The problem asks to find the average velocity of the function $s(t) = t^3 - 4t$ on the interval $[a, b]$, where $a = 2$ and $b = 3$.

Average Velocity Formula:

The average velocity over the interval $[a, b]$ is given by the formula: $v_{\text{avg}} = \frac{s(b) - s(a)}{b - a}$ where $s(t)$ is the position function.

Step-by-Step Solution:

1. Given function: $s(t) = t^3 - 4t$
2. Evaluate at $t = 3$: $s(3) = 3^3 - 4(3) = 27 - 12 = 15$
3. Evaluate at $t = 2$: $s(2) = 2^3 - 4(2) = 8 - 8 = 0$
4. Find the average velocity: $v_{\text{avg}} = \frac{s(3) - s(2)}{3 - 2} = \frac{15 - 0}{3 - 2} = \frac{15}{1} = 15$

Final Answer:

The average velocity over the interval $[2, 3]$ is $15$ units per time unit.

Would you like further details on this solution or have any questions?

Here are five related questions to expand your understanding:

1. How does average velocity differ from instantaneous velocity?
2. How would you find the instantaneous velocity at $t = 2$?
3. What is the significance of the position function in physics?
4. Can you apply the average velocity formula to non-linear motion?
5. How do you interpret the result of average velocity in real-world scenarios?

Tip: When calculating average velocity, remember that it's simply the total displacement over the time interval, not influenced by variations in speed during that interval.