Find the average rate of Change of s(t)=-9.8t^2+50t-14 between t=1 and t=3

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.

Find the average rate of change of s(t) = -9.8t^2 + 50t - 14 between t = 1 and t = 3.

Learn how to calculate the average rate of change for the quadratic function s(t) = -9.8t^2 + 50t - 14 over the interval [1, 3]. This solution provides a detailed step-by-step guide, applying core calculus concepts. Ideal for high school students in Grades 10-12.

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