The average rate of change of a function $s(t)$ over an interval $[t_1, t_2]$ is given by the formula:

$\text{Average rate of change} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$

For the function $s(t) = -9.8t^2 + 50t - 14$, and the interval $t_1 = 1$ and $t_2 = 3$, follow these steps:

Step 1: Find $s(1)$

$s(1) = -9.8(1)^2 + 50(1) - 14 = -9.8 + 50 - 14 = 26.2$

Step 2: Find $s(3)$

$s(3) = -9.8(3)^2 + 50(3) - 14 = -9.8(9) + 150 - 14 = -88.2 + 150 - 14 = 47.8$

Step 3: Calculate the average rate of change

$\text{Average rate of change} = \frac{s(3) - s(1)}{3 - 1} = \frac{47.8 - 26.2}{2} = \frac{21.6}{2} = 10.8$

Thus, the average rate of change of $s(t)$ between $t = 1$ and $t = 3$ is 10.8 units per second.

Would you like further details or have any questions about this solution?

Here are 5 related questions you might find useful:

1. How would you find the instantaneous rate of change of $s(t)$ at a specific point, say $t = 2$?
2. What is the physical interpretation of this average rate of change in a motion scenario?
3. How would you determine if the object’s velocity is increasing or decreasing between $t = 1$ and $t = 3$?
4. What would happen to the average rate of change if the interval were $t = 0$ to $t = 4$?
5. Can you apply this method to other types of functions, such as exponential or logarithmic functions?

Tip: The average rate of change of a quadratic function on a symmetric interval gives a clue about the function's behavior (increasing or decreasing) around the midpoint of the interval.