If an arrow is shot upward on Mars with a speed of 64 m/s, its height in meters t seconds later is given by
y = 64t − 1.86t^2.
(Round your answers to two decimal places.)
(a) Find the average speed over the given time intervals. [1, 2]

(ii)    [1, 1.5]

(iii)    [1, 1.1]

(iv)    [1, 1.01]

(v)    [1, 1.001]

To find the average speed of the arrow over the given time intervals, we need to calculate the change in height \( \Delta y \) divided by the change in time \( \Delta t \) for each interval. The height function is given by:

\[
y(t) = 64t - 1.86t^2
\]

The formula for average speed over a time interval \([t_1, t_2]\) is:

\[
\text{Average speed} = \frac{y(t_2) - y(t_1)}{t_2 - t_1}
\]

### (a) Calculate the average speed over the given intervals:

1. **Interval [1, 2]**
   - \( t_1 = 1 \), \( t_2 = 2 \)
   - \( y(1) = 64(1) - 1.86(1)^2 = 64 - 1.86 = 62.14 \) m
   - \( y(2) = 64(2) - 1.86(2)^2 = 128 - 7.44 = 120.56 \) m
   - Average speed = \( \frac{120.56 - 62.14}{2 - 1} = \frac{58.42}{1} = 58.42 \) m/s

2. **Interval [1, 1.5]**
   - \( t_1 = 1 \), \( t_2 = 1.5 \)
   - \( y(1.5) = 64(1.5) - 1.86(1.5)^2 = 96 - 4.185 = 91.815 \) m
   - Average speed = \( \frac{91.815 - 62.14}{1.5 - 1} = \frac{29.675}{0.5} = 59.35 \) m/s

3. **Interval [1, 1.1]**
   - \( t_1 = 1 \), \( t_2 = 1.1 \)
   - \( y(1.1) = 64(1.1) - 1.86(1.1)^2 = 70.4 - 2.2546 = 68.1454 \) m
   - Average speed = \( \frac{68.1454 - 62.14}{1.1 - 1} = \frac{6.0054}{0.1} = 60.05 \) m/s

4. **Interval [1, 1.01]**
   - \( t_1 = 1 \), \( t_2 = 1.01 \)
   - \( y(1.01) = 64(1.01) - 1.86(1.01)^2 = 64.64 - 1.890846 = 62.749154 \) m
   - Average speed = \( \frac{62.749154 - 62.14}{1.01 - 1} = \frac{0.609154}{0.01} = 60.92 \) m/s

5. **Interval [1, 1.001]**
   - \( t_1 = 1 \), \( t_2 = 1.001 \)
   - \( y(1.001) = 64(1.001) - 1.86(1.001)^2 = 64.064 - 1.86372086 = 62.20027914 \) m
   - Average speed = \( \frac{62.20027914 - 62.14}{1.001 - 1} = \frac{0.06027914}{0.001} = 60.28 \) m/s

### Summary of Average Speeds
- **[1, 2]**: \( 58.42 \) m/s
- **[1, 1.5]**: \( 59.35 \) m/s
- **[1, 1.1]**: \( 60.05 \) m/s
- **[1, 1.01]**: \( 60.92 \) m/s
- **[1, 1.001]**: \( 60.28 \) m/s

These results show that as the time interval becomes smaller, the average speed approaches a value closer to the instantaneous speed at \( t = 1 \) second.

Would you like details on how the height was calculated, or do you have any other questions?

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Here are 5 related questions for further exploration:

1. How would you calculate the instantaneous speed at \( t = 1 \) second?
2. What would the average speed be over a larger interval, such as [0, 2]?
3. How does the acceleration on Mars compare to that on Earth, given the height function?
4. How would you determine the maximum height reached by the arrow on Mars?
5. Can you derive a general expression for the average speed over any time interval \( [t_1, t_2] \)?

**Tip:** The average speed over very small intervals approaches the instantaneous speed, which can be found by differentiating the height function with respect to time.

Explore how to calculate the average speed of an arrow shot upward on Mars over various time intervals using the height function y(t) = 64t - 1.86t^2. Find detailed solutions for intervals [1, 2], [1, 1.5], [1, 1.1], [1, 1.01], and [1, 1.001].

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