The problem given is about an object dropped from rest, and its distance is described by the function $d(t) = 14t^2$ feet, where $t$ represents time in seconds. The task is to find the average rate of change of distance (which is the average velocity) between $t_1 = 3$ seconds and $t_2 = 6$ seconds.

Step-by-Step Solution:

1. Distance function:
   $d(t) = 14t^2$

2. Average velocity formula:
   The average velocity between times $t_1$ and $t_2$ is given by: $\text{Average velocity} = \frac{d(t_2) - d(t_1)}{t_2 - t_1}$

3. Calculate $d(t_1)$ and $d(t_2)$:

   • At $t_1 = 3$ seconds:
     $d(3) = 14(3^2) = 14(9) = 126 \, \text{feet}$
   • At $t_2 = 6$ seconds:
     $d(6) = 14(6^2) = 14(36) = 504 \, \text{feet}$

4. Compute the average velocity: $\text{Average velocity} = \frac{d(6) - d(3)}{6 - 3} = \frac{504 - 126}{3} = \frac{378}{3} = 126 \, \text{feet per second}$

Thus, the average velocity as the object falls between 3 seconds and 6 seconds is 126 feet per second, which matches the value shown in your image.

Do you want further details or have any questions? Here are some related questions for further exploration:

1. What would the average velocity be if the interval changed to $t_1 = 2$ and $t_2 = 5$?
2. How does the object's velocity change as time increases?
3. What does the derivative of the distance function represent in this context?
4. Can we find the object's instantaneous velocity at $t = 4$ seconds?
5. How would the average velocity change if the gravitational constant on Planet X were different?

Tip: The average velocity formula is essentially the slope of the secant line between two points on the graph of the distance function.