The problem is asking to calculate average velocities over different time intervals for a given position function $s(t) = -4.9t^2 + 32t + 21$, and then make a conjecture about the instantaneous velocity at $t = 3$.

Steps to solve:

1. Average velocity formula: The average velocity over a time interval $[a, b]$ is given by: $\text{Average Velocity} = \frac{s(b) - s(a)}{b - a}$ Here, $a = 3$ and $b$ takes different values like 4, 3.1, 3.01, etc.

2. Position function evaluation: You'll need to plug in $t = a$ and $t = b$ into the position function $s(t)$ for each interval.

I'll compute the values for you. Let's start with the first interval $[3, 4]$ and work through each interval step by step.Here are the calculated average velocities for each time interval:

• For the interval $[3, 4]$, the average velocity is approximately $-2.3$.
• For the interval $[3, 3.1]$, the average velocity is approximately $2.11$.
• For the interval $[3, 3.01]$, the average velocity is approximately $2.551$.
• For the interval $[3, 3.001]$, the average velocity is approximately $2.5951$.
• For the interval $[3, 3.0001]$, the average velocity is approximately $2.5995$.

Conjecture:

As the time intervals become smaller and closer to $t = 3$, the average velocity approaches approximately $2.6$. Therefore, the instantaneous velocity at $t = 3$ is likely around $2.6$.

Would you like further details or clarifications on these calculations?

Here are 5 follow-up questions to deepen your understanding:

1. How do we compute instantaneous velocity from a position function?
2. Why does the average velocity approach a specific value as the time interval shrinks?
3. Can we find the exact instantaneous velocity at $t = 3$ using differentiation?
4. How would the velocity change if the position function had a different quadratic coefficient?
5. What happens to the average velocity if we extend the time intervals instead of shrinking them?

Tip: The instantaneous velocity is the derivative of the position function at a specific time.