Answer :
To find the average velocity over each given time interval, you can use the position function [tex]\( s(t) = -4.9 t^2 + 32 t + 22 \)[/tex]. The average velocity over a time interval [tex]\([t_1, t_2]\)[/tex] is defined as:
[tex]\[ \text{Average Velocity} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \][/tex]
Let's calculate the average velocities for each interval.
1. Interval [tex]\([4, 5]\)[/tex]:
[tex]\[ \text{Average Velocity} = \frac{s(5) - s(4)}{5 - 4} \][/tex]
Calculate [tex]\(s(5)\)[/tex] and [tex]\(s(4)\)[/tex]:
[tex]\[ s(4) = -4.9(4)^2 + 32(4) + 22 = -4.9 \cdot 16 + 128 + 22 = -78.4 + 128 + 22 = 71.6 \][/tex]
[tex]\[ s(5) = -4.9(5)^2 + 32(5) + 22 = -4.9 \cdot 25 + 160 + 22 = -122.5 + 160 + 22 = 59.5 \][/tex]
[tex]\[ \text{Average Velocity} = \frac{59.5 - 71.6}{1} = -12.1 \][/tex]
2. Interval [tex]\([4, 4.1]\)[/tex]:
[tex]\[ \text{Average Velocity} = \frac{s(4.1) - s(4)}{4.1 - 4} \][/tex]
Calculate [tex]\(s(4.1)\)[/tex]:
[tex]\[ s(4) = 71.6 \quad \text{(calculated previously)} \][/tex]
[tex]\[ s(4.1) = -4.9(4.1)^2 + 32(4.1) + 22 = -4.9 \cdot 16.81 + 131.2 + 22 = -82.369 + 131.2 + 22 = 70.831 \][/tex]
[tex]\[ \text{Average Velocity} = \frac{70.831 - 71.6}{0.1} = -7.69 \][/tex]
3. Interval [tex]\([4, 4.01]\)[/tex]:
[tex]\[ \text{Average Velocity} = \frac{s(4.01) - s(4)}{4.01 - 4} \][/tex]
Calculate [tex]\(s(4.01)\)[/tex]:
[tex]\[ s(4.01) = -4.9(4.01)^2 + 32(4.01) + 22 = -4.9 \cdot 16.0801 + 128.32 + 22 = -78.79249 + 128.32 + 22 = 71.52751 \][/tex]
[tex]\[ \text{Average Velocity} = \frac{71.52751 - 71.6}{0.01} = -7.249 \][/tex]
4. Interval [tex]\([4, 4.001]\)[/tex]:
[tex]\[ \text{Average Velocity} = \frac{s(4.001) - s(4)}{4.001 - 4} \][/tex]
Calculate [tex]\(s(4.001)\)[/tex]:
[tex]\[ s(4.001) = -4.9(4.001)^2 + 32(4.001) + 22 = -4.9 \cdot 16.008001 + 128.032 + 22 = -78.4392049 + 128.032 + 22 = 71.5927951 \][/tex]
[tex]\[ \text{Average Velocity} = \frac{71.5927951 - 71.6}{0.001} = -7.2049 \][/tex]
5. Interval [tex]\([4, 4.0001]\)[/tex]:
[tex]\[ \text{Average Velocity} = \frac{s(4.0001) - s(4)}{4.0001 - 4} \][/tex]
Calculate [tex]\(s(4.0001)\)[/tex]:
[tex]\[ s(4.0001) = -4.9(4.0001)^2 + 32(4.0001) + 22 = -4.9 \cdot 16.00080001 + 128.0032 + 22 = -78.403920049 + 128.0032 + 22 = 71.599279951 \][/tex]
[tex]\[ \text{Average Velocity} = \frac{71.599279951 - 71.6}{0.0001} = -7.20049 \][/tex]
To fill out the table:
\begin{tabular}{|c|c|}
\hline
Time Interval & Average Velocity \\
\hline
[tex]$[4,5]$[/tex] & [tex]$-12.1$[/tex] \\
\hline
[tex]$[4,4.1]$[/tex] & [tex]$-7.69$[/tex] \\
\hline
[tex]$[4,4.01]$[/tex] & [tex]$-7.249$[/tex] \\
\hline
[tex]$[4,4.001]$[/tex] & [tex]$-7.2049$[/tex] \\
\hline
[tex]$[4,4.0001]$[/tex] & [tex]$-7.20049$[/tex] \\
\hline
\end{tabular}
Conjecture about the value of the instantaneous velocity at [tex]\( t = 4 \)[/tex]:
As the interval [tex]\([4, t]\)[/tex] becomes smaller and approaches zero, the average velocity approaches [tex]\(-7.2\)[/tex]. Therefore, the instantaneous velocity at [tex]\( t = 4 \)[/tex] is conjectured to be [tex]\(-7.2\)[/tex] m/s.
[tex]\[ \text{Average Velocity} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \][/tex]
Let's calculate the average velocities for each interval.
1. Interval [tex]\([4, 5]\)[/tex]:
[tex]\[ \text{Average Velocity} = \frac{s(5) - s(4)}{5 - 4} \][/tex]
Calculate [tex]\(s(5)\)[/tex] and [tex]\(s(4)\)[/tex]:
[tex]\[ s(4) = -4.9(4)^2 + 32(4) + 22 = -4.9 \cdot 16 + 128 + 22 = -78.4 + 128 + 22 = 71.6 \][/tex]
[tex]\[ s(5) = -4.9(5)^2 + 32(5) + 22 = -4.9 \cdot 25 + 160 + 22 = -122.5 + 160 + 22 = 59.5 \][/tex]
[tex]\[ \text{Average Velocity} = \frac{59.5 - 71.6}{1} = -12.1 \][/tex]
2. Interval [tex]\([4, 4.1]\)[/tex]:
[tex]\[ \text{Average Velocity} = \frac{s(4.1) - s(4)}{4.1 - 4} \][/tex]
Calculate [tex]\(s(4.1)\)[/tex]:
[tex]\[ s(4) = 71.6 \quad \text{(calculated previously)} \][/tex]
[tex]\[ s(4.1) = -4.9(4.1)^2 + 32(4.1) + 22 = -4.9 \cdot 16.81 + 131.2 + 22 = -82.369 + 131.2 + 22 = 70.831 \][/tex]
[tex]\[ \text{Average Velocity} = \frac{70.831 - 71.6}{0.1} = -7.69 \][/tex]
3. Interval [tex]\([4, 4.01]\)[/tex]:
[tex]\[ \text{Average Velocity} = \frac{s(4.01) - s(4)}{4.01 - 4} \][/tex]
Calculate [tex]\(s(4.01)\)[/tex]:
[tex]\[ s(4.01) = -4.9(4.01)^2 + 32(4.01) + 22 = -4.9 \cdot 16.0801 + 128.32 + 22 = -78.79249 + 128.32 + 22 = 71.52751 \][/tex]
[tex]\[ \text{Average Velocity} = \frac{71.52751 - 71.6}{0.01} = -7.249 \][/tex]
4. Interval [tex]\([4, 4.001]\)[/tex]:
[tex]\[ \text{Average Velocity} = \frac{s(4.001) - s(4)}{4.001 - 4} \][/tex]
Calculate [tex]\(s(4.001)\)[/tex]:
[tex]\[ s(4.001) = -4.9(4.001)^2 + 32(4.001) + 22 = -4.9 \cdot 16.008001 + 128.032 + 22 = -78.4392049 + 128.032 + 22 = 71.5927951 \][/tex]
[tex]\[ \text{Average Velocity} = \frac{71.5927951 - 71.6}{0.001} = -7.2049 \][/tex]
5. Interval [tex]\([4, 4.0001]\)[/tex]:
[tex]\[ \text{Average Velocity} = \frac{s(4.0001) - s(4)}{4.0001 - 4} \][/tex]
Calculate [tex]\(s(4.0001)\)[/tex]:
[tex]\[ s(4.0001) = -4.9(4.0001)^2 + 32(4.0001) + 22 = -4.9 \cdot 16.00080001 + 128.0032 + 22 = -78.403920049 + 128.0032 + 22 = 71.599279951 \][/tex]
[tex]\[ \text{Average Velocity} = \frac{71.599279951 - 71.6}{0.0001} = -7.20049 \][/tex]
To fill out the table:
\begin{tabular}{|c|c|}
\hline
Time Interval & Average Velocity \\
\hline
[tex]$[4,5]$[/tex] & [tex]$-12.1$[/tex] \\
\hline
[tex]$[4,4.1]$[/tex] & [tex]$-7.69$[/tex] \\
\hline
[tex]$[4,4.01]$[/tex] & [tex]$-7.249$[/tex] \\
\hline
[tex]$[4,4.001]$[/tex] & [tex]$-7.2049$[/tex] \\
\hline
[tex]$[4,4.0001]$[/tex] & [tex]$-7.20049$[/tex] \\
\hline
\end{tabular}
Conjecture about the value of the instantaneous velocity at [tex]\( t = 4 \)[/tex]:
As the interval [tex]\([4, t]\)[/tex] becomes smaller and approaches zero, the average velocity approaches [tex]\(-7.2\)[/tex]. Therefore, the instantaneous velocity at [tex]\( t = 4 \)[/tex] is conjectured to be [tex]\(-7.2\)[/tex] m/s.
