To find the average velocity as $t$ changes from $t_1 = 3$ seconds to $t_2 = 6$ seconds, we can use the formula for average rate of change:

$\text{Average Velocity} = \frac{d(t_2) - d(t_1)}{t_2 - t_1}$

Given the distance function:

$d(t) = 17t^2$

we first calculate $d(3)$ and $d(6)$:

1. Calculate $d(3)$: $d(3) = 17(3^2) = 17(9) = 153 \text{ feet}$

2. Calculate $d(6)$: $d(6) = 17(6^2) = 17(36) = 612 \text{ feet}$

Now, substitute these values into the average velocity formula:

$\text{Average Velocity} = \frac{612 - 153}{6 - 3} = \frac{459}{3} = 153 \text{ feet/second}$

Thus, the average velocity as $t$ changes from 3 to 6 seconds is

$\boxed{153 \text{ feet/second}}$

Do you want more details or have any questions? Here are some related questions you might consider:

1. How does the average velocity relate to the instantaneous velocity?
2. What would be the distance covered if the object fell for 10 seconds?
3. How would the average velocity change if the distance function were different?
4. What factors could affect the distance fallen on different planets?
5. Can you explain the concept of acceleration in relation to this problem?

Tip: Always check if the function you're working with is linear or quadratic, as it affects the calculations of average and instantaneous rates.