To solve the problem, we need to compute both the average velocity over the time interval $[2, 4]$ and the instantaneous velocity at $t = 2$.

1. Average Velocity Over the Interval $[2, 4]$

The average velocity over the time interval $[t_1, t_2]$ is given by:

$v_{\text{avg}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$

Here, $s(t) = 3t^3$. Let's calculate $s(2)$ and $s(4)$:

$s(2) = 3(2)^3 = 3 \times 8 = 24$ $s(4) = 3(4)^3 = 3 \times 64 = 192$

Now, calculate the average velocity:

$v_{\text{avg}} = \frac{192 - 24}{4 - 2} = \frac{168}{2} = 84$

So, the average velocity over the interval $[2, 4]$ is 84.

2. Instantaneous Velocity at $t = 2$

The instantaneous velocity is found by taking the derivative of $s(t)$ with respect to $t$, which gives us the velocity function $v(t)$:

$v(t) = \frac{d}{dt}[3t^3] = 9t^2$

Now, substitute $t = 2$ into the velocity function:

$v(2) = 9(2)^2 = 9 \times 4 = 36$

So, the instantaneous velocity at $t = 2$ is 36.

Summary:

• Average Velocity over $[2, 4]$: 84
• Instantaneous Velocity at $t = 2$: 36

Would you like any additional details or have any questions?

Questions to Expand Understanding:

1. How would the average velocity change if the interval were $[1, 3]$?
2. What is the general formula for average velocity for a function $s(t) = at^n$?
3. How do you interpret the difference between average and instantaneous velocity?
4. How would the instantaneous velocity change if the position function were $s(t) = 3t^4$?
5. What is the physical significance of the derivative in velocity calculations?
6. How would you find the position function if the velocity function were given instead?
7. Can the average velocity ever be negative? Under what conditions?
8. How would you calculate the acceleration of the particle at $t = 2$?

Tip:

Remember, the average velocity depends on the change in position over the entire interval, while the instantaneous velocity is the slope of the position function at a specific point in time.