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 62 m/s, then its height in meters after t seconds is given by $h(t)=62t - 0.83t^{2}$. find the average velocity over the given time intervals.
7, 8:
7, 7.5:
7, 7.1:
7, 7.01:
7, 7.001:
question help: video

Explanation:

Step1: Recall average - velocity formula

The average velocity over the interval $[a,b]$ is $\frac{h(b)-h(a)}{b - a}$, where $h(t)=62t-0.83t^{2}$.

Step2: For $[7,8]$

$h(8)=62\times8 - 0.83\times8^{2}=496-53.12 = 442.88$, $h(7)=62\times7-0.83\times7^{2}=434 - 40.67=393.33$. Then $\frac{h(8)-h(7)}{8 - 7}=\frac{442.88 - 393.33}{1}=49.5500$.

Step3: For $[7,7.5]$

$h(7.5)=62\times7.5-0.83\times7.5^{2}=465 - 46.6875 = 418.3125$. $\frac{h(7.5)-h(7)}{7.5 - 7}=\frac{418.3125 - 393.33}{0.5}=50.0050$.

Step4: For $[7,7.1]$

$h(7.1)=62\times7.1-0.83\times7.1^{2}=440.2-41.7763 = 398.4237$. $\frac{h(7.1)-h(7)}{7.1 - 7}=\frac{398.4237 - 393.33}{0.1}=50.9370$.

Step5: For $[7,7.01]$

$h(7.01)=62\times7.01-0.83\times7.01^{2}=434.62-40.858783 = 393.761217$. $\frac{h(7.01)-h(7)}{7.01 - 7}=\frac{393.761217 - 393.33}{0.01}=43.1217$.

Step6: For $[7,7.001]$

$h(7.001)=62\times7.001-0.83\times7.001^{2}=434.062-40.67908303 = 393.38291697$. $\frac{h(7.001)-h(7)}{7.001 - 7}=\frac{393.38291697 - 393.33}{0.001}=52.91697$.

Answer:

$[7,8]:$ $49.5500$
$[7,7.5]:$ $50.0050$
$[7,7.1]:$ $50.9370$
$[7,7.01]:$ $43.1217$
$[7,7.001]:$ $52.9170$