Question

question
the position in feet of a race car along a straight track after t seconds is modeled by the function s(t)=5t² + 4. find the average velocity of the vehicle over the following time intervals: 2, 2.1, 2, 2.01, 2, 2.001, and 2, 2.0001. use the calculated averages to estimate the instantaneous velocity v_inst of the vehicle at t = 2 seconds.
round each answer in the table to six decimal places, but enter an integer value for v_inst.
provide your answer below:

t	v_ave

|2.1|
|2.01|
|2.001|
|2.0001|

Explanation:

Step1: Recall average - velocity formula

The average velocity $v_{ave}$ over the interval $[a,b]$ for a position - function $s(t)$ is given by $v_{ave}=\frac{s(b)-s(a)}{b - a}$. Here $a = 2$ and $s(t)=5t^{2}+4$.

Step2: Calculate $s(2)$

$s(2)=5\times(2)^{2}+4=5\times4 + 4=20 + 4=24$.

Step3: Calculate average velocity for $[2,2.1]$

First, find $s(2.1)=5\times(2.1)^{2}+4=5\times4.41+4=22.05 + 4=26.05$. Then $v_{ave}=\frac{s(2.1)-s(2)}{2.1 - 2}=\frac{26.05 - 24}{0.1}=\frac{2.05}{0.1}=20.500000$.

Step4: Calculate average velocity for $[2,2.01]$

Find $s(2.01)=5\times(2.01)^{2}+4=5\times4.0401+4=20.2005+4=24.2005$. Then $v_{ave}=\frac{s(2.01)-s(2)}{2.01 - 2}=\frac{24.2005 - 24}{0.01}=\frac{0.2005}{0.01}=20.050000$.

Step5: Calculate average velocity for $[2,2.001]$

Find $s(2.001)=5\times(2.001)^{2}+4=5\times4.004001+4=20.020005+4=24.020005$. Then $v_{ave}=\frac{s(2.001)-s(2)}{2.001 - 2}=\frac{24.020005 - 24}{0.001}=\frac{0.020005}{0.001}=20.005000$.

Step6: Calculate average velocity for $[2,2.0001]$

Find $s(2.0001)=5\times(2.0001)^{2}+4=5\times4.00040001+4=20.00200005+4=24.00200005$. Then $v_{ave}=\frac{s(2.0001)-s(2)}{2.0001 - 2}=\frac{24.00200005 - 24}{0.0001}=\frac{0.00200005}{0.0001}=20.000500$.

Step7: Estimate instantaneous velocity

As the time - intervals get smaller and smaller around $t = 2$, the average velocities approach the instantaneous velocity. The instantaneous velocity $v_{inst}$ at $t = 2$ is estimated to be $20$.

Answer:

$t$	$v_{ave}$
$2.01$	$20.050000$
$2.001$	$20.005000$
$2.0001$	$20.000500$
$v_{inst}$	$20$