 16 meters is \( t = 2 \, \text{s} \).