Question

t 17. instantaneous velocity consider the position function ( s(t) = -16t^2 + 128t ) (exercise 13). complete the following table with the appropriate average velocities. then make a conjecture about the value of the instantaneous velocity at ( t = 1 ).

time interval	1, 2	1, 1.5	1, 1.1	1, 1.01	1, 1.001

Explanation:

The formula for average velocity over the interval \([a, b]\) is \(\frac{s(b) - s(a)}{b - a}\), where \(s(t)\) is the position function. We will calculate the average velocity for each interval \([1, b]\) by substituting \(a = 1\) and the respective \(b\) values into this formula.

Step 1: Calculate for interval \([1, 2]\)

First, find \(s(1)\) and \(s(2)\):
\(s(1)=-16(1)^2 + 128(1)=-16 + 128 = 112\)
\(s(2)=-16(2)^2 + 128(2)=-64 + 256 = 192\)
Average velocity \(=\frac{s(2)-s(1)}{2 - 1}=\frac{192 - 112}{1}=80\)

Step 2: Calculate for interval \([1, 1.5]\)

Find \(s(1.5)\):
\(s(1.5)=-16(1.5)^2 + 128(1.5)=-16(2.25)+192=-36 + 192 = 156\)
Average velocity \(=\frac{s(1.5)-s(1)}{1.5 - 1}=\frac{156 - 112}{0.5}=\frac{44}{0.5}=88\)

Step 3: Calculate for interval \([1, 1.1]\)

Find \(s(1.1)\):
\(s(1.1)=-16(1.1)^2 + 128(1.1)=-16(1.21)+140.8=-19.36 + 140.8 = 121.44\)
Average velocity \(=\frac{s(1.1)-s(1)}{1.1 - 1}=\frac{121.44 - 112}{0.1}=\frac{9.44}{0.1}=94.4\)

Step 4: Calculate for interval \([1, 1.01]\)

Find \(s(1.01)\):
\(s(1.01)=-16(1.01)^2 + 128(1.01)=-16(1.0201)+129.28=-16.3216 + 129.28 = 112.9584\)
Average velocity \(=\frac{s(1.01)-s(1)}{1.01 - 1}=\frac{112.9584 - 112}{0.01}=\frac{0.9584}{0.01}=95.84\)

Step 5: Calculate for interval \([1, 1.001]\)

Find \(s(1.001)\):
\(s(1.001)=-16(1.001)^2 + 128(1.001)=-16(1.002001)+128.128=-16.032016 + 128.128 = 112.095984\)
Average velocity \(=\frac{s(1.001)-s(1)}{1.001 - 1}=\frac{112.095984 - 112}{0.001}=\frac{0.095984}{0.001}=95.984\)

Now, as the interval \([1, b]\) gets closer to \(t = 1\) (i.e., \(b\) approaches \(1\)), the average velocity approaches \(96\). So we conjecture that the instantaneous velocity at \(t = 1\) is \(96\).

Answer:

The average velocities for the intervals \([1, 2]\), \([1, 1.5]\), \([1, 1.1]\), \([1, 1.01]\), \([1, 1.001]\) are \(80\), \(88\), \(94.4\), \(95.84\), \(95.984\) respectively. The conjectured instantaneous velocity at \(t = 1\) is \(\boldsymbol{96}\).