Question

t 7. 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 ( 1, 1.01 ), ( 1, 1.001 ), and ( 1, 1.0001 ). then make a conjecture about the value of the instantaneous velocity at ( t = 1 ).

( t )	1	1.0001	1.001	1.01

Explanation:

The formula for average velocity over the interval \([a, b]\) is \(\frac{s(b)-s(a)}{b - a}\). We will apply this formula to each of the given intervals.

Step 1: Average velocity over \([1, 1.01]\)

We know that \(a = 1\), \(b=1.01\), \(s(1)=64\) and \(s(1.01) = 64.4784\).
Using the average velocity formula:
\[
\frac{s(1.01)-s(1)}{1.01 - 1}=\frac{64.4784 - 64}{0.01}=\frac{0.4784}{0.01} = 47.84
\]

Step 2: Average velocity over \([1, 1.001]\)

Here, \(a = 1\), \(b = 1.001\), \(s(1)=64\) and \(s(1.001)=64.047984\)
Using the formula:
\[
\frac{s(1.001)-s(1)}{1.001 - 1}=\frac{64.047984 - 64}{0.001}=\frac{0.047984}{0.001}=47.984
\]

Step 3: Average velocity over \([1, 1.0001]\)

Here, \(a = 1\), \(b=1.0001\), \(s(1) = 64\) and \(s(1.0001)=64.00479984\)
Using the formula:
\[
\frac{s(1.0001)-s(1)}{1.0001 - 1}=\frac{64.00479984 - 64}{0.0001}=\frac{0.00479984}{0.0001}=47.9984
\]

Step 4: Conjecture about instantaneous velocity at \(t = 1\)

As the length of the time interval \([1, b]\) (where \(b\) is \(1.01\), \(1.001\), \(1.0001\)) gets smaller (approaches \(0\)), the average velocity approaches \(48\) (since \(47.84\), \(47.984\), \(47.9984\) are getting closer to \(48\)). So we conjecture that the instantaneous velocity at \(t = 1\) is \(48\).

Answer:

• Average velocity over \([1, 1.01]\): \(47.84\)
• Average velocity over \([1, 1.001]\): \(47.984\)
• Average velocity over \([1, 1.0001]\): \(47.9984\)
• Conjecture for instantaneous velocity at \(t = 1\): \(48\)