Question

the position of an object moving along a straight line after t seconds is modeled by the function s(t)=-4t^3 - 1. find the average velocity of the object 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_instant of the object at t = 2 seconds. round each answer in the table to six decimal places, but enter an integer value for v_instant. provide your answer below. t v_ave 2.1 2.01 2.001 2.0001 v_instant≈

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 $v_{ave}=\frac{s(b)-s(a)}{b - a}$.

Step2: Calculate $s(t)$ at $t = 2$

$s(2)=-4\times2^{3}-1=-4\times8 - 1=-32 - 1=-33$.

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

$s(2.1)=-4\times(2.1)^{3}-1=-4\times9.261-1=-37.044 - 1=-38.044$.
$v_{ave}=\frac{s(2.1)-s(2)}{2.1 - 2}=\frac{-38.044+33}{0.1}=\frac{-5.044}{0.1}=-50.440000$.

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

$s(2.01)=-4\times(2.01)^{3}-1=-4\times8.120601-1=-32.482404 - 1=-33.482404$.
$v_{ave}=\frac{s(2.01)-s(2)}{2.01 - 2}=\frac{-33.482404 + 33}{0.01}=\frac{-0.482404}{0.01}=-48.240400$.

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

$s(2.001)=-4\times(2.001)^{3}-1=-4\times8.012006001-1=-32.048024004 - 1=-33.048024004$.
$v_{ave}=\frac{s(2.001)-s(2)}{2.001 - 2}=\frac{-33.048024004+33}{0.001}=\frac{-0.048024004}{0.001}=-48.024004$.

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

$s(2.0001)=-4\times(2.0001)^{3}-1=-4\times8.001200060001-1=-32.004800240004 - 1=-33.004800240004$.
$v_{ave}=\frac{s(2.0001)-s(2)}{2.0001 - 2}=\frac{-33.004800240004 + 33}{0.0001}=\frac{-0.004800240004}{0.0001}=-48.002400$.

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 $v_{inst}\approx - 48$.

Answer:

$t$	$v_{ave}$
$2.01$	$-48.240400$
$2.001$	$-48.024004$
$2.0001$	$-48.002400$
$v_{inst}$	$-48$