The height of a bottle rocket, in meters, is given by the function
s
(
t
)
=
−
15
t
2
+
64
t
+
554
. Find the average velocity of the bottle rocket over the intervals
[
4.999
,
5.0
]
and
[
5.0
,
5.001
]
. Use this information to approximate the instantaneous velocity of the bottle rocket at time
t
=
5.0
.

(Round your answer to the nearest integer if necessary.)

To solve this problem, we'll use the formula for the average velocity over an interval \([t_1, t_2]\) for the function \(s(t)\):

\[
\text{Average Velocity} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}
\]

We are asked to find the average velocity over two intervals: \([4.999, 5.0]\) and \([5.0, 5.001]\).

Given the height function:

\[
s(t) = -15t^2 + 64t + 554
\]

### Step 1: Calculate \(s(t)\) at the relevant points
1. \(s(4.999)\)
2. \(s(5.0)\)
3. \(s(5.001)\)

### Step 2: Calculate the average velocities over the given intervals
1. Interval \([4.999, 5.0]\)
   \[
   \text{Average Velocity} = \frac{s(5.0) - s(4.999)}{5.0 - 4.999}
   \]
2. Interval \([5.0, 5.001]\)
   \[
   \text{Average Velocity} = \frac{s(5.001) - s(5.0)}{5.001 - 5.0}
   \]

### Step 3: Use the average velocities to approximate the instantaneous velocity at \(t = 5.0\)

Let's calculate this step by step. 

#### Step 1: Evaluate \(s(t)\) at the relevant points
1. \(s(4.999) = -15(4.999)^2 + 64(4.999) + 554\)
2. \(s(5.0) = -15(5.0)^2 + 64(5.0) + 554\)
3. \(s(5.001) = -15(5.001)^2 + 64(5.001) + 554\)

I'll calculate these values and then use them to find the average velocities.The function \(s(t)\) evaluated at the relevant points yields:

- \(s(4.999) = 499.086\) (approximately)
- \(s(5.0) = 499.0\)
- \(s(5.001) = 498.914\) (approximately)

### Step 2: Calculate the average velocities

1. **Interval \([4.999, 5.0]\):**
   \[
   \text{Average Velocity} = \frac{499.0 - 499.086}{5.0 - 4.999} = \frac{-0.086}{0.001} = -86 \text{ m/s}
   \]

2. **Interval \([5.0, 5.001]\):**
   \[
   \text{Average Velocity} = \frac{498.914 - 499.0}{5.001 - 5.0} = \frac{-0.086}{0.001} = -86 \text{ m/s}
   \]

### Step 3: Approximate the instantaneous velocity at \(t = 5.0\)

Since the average velocities over the intervals \([4.999, 5.0]\) and \([5.0, 5.001]\) are both \(-86\) m/s, we can approximate the instantaneous velocity at \(t = 5.0\) as:

\[
\text{Instantaneous Velocity at } t = 5.0 \text{ is } -86 \text{ m/s}.
\]

Would you like more details on any of these steps or have further questions?

---

**Related Questions:**
1. What is the formula for instantaneous velocity?
2. How does average velocity differ from instantaneous velocity?
3. What is the significance of negative velocity in this context?
4. Can we find the exact derivative of the height function to confirm the instantaneous velocity?
5. How does changing the interval affect the average velocity approximation?

**Tip:** When approximating instantaneous velocity, using smaller intervals around the point of interest usually yields more accurate results.

Explore how to calculate the average velocity of a bottle rocket over two intervals and approximate its instantaneous velocity using a quadratic height function. Detailed steps and formulas included.

Math Problem Statement

The height of a bottle rocket, in meters, is given by the function s ( t )

− 15 t 2 + 64 t + 554 . Find the average velocity of the bottle rocket over the intervals [ 4.999 , 5.0 ] and [ 5.0 , 5.001 ] . Use this information to approximate the instantaneous velocity of the bottle rocket at time t

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