To determine when the particle is moving toward the origin, we need to examine the position function $s(t) = t^3 - 3t^2 - 6$ and find when the velocity is negative while the position is positive (indicating it’s approaching zero) or vice versa.

1. Find the derivative $s'(t)$ to determine the velocity of the particle: $s'(t) = 3t^2 - 6t$

2. Factor the derivative to find when the velocity is zero (which indicates potential changes in direction): $s'(t) = 3t(t - 2) = 0$ This gives $t = 0$ and $t = 2$ as critical points.

3. Analyze intervals around the critical points to determine when $s'(t)$ is positive or negative:

   • For $t < 0$: $s'(t)$ is positive (moving away from the origin).
   • For $0 < t < 2$: $s'(t)$ is negative (moving toward the origin).
   • For $t > 2$: $s'(t)$ is positive again (moving away from the origin).

4. Evaluate the position at $t = 1.5$ (since it falls in the interval $0 < t < 2$):

   • If the particle is moving toward the origin within this interval, then $t = 1.5$ would be the correct answer.

Given this reasoning, the correct answer is: (B) $t = 1.5$ At a Dodger’s game.

Would you like further details, or do you have any questions?

Here are five related questions to explore this topic further:

1. How would we determine if the particle ever stops moving entirely?
2. What does the concavity of $s(t)$ tell us about the particle’s movement?
3. How would we determine the distance traveled by the particle over a specific interval?
4. Can we determine when the particle is at its farthest point from the origin?
5. How does this approach apply if we want to find when the particle is moving in the opposite direction?

Tip: Analyzing derivatives helps you understand the behavior of functions, particularly in physics problems involving motion.