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The problem involves an arrow shot upward on the moon with an initial velocity of \( 58 \, \text{m/s} \). The height of the arrow at any time \( t \) (in seconds) is given by the equation:

\[
y(t) = 58t - 0.83t^2
\]

### Part (a): Average Velocity Over Time Intervals
To find the **average velocity** over a given time interval \([t_1, t_2]\), we use the formula:

\[
v_{\text{avg}} = \frac{y(t_2) - y(t_1)}{t_2 - t_1}
\]

We are asked to calculate the average velocity over the following intervals:
1. \([1, 1.5]\)
2. \([1, 1.1]\)
3. \([1, 1.01]\)
4. \([1, 1.001]\)

I'll compute the average velocity for each interval now.

### Part (b): Instantaneous Velocity at \( t = 1 \)
The **instantaneous velocity** is the derivative of the height function \( y(t) \). Taking the derivative:

\[
v(t) = \frac{dy}{dt} = 58 - 2(0.83)t = 58 - 1.66t
\]

We need to find \( v(1) \), the instantaneous velocity after 1 second.

Let me calculate both parts (a) and (b).### Results:

#### Part (a): Average Velocity Over the Time Intervals
1. \([1, 1.5]\): \( 55.93 \, \text{m/s} \)
2. \([1, 1.1]\): \( 56.26 \, \text{m/s} \)
3. \([1, 1.01]\): \( 56.33 \, \text{m/s} \)
4. \([1, 1.001]\): \( 56.34 \, \text{m/s} \)

#### Part (b): Instantaneous Velocity at \( t = 1 \)
The instantaneous velocity after 1 second is \( 56.34 \, \text{m/s} \) (rounded to the nearest hundredth).

Would you like further details or clarification on any step?

Here are some additional questions you might want to explore:
1. What is the general formula for calculating average velocity?
2. How does the average velocity compare to the instantaneous velocity as the interval narrows?
3. What is the physical significance of the instantaneous velocity?
4. How would the problem change if the object were launched on Earth?
5. Can you find the time when the arrow reaches its highest point?

**Tip**: Remember that instantaneous velocity is essentially the slope of the tangent to the position-time curve at any given point.

If an arrow is shot upward on the moon with a velocity of 58 m/s, its height in meters after t seconds is given by y = 58t - 0.83t^2. (a) Find the average velocity over the given time intervals: (1) [1,1.5], (2) [1,1.1], (3) [1,1.01], (4) [1,1.001]. (b) Find the instantaneous velocity after one second (to the nearest hundredth).

Learn how to calculate the average velocity of an arrow shot upward on the moon using y = 58t - 0.83t^2. Understand how to compute instantaneous velocity after 1 second using derivatives. This problem covers kinematic concepts and calculus.

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