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To solve this problem, we need to calculate the average velocity of the particle between times \( t_1 = 2 \, s \) and \( t_2 = 4 \, s \). 

### Step-by-step Solution:

1. **Understand the given equation:**
   The equation of motion provided is:
   \[
   x(t) = 15 \, m - (8.5 \, m/s) \, t + \frac{(1.8 \, m/s^2) \, t^2}{2}
   \]
   This is a quadratic equation representing the position \( x(t) \) of the particle as a function of time \( t \).

2. **Calculate \( x(t_1) \) and \( x(t_2) \):**
   - For \( t_1 = 2 \, s \):
     \[
     x(2) = 15 - (8.5)(2) + \frac{1.8}{2} (2^2)
     \]
   - For \( t_2 = 4 \, s \):
     \[
     x(4) = 15 - (8.5)(4) + \frac{1.8}{2} (4^2)
     \]

3. **Compute the average velocity \( v_m \):**
   The average velocity between two points in time is given by:
   \[
   v_m = \frac{x(t_2) - x(t_1)}{t_2 - t_1}
   \]

Let's compute these values step-by-step.The calculations provide the following results:

- Position at \( t_1 = 2 \, s \): \( x(2) = 1.6 \, m \)
- Position at \( t_2 = 4 \, s \): \( x(4) = -4.6 \, m \)

The average velocity \( v_m \) between \( t_1 = 2 \, s \) and \( t_2 = 4 \, s \) is:
\[
v_m = -3.1 \, m/s
\]

### Conclusion:
The correct answer is **C) \( v_m = -3.1 \, m/s \)**.

Would you like more details on this solution, or do you have any questions? Here are some related questions to consider:

1. What is the significance of a negative average velocity in this context?
2. How would the answer change if the acceleration was different?
3. How can we derive instantaneous velocity from the given position equation?
4. What if the initial velocity \( v_0 \) were positive? How would it affect the motion?
5. How do we calculate the displacement instead of the average velocity over a time interval?

**Tip:** Remember that average velocity over an interval can differ from instantaneous velocities at the boundaries due to varying acceleration.

La ecuación fundamental de movimiento de una partícula que se mueve a lo largo del eje x está dada por: x(t) = 15 m - (8.5 m/s) t + (1.8 m/s²) t² / 2. ¿Cuál de las siguientes opciones corresponde a la velocidad media de la partícula entre los tiempos t1 = 2 s y t2 = 4 s?

Learn how to calculate the average velocity of a particle given its quadratic position equation x(t) = 15 m - (8.5 m/s) t + (1.8 m/s²) t² / 2 between times t1 = 2 s and t2 = 4 s. The correct answer is v_m = -3.1 m/s, and this solution involves kinematic concepts and the average velocity formula.

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