The problem you provided involves the movement of a particle along the x-axis, where the position is given by the function $s(t)$ and several conditions on its velocity $s'(t)$ are given. Let's analyze the details:

• $s'(t) > 0$: The velocity is positive on the intervals $(-2, 0) \cup (3, 5) \cup (5, \infty)$. This means the particle is moving to the right in these intervals.

• $s'(t) < 0$: The velocity is negative on the intervals $(-\infty, -2) \cup (0, 3)$. This means the particle is moving to the left in these intervals.

• $s'(t) = 0$ at $t = -2, 0, 3, 5$. These are critical points where the particle's velocity is zero, which could indicate a change in direction or a momentary stop.

To determine which statements are true, we need to evaluate the behavior of $s(t)$ and $s'(t)$ based on the information provided:

Critical points:

• At $t = -2$, $t = 0$, $t = 3$, and $t = 5$, $s'(t) = 0$, meaning the particle is momentarily at rest.
• From the velocity sign changes around these points, we can determine whether the particle is speeding up, slowing down, or reversing direction at these points.

If you want to specify the answer options or clarify further questions, I can help you analyze each one in more detail. Would you like a more detailed explanation or analysis?

Here are 5 questions to expand understanding:

1. What does it mean for the particle’s velocity to be positive or negative in terms of its motion?
2. How can the critical points where $s'(t) = 0$ help determine the behavior of the particle's position?
3. What is the significance of changes in the sign of $s'(t)$ between intervals?
4. How would you identify when the particle is moving at its fastest or slowest?
5. How does the concept of concavity relate to the intervals where the particle is speeding up or slowing down?

Tip: To fully understand particle motion, always relate velocity $s'(t)$ to speed and direction, and acceleration $s''(t)$ to changes in velocity.