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For each value below, enter the number correct to four decimal places.

Suppose an arrow is shot upward on the moon with a velocity of [tex]$34 \, m/s$[/tex], then its height in meters after [tex]$t$[/tex] seconds is given by [tex]$h(t) = 34t - 0.83t^2$[/tex].

Find the average velocity over the given time intervals.

[tex][5,6]:[/tex] [tex]\square[/tex]
[tex][5,5.5]:[/tex] [tex]\square[/tex]
[tex][5,5.1]:[/tex] [tex]\square[/tex]
[tex][5,5.01]:[/tex] [tex]\square[/tex]
[tex][5,5.001]:[/tex] [tex]\square[/tex]

Question Help: [tex]\square[/tex] Video


Sagot :

To find the average velocity of the arrow over specific time intervals, we need to follow these steps:

1. Identify the function for height: Given [tex]\( h(t) = 34t - 0.83t^2 \)[/tex], this function describes how the height of the arrow changes over time.

2. Understand the formula for average velocity: The average velocity [tex]\( v_{avg} \)[/tex] over a time interval [tex]\([t_1, t_2]\)[/tex] is given by:
[tex]\[ v_{avg} = \frac{h(t_2) - h(t_1)}{t_2 - t_1} \][/tex]

3. Calculate the height values at specific times: Evaluate [tex]\( h(t) \)[/tex] at the given points [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex].

4. Apply the average velocity formula for each interval.

Let's calculate these step-by-step for each interval:

### Interval [5, 6]:
- [tex]\( t_1 = 5 \)[/tex], [tex]\( t_2 = 6 \)[/tex]
- Calculate [tex]\( h(5) \)[/tex]:
[tex]\[ h(5) = 34 \cdot 5 - 0.83 \cdot 5^2 = 170 - 20.75 = 149.25 \][/tex]
- Calculate [tex]\( h(6) \)[/tex]:
[tex]\[ h(6) = 34 \cdot 6 - 0.83 \cdot 6^2 = 204 - 29.88 = 174.12 \][/tex]
- Average velocity:
[tex]\[ v_{avg} = \frac{h(6) - h(5)}{6 - 5} = \frac{174.12 - 149.25}{1} = 24.87 \][/tex]

So, for the interval [tex]\([5, 6]\)[/tex], the average velocity is [tex]\( 24.8700 \)[/tex] meters/second.

### Interval [5, 5.5]:
- [tex]\( t_1 = 5 \)[/tex], [tex]\( t_2 = 5.5 \)[/tex]
- Calculate [tex]\( h(5.5) \)[/tex]:
[tex]\[ h(5.5) = 34 \cdot 5.5 - 0.83 \cdot 5.5^2 = 187 - 25.1475 = 161.8525 \][/tex]
- Average velocity:
[tex]\[ v_{avg} = \frac{h(5.5) - h(5)}{5.5 - 5} = \frac{161.8525 - 149.25}{0.5} = 25.285 \][/tex]

So, for the interval [tex]\([5, 5.5]\)[/tex], the average velocity is [tex]\( 25.2850 \)[/tex] meters/second.

### Interval [5, 5.1]:
- [tex]\( t_1 = 5 \)[/tex], [tex]\( t_2 = 5.1 \)[/tex]
- Calculate [tex]\( h(5.1) \)[/tex]:
[tex]\[ h(5.1) = 34 \cdot 5.1 - 0.83 \cdot 5.1^2 = 173.4 - 21.8325 = 151.5675 \][/tex]
- Average velocity:
[tex]\[ v_{avg} = \frac{h(5.1) - h(5)}{5.1 - 5} = \frac{151.5675 - 149.25}{0.1} = 25.617 \][/tex]

So, for the interval [tex]\([5, 5.1]\)[/tex], the average velocity is [tex]\( 25.6170 \)[/tex] meters/second.

### Interval [5, 5.01]:
- [tex]\( t_1 = 5 \)[/tex], [tex]\( t_2 = 5.01 \)[/tex]
- Calculate [tex]\( h(5.01) \)[/tex]:
[tex]\[ h(5.01) = 34 \cdot 5.01 - 0.83 \cdot 5.01^2 = 170.34 - 20.902501 = 149.437499 \][/tex]
- Average velocity:
[tex]\[ v_{avg} = \frac{h(5.01) - h(5)}{5.01 - 5} = \frac{149.437499 - 149.25}{0.01} = 25.6917 \][/tex]

So, for the interval [tex]\([5, 5.01]\)[/tex], the average velocity is [tex]\( 25.6917 \)[/tex] meters/second.

### Interval [5, 5.001]:
- [tex]\( t_1 = 5 \)[/tex], [tex]\( t_2 = 5.001 \)[/tex]
- Calculate [tex]\( h(5.001) \)[/tex]:
[tex]\[ h(5.001) = 34 \cdot 5.001 - 0.83 \cdot 5.001^2 = 170.034 - 20.83000833 = 149.20399167 \][/tex]
- Average velocity:
[tex]\[ v_{avg} = \frac{h(5.001) - h(5)}{5.001 - 5} = \frac{149.20399167 - 149.25}{0.001} = 25.6992 \][/tex]

So, for the interval [tex]\([5, 5.001]\)[/tex], the average velocity is [tex]\( 25.6992 \)[/tex] meters/second.

To summarize, the average velocities over the given time intervals are:

[tex]\[ [5,6]: 24.8700 \][/tex]
[tex]\[ [5,5.5]: 25.2850 \][/tex]
[tex]\[ [5,5.1]: 25.6170 \][/tex]
[tex]\[ [5,5.01]: 25.6917 \][/tex]
[tex]\[ [5,5.001]: 25.6992 \][/tex]
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