We are tasked with analyzing the motion of a particle given $s(t) = t^3 - 13.5t^2 + 54t$. To determine when the particle is speeding up and slowing down, we will analyze the velocity $v(t)$ and acceleration $a(t)$. The particle is:

1. Speeding up when the velocity and acceleration have the same sign (both positive or both negative).
2. Slowing down when the velocity and acceleration have opposite signs.

Step 1: Compute the velocity $v(t)$ and acceleration $a(t)$

The velocity is the first derivative of $s(t)$, and the acceleration is the second derivative:

$v(t) = \frac{ds(t)}{dt} = 3t^2 - 27t + 54$

$a(t) = \frac{dv(t)}{dt} = 6t - 27$

Step 2: Analyze the velocity $v(t)$

We find the critical points of $v(t)$ by setting $v(t) = 0$:

$3t^2 - 27t + 54 = 0$

Dividing through by 3:

$t^2 - 9t + 18 = 0$

Factoring:

$(t - 3)(t - 6) = 0$

Thus, $v(t) = 0$ at $t = 3$ and $t = 6$.

To analyze the sign of $v(t)$, test intervals around these critical points: $(0, 3)$, $(3, 6)$, and $(6, \infty)$.

• For $t \in (0, 3)$, choose $t = 1$: $v(1) = 3(1)^2 - 27(1) + 54 = 30$ (positive).
• For $t \in (3, 6)$, choose $t = 4$: $v(4) = 3(4)^2 - 27(4) + 54 = -6$ (negative).
• For $t \in (6, \infty)$, choose $t = 7$: $v(7) = 3(7)^2 - 27(7) + 54 = 30$ (positive).

Thus, $v(t) > 0$ on $(0, 3) \cup (6, \infty)$ and $v(t) < 0$ on $(3, 6)$.

Step 3: Analyze the acceleration $a(t)$

We find the critical point of $a(t)$ by setting $a(t) = 0$:

$6t - 27 = 0 \implies t = 4.5$

To analyze the sign of $a(t)$, test intervals around $t = 4.5$:

• For $t \in (0, 4.5)$, choose $t = 1$: $a(1) = 6(1) - 27 = -21$ (negative).
• For $t \in (4.5, \infty)$, choose $t = 5$: $a(5) = 6(5) - 27 = 3$ (positive).

Thus, $a(t) < 0$ on $(0, 4.5)$ and $a(t) > 0$ on $(4.5, \infty)$.

Step 4: Determine when the particle is speeding up or slowing down

• Speeding up: When $v(t)$ and $a(t)$ have the same sign:

  • $t \in (0, 3)$: $v(t) > 0$ and $a(t) < 0$ (not speeding up).
  • $t \in (3, 4.5)$: $v(t) < 0$ and $a(t) < 0$ (speeding up).
  • $t \in (6, \infty)$: $v(t) > 0$ and $a(t) > 0$ (speeding up).

  Therefore, the particle is speeding up on $(3, 4.5) \cup (6, \infty)$.

• Slowing down: When $v(t)$ and $a(t)$ have opposite signs:

  • $t \in (0, 3)$: $v(t) > 0$ and $a(t) < 0$ (slowing down).
  • $t \in (4.5, 6)$: $v(t) < 0$ and $a(t) > 0$ (slowing down).

  Therefore, the particle is slowing down on $(0, 3) \cup (4.5, 6)$.

Final Answer

• Speeding up: $(3, 4.5) \cup (6, \infty)$
• Slowing down: $(0, 3) \cup (4.5, 6)$

Let me know if you'd like further clarification or details on any step!

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Here are 5 related questions to expand your understanding:

1. How do we determine the intervals of increasing and decreasing velocity?
2. What is the significance of critical points in $v(t)$ and $a(t)$?
3. How do the signs of velocity and acceleration relate to the concept of speeding up or slowing down?
4. How would the analysis change if the acceleration equation had more roots?
5. Can a particle simultaneously have zero velocity and be speeding up?

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