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 52 m/s, then its height in meters after t seconds is given by h(t)=52t - 0.83t². find the average velocity over the given time intervals. 6, 7: 6, 6.5: 6, 6.1: 6, 6.01: 6, 6.001: question help: message instructor submit question jump to answer

Explanation:

Step1: Recall average - velocity formula

The average velocity $v_{avg}$ over the interval $[a,b]$ is given by $v_{avg}=\frac{h(b)-h(a)}{b - a}$, where $h(t)=52t-0.83t^{2}$.

Step2: For the interval $[6,7]$

First, find $h(7)$ and $h(6)$:
$h(7)=52\times7 - 0.83\times7^{2}=364-0.83\times49 = 364 - 40.67=323.33$.
$h(6)=52\times6-0.83\times6^{2}=312 - 0.83\times36=312 - 29.88 = 282.12$.
Then $v_{avg}=\frac{h(7)-h(6)}{7 - 6}=\frac{323.33 - 282.12}{1}=41.2100$.

Step3: For the interval $[6,6.5]$

$h(6.5)=52\times6.5-0.83\times6.5^{2}=338-0.83\times42.25=338 - 35.0675 = 302.9325$.
$v_{avg}=\frac{h(6.5)-h(6)}{6.5 - 6}=\frac{302.9325 - 282.12}{0.5}=\frac{20.8125}{0.5}=41.6250$.

Step4: For the interval $[6,6.1]$

$h(6.1)=52\times6.1-0.83\times6.1^{2}=317.2-0.83\times37.21=317.2 - 30.8843 = 286.3157$.
$v_{avg}=\frac{h(6.1)-h(6)}{6.1 - 6}=\frac{286.3157 - 282.12}{0.1}=\frac{4.1957}{0.1}=41.9570$.

Step5: For the interval $[6,6.01]$

$h(6.01)=52\times6.01-0.83\times6.01^{2}=312.52-0.83\times36.1201=312.52 - 29.9797 = 282.5403$.
$v_{avg}=\frac{h(6.01)-h(6)}{6.01 - 6}=\frac{282.5403 - 282.12}{0.01}=\frac{0.4203}{0.01}=42.0300$.

Step6: For the interval $[6,6.001]$

$h(6.001)=52\times6.001-0.83\times6.001^{2}=312.052-0.83\times36.012001=312.052 - 29.8990 = 282.1530$.
$v_{avg}=\frac{h(6.001)-h(6)}{6.001 - 6}=\frac{282.1530 - 282.12}{0.001}=\frac{0.033}{0.001}=42.0030$.

Answer:

$[6,7]:$ 41.2100
$[6,6.5]:$ 41.6250
$[6,6.1]:$ 41.9570
$[6,6.01]:$ 42.0300
$[6,6.001]:$ 42.0030