Question

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 25 m/s, then its height in meters after t seconds is given by $h(t) = 25t - 0.83t^2$. find the average velocity over the given time intervals. 10, 11: 10, 10.5: 10, 10.1: 10, 10.01: 10, 10.001: question help: video message instructor

Explanation:

To find the average velocity over a time interval \([a, b]\), we use the formula for average velocity:

\[
\text{Average Velocity} = \frac{h(b) - h(a)}{b - a}
\]

where \(h(t) = 25t - 0.83t^2\).

For the interval \([10, 11]\):

Step 1: Calculate \(h(11)\) and \(h(10)\)

• \(h(11) = 25(11) - 0.83(11)^2 = 275 - 0.83(121) = 275 - 100.43 = 174.57\)
• \(h(10) = 25(10) - 0.83(10)^2 = 250 - 83 = 167\)

Step 2: Apply the average velocity formula

\[
\text{Average Velocity} = \frac{h(11) - h(10)}{11 - 10} = \frac{174.57 - 167}{1} = 7.57
\]
(Wait, let's recalculate \(h(11)\) correctly: \(0.83 \times 121 = 100.43\), so \(25 \times 11 = 275\), \(275 - 100.43 = 174.57\). Then \(174.57 - 167 = 7.57\). So average velocity is \(7.57\). But let's check with more precise calculation. Wait, maybe I made a mistake. Wait, \(h(t) = 25t - 0.83t^2\). So \(h(10) = 2510 - 0.83(10)^2 = 250 - 83 = 167\). \(h(11) = 2511 - 0.83(121) = 275 - 100.43 = 174.57\). Then the difference is \(174.57 - 167 = 7.57\), divided by \(1\) is \(7.57\). So average velocity is \(7.57\). But let's do it more accurately. Wait, maybe the formula is correct. Let's proceed.

For the interval \([10, 10.5]\):

Step 1: Calculate \(h(10.5)\) and \(h(10)\)

• \(h(10.5) = 25(10.5) - 0.83(10.5)^2 = 262.5 - 0.83(110.25) = 262.5 - 91.5075 = 170.9925\)
• \(h(10) = 167\) (as before)

Step 2: Apply the average velocity formula

\[
\text{Average Velocity} = \frac{h(10.5) - h(10)}{10.5 - 10} = \frac{170.9925 - 167}{0.5} = \frac{3.9925}{0.5} = 7.985
\]

For the interval \([10, 10.1]\):

Step 1: Calculate \(h(10.1)\) and \(h(10)\)

• \(h(10.1) = 25(10.1) - 0.83(10.1)^2 = 252.5 - 0.83(102.01) = 252.5 - 84.6683 = 167.8317\)
• \(h(10) = 167\)

Step 2: Apply the average velocity formula

\[
\text{Average Velocity} = \frac{h(10.1) - h(10)}{10.1 - 10} = \frac{167.8317 - 167}{0.1} = \frac{0.8317}{0.1} = 8.317
\]

For the interval \([10, 10.01]\):

Step 1: Calculate \(h(10.01)\) and \(h(10)\)

• \(h(10.01) = 25(10.01) - 0.83(10.01)^2 = 250.25 - 0.83(100.2001) = 250.25 - 83.166083 = 167.083917\)
• \(h(10) = 167\)

Step 2: Apply the average velocity formula

\[
\text{Average Velocity} = \frac{h(10.01) - h(10)}{10.01 - 10} = \frac{167.083917 - 167}{0.01} = \frac{0.083917}{0.01} = 8.3917
\]

For the interval \([10, 10.001]\):

Step 1: Calculate \(h(10.001)\) and \(h(10)\)

• \(h(10.001) = 25(10.001) - 0.83(10.001)^2 = 250.025 - 0.83(100.020001) = 250.025 - 83.01660083 = 167.00839917\)
• \(h(10) = 167\)

Step 2: Apply the average velocity formula

\[
\text{Average Velocity} = \frac{h(10.001) - h(10)}{10.001 - 10} = \frac{167.00839917 - 167}{0.001} = \frac{0.00839917}{0.001} = 8.39917 \approx 8.3992
\]

Final Answers:

• \([10, 11]\): \(7.5700\)
• \([10, 10.5]\): \(7.9850\)
• \([10, 10.1]\): \(8.3170\)
• \([10, 10.01]\): \(8.3917\)
• \([10, 10.001]\): \(8.3992\)

Answer:

To find the average velocity over a time interval \([a, b]\), we use the formula for average velocity:

\[
\text{Average Velocity} = \frac{h(b) - h(a)}{b - a}
\]

where \(h(t) = 25t - 0.83t^2\).

For the interval \([10, 11]\):

Step 1: Calculate \(h(11)\) and \(h(10)\)

• \(h(11) = 25(11) - 0.83(11)^2 = 275 - 0.83(121) = 275 - 100.43 = 174.57\)
• \(h(10) = 25(10) - 0.83(10)^2 = 250 - 83 = 167\)

Step 2: Apply the average velocity formula

\[
\text{Average Velocity} = \frac{h(11) - h(10)}{11 - 10} = \frac{174.57 - 167}{1} = 7.57
\]
(Wait, let's recalculate \(h(11)\) correctly: \(0.83 \times 121 = 100.43\), so \(25 \times 11 = 275\), \(275 - 100.43 = 174.57\). Then \(174.57 - 167 = 7.57\). So average velocity is \(7.57\). But let's check with more precise calculation. Wait, maybe I made a mistake. Wait, \(h(t) = 25t - 0.83t^2\). So \(h(10) = 2510 - 0.83(10)^2 = 250 - 83 = 167\). \(h(11) = 2511 - 0.83(121) = 275 - 100.43 = 174.57\). Then the difference is \(174.57 - 167 = 7.57\), divided by \(1\) is \(7.57\). So average velocity is \(7.57\). But let's do it more accurately. Wait, maybe the formula is correct. Let's proceed.

For the interval \([10, 10.5]\):

Step 1: Calculate \(h(10.5)\) and \(h(10)\)

• \(h(10.5) = 25(10.5) - 0.83(10.5)^2 = 262.5 - 0.83(110.25) = 262.5 - 91.5075 = 170.9925\)
• \(h(10) = 167\) (as before)

Step 2: Apply the average velocity formula

\[
\text{Average Velocity} = \frac{h(10.5) - h(10)}{10.5 - 10} = \frac{170.9925 - 167}{0.5} = \frac{3.9925}{0.5} = 7.985
\]

For the interval \([10, 10.1]\):

Step 1: Calculate \(h(10.1)\) and \(h(10)\)

• \(h(10.1) = 25(10.1) - 0.83(10.1)^2 = 252.5 - 0.83(102.01) = 252.5 - 84.6683 = 167.8317\)
• \(h(10) = 167\)

Step 2: Apply the average velocity formula

\[
\text{Average Velocity} = \frac{h(10.1) - h(10)}{10.1 - 10} = \frac{167.8317 - 167}{0.1} = \frac{0.8317}{0.1} = 8.317
\]

For the interval \([10, 10.01]\):

Step 1: Calculate \(h(10.01)\) and \(h(10)\)

• \(h(10.01) = 25(10.01) - 0.83(10.01)^2 = 250.25 - 0.83(100.2001) = 250.25 - 83.166083 = 167.083917\)
• \(h(10) = 167\)

Step 2: Apply the average velocity formula

\[
\text{Average Velocity} = \frac{h(10.01) - h(10)}{10.01 - 10} = \frac{167.083917 - 167}{0.01} = \frac{0.083917}{0.01} = 8.3917
\]

For the interval \([10, 10.001]\):

Step 1: Calculate \(h(10.001)\) and \(h(10)\)

• \(h(10.001) = 25(10.001) - 0.83(10.001)^2 = 250.025 - 0.83(100.020001) = 250.025 - 83.01660083 = 167.00839917\)
• \(h(10) = 167\)

Step 2: Apply the average velocity formula

\[
\text{Average Velocity} = \frac{h(10.001) - h(10)}{10.001 - 10} = \frac{167.00839917 - 167}{0.001} = \frac{0.00839917}{0.001} = 8.39917 \approx 8.3992
\]

Final Answers:

• \([10, 11]\): \(7.5700\)
• \([10, 10.5]\): \(7.9850\)
• \([10, 10.1]\): \(8.3170\)
• \([10, 10.01]\): \(8.3917\)
• \([10, 10.001]\): \(8.3992\)