The position function of the rock is given as $s(t) = -5t^2 + 100$. To find the average velocity over the intervals $[2.999, 3.0]$ and $[3.0, 3.001]$, we use the formula for average velocity:

$\text{Average velocity} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$

Interval $[2.999, 3.0]$:

• $t_1 = 2.999$, $t_2 = 3.0$
• $s(2.999) = -5(2.999)^2 + 100$
• $s(3.0) = -5(3.0)^2 + 100$

Let's calculate:

$s(2.999) = -5(2.999)^2 + 100 = -5(8.994001) + 100 = -44.970005 + 100 = 55.029995$ $s(3.0) = -5(9.0) + 100 = -45 + 100 = 55$ Now, the average velocity for this interval is:

$\text{Average velocity} = \frac{55 - 55.029995}{3.0 - 2.999} = \frac{-0.029995}{0.001} = -29.995 \approx -30$

Interval $[3.0, 3.001]$:

• $t_1 = 3.0$, $t_2 = 3.001$
• $s(3.001) = -5(3.001)^2 + 100$
• $s(3.0) = -5(3.0)^2 + 100 = 55$

$s(3.001) = -5(3.001)^2 + 100 = -5(9.006001) + 100 = -45.030005 + 100 = 54.969995$ Now, the average velocity for this interval is:

$\text{Average velocity} = \frac{54.969995 - 55}{3.001 - 3.0} = \frac{-0.030005}{0.001} = -30.005 \approx -30$

Instantaneous velocity at $t = 3.0$:

The instantaneous velocity is approximated by the average velocity over these small intervals. Since both intervals give approximately $-30$, we conclude that the instantaneous velocity at $t = 3.0$ is approximately:

$\boxed{-30 \text{ m/s}}$

Would you like more details or have any further questions? Here are some related questions for practice:

1. What is the velocity of the rock at $t = 2.0$?
2. Find the position of the rock at $t = 4.0$.
3. What is the total distance traveled by the rock in the first 3 seconds?
4. How do you interpret the negative velocity in terms of the rock's motion?
5. At what time does the rock hit the ground?

Tip: Instantaneous velocity can be found as the derivative of the position function $s(t)$, which represents the rate of change of position at any specific time $t$.