Given that the position of the particle $x(t)$ is a twice-differentiable function, we are provided with the following conditions at $t = 2$:

• $x(2) < 0$: The position of the particle at $t = 2$ is negative, meaning the particle is located to the left of the origin on the $x$-axis.
• $x'(2) < 0$: The velocity of the particle at $t = 2$ is negative, meaning the particle is moving to the left (in the negative direction along the $x$-axis).
• $x''(2) < 0$: The acceleration of the particle at $t = 2$ is negative, meaning the velocity of the particle is decreasing at that moment, or the particle is decelerating in the negative direction.

Statements that must be true:

1. The particle is to the left of the origin.
   Since $x(2) < 0$, the particle is positioned to the left of the origin (on the negative side of the $x$-axis).

2. The particle is moving to the left.
   Since $x'(2) < 0$, the particle’s velocity is negative, meaning it is moving to the left.

3. The particle is slowing down.
   Since $x'(2) < 0$ and $x''(2) < 0$, the particle is moving in the negative direction and is decelerating, which implies it is slowing down.

Conclusion:

The following statements must be true about the particle at time $t = 2$:

1. The particle is to the left of the origin.
2. The particle is moving to the left.
3. The particle is slowing down.

Would you like further explanation or details on any of these conclusions?

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Here are 5 related questions to expand on this information:

1. What is the physical interpretation of $x'(t)$ and $x''(t)$?
2. How can we determine when a particle is speeding up or slowing down?
3. Can the particle have positive acceleration but still move to the left?
4. What happens when the velocity and acceleration have opposite signs?
5. How can the graph of $x(t)$ help visualize the particle’s motion?

Tip: A particle slows down when its velocity and acceleration have opposite signs. It speeds up when they have the same sign.