Question

t 8. the following table gives the position ( s(t) ) of an object moving along a line at time ( t ). determine the average velocities over the time intervals (2, 2.01), (2, 2.001), and (2, 2.0001). then make a conjecture about the value of the instantaneous velocity at ( t = 2 ).
\begin{tabular}{|l|c|c|c|c|} hline ( t ) & 2 & 2.0001 & 2.001 & 2.01 \\ hline ( s(t) ) & 56 & 55.99959984 & 55.995984 & 55.9584 \\ hline end{tabular}

Explanation:

Step1: Recall average velocity formula

The average velocity over the interval \([a, b]\) is given by \(\frac{s(b) - s(a)}{b - a}\).

Step2: Calculate for \([2, 2.01]\)

Here, \(a = 2\), \(b = 2.01\), \(s(2)=56\), \(s(2.01)=55.9584\).
\[
\frac{s(2.01)-s(2)}{2.01 - 2}=\frac{55.9584 - 56}{0.01}=\frac{-0.0416}{0.01}=- 4.16
\]

Step3: Calculate for \([2, 2.001]\)

Here, \(a = 2\), \(b = 2.001\), \(s(2)=56\), \(s(2.001)=55.995984\).
\[
\frac{s(2.001)-s(2)}{2.001 - 2}=\frac{55.995984 - 56}{0.001}=\frac{-0.004016}{0.001}=-4.016
\]

Step4: Calculate for \([2, 2.0001]\)

Here, \(a = 2\), \(b = 2.0001\), \(s(2)=56\), \(s(2.0001)=55.99959984\).
\[
\frac{s(2.0001)-s(2)}{2.0001 - 2}=\frac{55.99959984 - 56}{0.0001}=\frac{-0.00040016}{0.0001}=-4.0016
\]

Step5: Conjecture instantaneous velocity

As the time interval \([2, b]\) gets smaller ( \(b\) approaches \(2\) ), the average velocity approaches \(- 4\). So the instantaneous velocity at \(t = 2\) is likely \(-4\).

Answer:

Average velocities: \([2, 2.01]\): \(-4.16\), \([2, 2.001]\): \(-4.016\), \([2, 2.0001]\): \(-4.0016\); Conjecture: Instantaneous velocity at \(t = 2\) is \(-4\).