To determine when the particle is speeding up, we need to evaluate the velocity $v(t)$ and acceleration $a(t)$ of the particle and find where their signs are the same (both positive or both negative). Here's the step-by-step process:

Step 1: Find the velocity $v(t)$

The velocity is the derivative of the position function $y(t)$: $v(t) = \frac{dy}{dt} = \frac{d}{dt}(2t^3 - 3t^2) = 6t^2 - 6t$ $v(t) = 6t(t - 1)$

Step 2: Find the acceleration $a(t)$

The acceleration is the derivative of the velocity function $v(t)$: $a(t) = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 6t) = 12t - 6$

Step 3: Analyze the velocity and acceleration signs

• Velocity $v(t) = 6t(t - 1)$:

  • $v(t) = 0$ when $t = 0$ or $t = 1$.
  • The sign of $v(t)$ depends on the intervals:
    • For $t \in (0, 1)$, $v(t) < 0$.
    • For $t > 1$, $v(t) > 0$.

• Acceleration $a(t) = 12t - 6$:

  • $a(t) = 0$ when $t = 0.5$.
  • The sign of $a(t)$ depends on the intervals:
    • For $t < 0.5$, $a(t) < 0$.
    • For $t > 0.5$, $a(t) > 0$.

Step 4: Determine when the particle is speeding up

The particle is speeding up when both velocity $v(t)$ and acceleration $a(t)$ have the same sign:

1. For $t \in (0, 0.5)$:

   • $v(t) < 0$ and $a(t) < 0$ → Speeding up.

2. For $t \in (1, \infty)$:

   • $v(t) > 0$ and $a(t) > 0$ → Speeding up.

Final Answer:

The particle is speeding up for: $t \in (0, 0.5) \cup (1, \infty)$

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Would you like further details or clarification?
Here are 5 related questions for practice:

1. How do you determine when a particle is slowing down?
2. What is the particle's position at $t = 2$?
3. Can a particle have zero velocity but nonzero acceleration?
4. What are the intervals when the particle is moving downward?
5. How would you graph $v(t)$ and $a(t)$?

Tip: Always check the signs of velocity and acceleration to determine motion characteristics like speeding up or slowing down!