To find the average velocity of the rock over the given time intervals, we'll use the formula for average velocity:

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

The height function is given by:

$y(t) = 16t - 1.86t^2$

We will compute the average velocity for the following intervals:

1. $[1, 2]$
2. $[1, 1.5]$
3. $[1, 1.1]$
4. $[1, 1.01]$
5. $[1, 1.001]$

Let's calculate these now.The average velocity of the rock over the given time intervals is:

1. $[1, 2]$: $v_{\text{avg}} = 10.42 \, \text{m/s}$
2. $[1, 1.5]$: $v_{\text{avg}} = 11.35 \, \text{m/s}$
3. $[1, 1.1]$: $v_{\text{avg}} = 12.09 \, \text{m/s}$
4. $[1, 1.01]$: $v_{\text{avg}} = 12.26 \, \text{m/s}$
5. $[1, 1.001]$: $v_{\text{avg}} = 12.28 \, \text{m/s}$

As the time interval narrows, the average velocity approaches a certain value, indicating the instantaneous velocity at $t = 1$.

Would you like further details or have any questions?

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Here are some related questions you might find interesting:

1. What is the significance of the instantaneous velocity at $t = 1$?
2. How does gravity on Mars affect the motion of the rock compared to Earth?
3. How would the height function change if the initial velocity was different?
4. What would be the effect of a longer time interval on the average velocity?
5. How would you find the exact instantaneous velocity at $t = 1$ using calculus?

Tip: As time intervals become smaller, the average velocity converges to the instantaneous velocity at a specific point.