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Part 1
The position of an object moving along a line is given by the function s left parenthesis t right parenthesis equals negative 4 t squared plus 20 t. Find the average velocity of the object over the following intervals.
 
​(a)​ [1, 7​]
​(b) ​[1, 6​]
​(c)​ [1, 5​]
​(d) ​[1, 1plus​h] where hgreater than0 is any real number.
Question content area bottom
Part 1
​(a) The average velocity of the object over the interval​ [1, 7​] is 
  
enter your response here.

To find the **average velocity** of the object over a given interval \([a, b]\), we use the formula: 

\[
v_{\text{avg}} = \frac{s(b) - s(a)}{b - a}
\]

The position function is given as:
\[
s(t) = -4t^2 + 20t
\]

We calculate each part step by step:

---

### **(a) Interval \([1, 7]\)**

1. Compute \(s(7)\) and \(s(1)\):
   \[
   s(7) = -4(7)^2 + 20(7) = -4(49) + 140 = -196 + 140 = -56
   \]
   \[
   s(1) = -4(1)^2 + 20(1) = -4(1) + 20 = -4 + 20 = 16
   \]

2. Substitute into the average velocity formula:
   \[
   v_{\text{avg}} = \frac{s(7) - s(1)}{7 - 1} = \frac{-56 - 16}{6} = \frac{-72}{6} = -12
   \]

Thus, the average velocity over \([1, 7]\) is:
\[
\boxed{-12}
\]

---

### **(b) Interval \([1, 6]\)**

1. Compute \(s(6)\) and \(s(1)\):
   \[
   s(6) = -4(6)^2 + 20(6) = -4(36) + 120 = -144 + 120 = -24
   \]
   \[
   s(1) = 16 \quad \text{(already computed above).}
   \]

2. Substitute into the average velocity formula:
   \[
   v_{\text{avg}} = \frac{s(6) - s(1)}{6 - 1} = \frac{-24 - 16}{5} = \frac{-40}{5} = -8
   \]

Thus, the average velocity over \([1, 6]\) is:
\[
\boxed{-8}
\]

---

### **(c) Interval \([1, 5]\)**

1. Compute \(s(5)\) and \(s(1)\):
   \[
   s(5) = -4(5)^2 + 20(5) = -4(25) + 100 = -100 + 100 = 0
   \]
   \[
   s(1) = 16 \quad \text{(already computed above).}
   \]

2. Substitute into the average velocity formula:
   \[
   v_{\text{avg}} = \frac{s(5) - s(1)}{5 - 1} = \frac{0 - 16}{4} = \frac{-16}{4} = -4
   \]

Thus, the average velocity over \([1, 5]\) is:
\[
\boxed{-4}
\]

---

### **(d) Interval \([1, 1 + h]\)**

1. Compute \(s(1 + h)\) and \(s(1)\):
   \[
   s(1 + h) = -4(1 + h)^2 + 20(1 + h)
   \]
   Expand:
   \[
   s(1 + h) = -4(1 + 2h + h^2) + 20(1 + h) = -4 - 8h - 4h^2 + 20 + 20h
   \]
   Simplify:
   \[
   s(1 + h) = 16 + 12h - 4h^2
   \]
   \[
   s(1) = 16 \quad \text{(already computed above).}
   \]

2. Substitute into the average velocity formula:
   \[
   v_{\text{avg}} = \frac{s(1 + h) - s(1)}{(1 + h) - 1}
   \]
   Simplify:
   \[
   v_{\text{avg}} = \frac{(16 + 12h - 4h^2) - 16}{h} = \frac{12h - 4h^2}{h}
   \]
   Factor and simplify:
   \[
   v_{\text{avg}} = 12 - 4h
   \]

Thus, the average velocity over \([1, 1 + h]\) is:
\[
\boxed{12 - 4h}
\]

---

### Summary of Results:
- (a) \(-12\)
- (b) \(-8\)
- (c) \(-4\)
- (d) \(12 - 4h\)

Let me know if you'd like more details or have any further questions! Here are some related questions for exploration:

1. What is the instantaneous velocity of the object at \(t = 1\)?
2. How does the average velocity behave as \(h \to 0\) in part (d)?
3. Can you determine when the object changes direction based on the position function?
4. What is the displacement of the object over the interval \([1, 7]\)?
5. How would the results change if the position function had a constant term?

**Tip:** The average velocity formula closely resembles the slope of the secant line; interpreting it geometrically helps build intuition.

The position of an object moving along a line is given by the function s(t) = -4t^2 + 20t. Find the average velocity of the object over the following intervals.

(a) [1, 7]
(b) [1, 6]
(c) [1, 5]
(d) [1, 1+h] where h > 0 is any real number.

Learn how to calculate the average velocity of an object moving along a line, given the position function s(t) = -4t^2 + 20t. This problem covers the calculation of average velocity over different intervals, with detailed steps and solutions for intervals [1, 7], [1, 6], [1, 5], and [1, 1+h]. Suitable for students in Grades 9-11.

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