To address this problem, we’ll start by analyzing the position function $s(t) = 3t^4 - 16t^3 + 24t^2$. The steps will involve calculating the derivative of $s(t)$ to determine the velocity function, and then analyzing where this derivative is positive, negative, or zero.

Let’s go through each part of the problem:

1. Calculate the velocity function $v(t)$

Since velocity is the derivative of position $s(t)$, we find: $v(t) = \frac{ds}{dt} = 12t^3 - 48t^2 + 48t.$

We can factor $v(t)$ to simplify analysis: $v(t) = 12t(t^2 - 4t + 4) = 12t(t - 2)^2.$

2. Determine intervals where the object moves to the right or left

The object moves to the right where $v(t) > 0$ and to the left where $v(t) < 0$. Set $v(t) = 0$ to find critical points: $12t(t - 2)^2 = 0.$ This equation gives $t = 0$ and $t = 2$ as critical points.

We will test intervals around these points to determine the sign of $v(t)$:

• For $t < 0$: $v(t)$ is undefined because $t \geq 0$ by the problem's domain.
• For $0 < t < 2$: $v(t) > 0$, so the object moves to the right.
• For $t > 2$: $v(t) = 12t \cdot \text{positive quantity} > 0$, so the object continues moving to the right.

Answer:

• The object is moving to the right on $(0, 2) \cup (2, \infty)$.
• The object is not moving to the left on any interval (answer is $\varnothing$).

3. Determine when the object reverses direction

The object reverses direction when $v(t)$ changes sign. Since $v(t)$ does not change from positive to negative at any point in the domain $t \geq 0$, the object does not reverse direction.

Answer: $t = \text{DNE}$ (does not exist).

4. Find intervals when the velocity is increasing or decreasing

To determine where the velocity is increasing or decreasing, we analyze the acceleration, which is the derivative of velocity $v(t)$.

$a(t) = \frac{dv}{dt} = 36t^2 - 96t + 48 = 12(3t^2 - 8t + 4).$

We’ll factor or use the quadratic formula to solve $a(t) = 0$: $3t^2 - 8t + 4 = 0.$ Solving for $t$: $t = \frac{8 \pm \sqrt{64 - 48}}{6} = \frac{8 \pm 4}{6}.$ Thus, $t = 2$ or $t = \frac{2}{3}$.

Analyzing $a(t)$ in intervals around $t = \frac{2}{3}$ and $t = 2$:

• For $0 < t < \frac{2}{3}$: $a(t) > 0$, so $v(t)$ is increasing.
• For $\frac{2}{3} < t < 2$: $a(t) < 0$, so $v(t)$ is decreasing.
• For $t > 2$: $a(t) > 0$, so $v(t)$ is increasing again.

Answer:

• The velocity is increasing on $(0, \frac{2}{3}) \cup (2, \infty)$.
• The velocity is decreasing on $\left(\frac{2}{3}, 2\right)$.

Would you like further clarification on any of these steps or have additional questions?

Additional Related Questions

1. How is the sign of the velocity related to the direction of motion?
2. What does the acceleration tell us about the changes in velocity?
3. Why does the object not reverse direction in this problem?
4. How can we find the exact time when the velocity reaches zero?
5. What other information can we derive from the second derivative (acceleration)?

Tip: Always verify critical points by testing intervals, as this helps confirm the behavior of the function in each region.