Question

the position in feet of a race car along a straight track after t seconds is modeled by the function s(t)=5t^2 + 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.

Explanation:

Step1: Recall the average - velocity formula

The average - velocity formula over the interval $[a,b]$ is $v_{avg}=\frac{s(b)-s(a)}{b - a}$, where $s(t)$ is the position function. Given $s(t)=5t^{2}+4$.

Step2: Calculate average velocity for $[2,2.1]$

First, find $s(2)$ and $s(2.1)$.
$s(2)=5\times(2)^{2}+4=5\times4 + 4=20 + 4=24$.
$s(2.1)=5\times(2.1)^{2}+4=5\times4.41+4=22.05 + 4=26.05$.
Then $v_{avg}=\frac{s(2.1)-s(2)}{2.1 - 2}=\frac{26.05 - 24}{0.1}=\frac{2.05}{0.1}=20.5$.

Step3: Calculate average velocity for $[2,2.01]$

Find $s(2.01)=5\times(2.01)^{2}+4=5\times4.0401+4=20.2005 + 4=24.2005$.
$v_{avg}=\frac{s(2.01)-s(2)}{2.01 - 2}=\frac{24.2005 - 24}{0.01}=\frac{0.2005}{0.01}=20.05$.

Step4: Calculate average velocity for $[2,2.001]$

Find $s(2.001)=5\times(2.001)^{2}+4=5\times4.004001+4=20.020005 + 4=24.020005$.
$v_{avg}=\frac{s(2.001)-s(2)}{2.001 - 2}=\frac{24.020005 - 24}{0.001}=\frac{0.020005}{0.001}=20.005$.

Step5: Calculate average velocity for $[2,2.0001]$

Find $s(2.0001)=5\times(2.0001)^{2}+4=5\times4.00040001+4=20.00200005 + 4=24.00200005$.
$v_{avg}=\frac{s(2.0001)-s(2)}{2.0001 - 2}=\frac{24.00200005 - 24}{0.0001}=\frac{0.00200005}{0.0001}=20.0005$.
As the time - intervals get smaller and smaller around $t = 2$, the average velocities approach the instantaneous velocity at $t = 2$.
The derivative of $s(t)=5t^{2}+4$ using the power rule $(x^{n})^\prime=nx^{n - 1}$ is $s^\prime(t)=10t$.
Evaluating $s^\prime(t)$ at $t = 2$, we get $s^\prime(2)=10\times2=20$.

Answer:

20