The problem asks for the average velocity of an object with height given by the function:

$h(x) = -16x^2 + 149x + 94$

where $x$ is time in seconds and $h(x)$ is height in feet. You are tasked with finding the average velocity between $x = 1$ and $x = 1.3$.

Average Velocity Formula

The average velocity between two points $x_1$ and $x_2$ is given by the formula:

$v_{\text{avg}} = \frac{h(x_2) - h(x_1)}{x_2 - x_1}$

Step-by-Step Solution

1. Find $h(1)$: Substitute $x = 1$ into the function:

   $h(1) = -16(1)^2 + 149(1) + 94 = -16 + 149 + 94 = 227 \, \text{feet}$

2. Find $h(1.3)$: Substitute $x = 1.3$ into the function:

   $h(1.3) = -16(1.3)^2 + 149(1.3) + 94$ $h(1.3) = -16(1.69) + 193.7 + 94 = -27.04 + 193.7 + 94 = 260.66 \, \text{feet}$

3. Calculate the average velocity:

   $v_{\text{avg}} = \frac{h(1.3) - h(1)}{1.3 - 1} = \frac{260.66 - 227}{0.3} = \frac{33.66}{0.3} = 112.2 \, \text{feet per second}$

Thus, the average velocity between $x = 1$ and $x = 1.3$ is approximately 112.2 feet per second.

Let me know if you want further clarifications or details!

Related Questions:

1. How is the concept of average velocity different from instantaneous velocity?
2. Can you calculate the instantaneous velocity at $x = 1$ using derivatives?
3. How would the average velocity change if the interval was $x = 1.3$ to $x = 1.5$?
4. How does the height function reflect the motion of the object (e.g., does it represent free fall)?
5. What is the significance of the coefficients in the quadratic height function?

Tip:

When calculating average velocity over a small interval, you're approximating the instantaneous velocity at the midpoint of the interval!