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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.

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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.

Find the average velocity for the function s(t) = t^3 - 4t with a = 2 and b = 3.

This problem guides you through calculating the average velocity of the function s(t) = t^3 - 4t over the interval [2, 3]. Using the formula for average velocity, we find the displacement over the given time interval and solve step-by-step.

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