Question

t 21–24. instantaneous velocity for the following position functions, make a table of average velocities similar to those in exercises 19–20 and make a conjecture about the instantaneous velocity at the indicated time. 21. ( s(t) = -16t^2 + 80t + 60 ) at ( t = 3 )

Explanation:

Step1: Recall the formula for average velocity

The average velocity over the interval \([t, t + h]\) is given by \(\frac{s(t + h)-s(t)}{h}\), where \(s(t)\) is the position function. Here, \(t = 3\), so we need to compute \(\frac{s(3 + h)-s(3)}{h}\) for different values of \(h\) (both positive and negative, approaching 0) and then conjecture the instantaneous velocity as \(h ightarrow0\).

First, compute \(s(3)\):
\(s(3)=- 16(3)^{2}+80(3)+60=-16\times9 + 240+60=-144 + 300 = 156\)

Then, compute \(s(3 + h)\):
\(s(3 + h)=-16(3 + h)^{2}+80(3 + h)+60\)
Expand \((3 + h)^{2}=9 + 6h+h^{2}\)
\(s(3 + h)=-16(9 + 6h+h^{2})+240 + 80h+60\)
\(=-144-96h-16h^{2}+300 + 80h\)
\(=156-16h-16h^{2}\)

Now, the average velocity \(v_{avg}=\frac{s(3 + h)-s(3)}{h}=\frac{(156-16h-16h^{2})-156}{h}=\frac{-16h-16h^{2}}{h}=-16 - 16h\) (for \(h eq0\))

Step2: Create a table for different values of \(h\)

Let's choose values of \(h\) approaching 0 from both sides:

- When \(h = 1\): \(v_{avg}=-16-16(1)=-32\)
- When \(h = 0.5\): \(v_{avg}=-16-16(0.5)=-16 - 8=-24\)
- When \(h = 0.1\): \(v_{avg}=-16-16(0.1)=-16 - 1.6=-17.6\)
- When \(h = 0.01\): \(v_{avg}=-16-16(0.01)=-16 - 0.16=-16.16\)
- When \(h=-1\) (i.e., interval \([2,3]\)): \(v_{avg}=-16-16(-1)=-16 + 16 = 0\)? Wait, no, wait. Wait, our formula for \(v_{avg}\) when \(h\) is negative: let's re - check. Wait, when \(h\) is negative, say \(h=-k\) where \(k>0\), the interval is \([3 - k,3]\). Let's recast the average velocity formula as \(\frac{s(3)-s(3 - k)}{k}\). Let's recompute \(s(3 - k)\):

\(s(3 - k)=-16(3 - k)^{2}+80(3 - k)+60=-16(9 - 6k + k^{2})+240-80k + 60=-144 + 96k-16k^{2}+300-80k = 156 + 16k-16k^{2}\)

Then \(\frac{s(3)-s(3 - k)}{k}=\frac{156-(156 + 16k-16k^{2})}{k}=\frac{-16k + 16k^{2}}{k}=-16 + 16k\) (for \(k eq0\), \(h=-k\))

So when \(h=-1\) ( \(k = 1\)): \(v_{avg}=-16+16(1)=0\)? No, that can't be. Wait, I made a mistake in the first derivation. Let's redo the average velocity formula.

The correct formula for average velocity over \([a,b]\) is \(\frac{s(b)-s(a)}{b - a}\). So when \(t = 3\) and the interval is \([3,3 + h]\) ( \(h>0\)), then \(a = 3\), \(b = 3 + h\), so \(v_{avg}=\frac{s(3 + h)-s(3)}{h}\). When \(h<0\), say \(h=-k\) (\(k>0\)), the interval is \([3 - k,3]\), so \(a = 3 - k\), \(b = 3\), and \(v_{avg}=\frac{s(3)-s(3 - k)}{k}\)

Let's recalculate \(s(3 + h)-s(3)\) correctly:

\(s(3 + h)=-16(3 + h)^{2}+80(3 + h)+60=-16(9 + 6h+h^{2})+240 + 80h+60=-144-96h-16h^{2}+300 + 80h=156-16h-16h^{2}\)

\(s(3)=156\), so \(s(3 + h)-s(3)=-16h-16h^{2}\), so \(\frac{s(3 + h)-s(3)}{h}=\frac{-16h-16h^{2}}{h}=-16 - 16h\) (for \(h eq0\)), this is correct. When \(h=-1\), \(v_{avg}=\frac{s(2)-s(3)}{-1}\). Let's compute \(s(2)=-16(4)+160 + 60=-64 + 220 = 156\). So \(\frac{s(2)-s(3)}{-1}=\frac{156 - 156}{-1}=0\), and using the formula \(-16-16h\) with \(h=-1\), we get \(-16-16(-1)=0\), which matches.

When \(h=-0.5\) (interval \([2.5,3]\)): \(s(2.5)=-16(6.25)+200 + 60=-100 + 260 = 160\). \(\frac{s(3)-s(2.5)}{0.5}=\frac{156 - 160}{0.5}=\frac{-4}{0.5}=-8\). Using the formula \(-16-16h\) with \(h=-0.5\): \(-16-16(-0.5)=-16 + 8=-8\), correct.

When \(h=-0.1\) (interval \([2.9,3]\)): \(s(2.9)=-16(8.41)+80(2.9)+60=-134.56+232 + 60=157.44\). \(\frac{s(3)-s(2.9)}{0.1}=\frac{156 - 157.44}{0.1}=\frac{-1.44}{0.1}=-14.4\). Using the formula \(-16-16h\) with \(h=-0.1\): \(-16-16(-0.1)=-16 + 1.6=-14.4\), correct.

When \(h=-0.01\) (interval \([2.99,3]\)): \(s(2.99)=-16(2.99)^{2}+80(2.99)+60\). \(2.99^{2}=8.9401\), so \(-16\times8.9401=-143.0416\), \(80\times2.99 = 239.2\), so \(s(2.99)=-143.0416+239.2 + 60=156.1584\). \(\frac{s(3)-s(2.99)}{0.01}=\frac{156 - 156.1584}{0.01}=\frac{-0.1584}{0.01}=-15.84\). Using the formula \(-16-16h\) with \(h=-0.01\): \(-16-16(-0.01)=-16 + 0.16=-15.84\), correct.

Now, let's list the values:

| \(h\) | \(v_{avg}=\frac{s(3 + h)-s(3)}{h}\) |
|-------|--------------------------------------|
| 1 | -32 |
| 0.5 | -24 |
| 0.1 | -17.6 |
| 0.01 | -16.16 |
| -1 | 0 |
| -0.5 | -8 |
| -0.1 | -14.4 |
| -0.01 | -15.84 |

Step3: Conjecture the instantaneous velocity

As \(h\) approaches 0 (from both positive and negative sides), we look at the values of \(v_{avg}\). For positive \(h\) approaching 0: when \(h = 0.1\), \(v_{avg}=-17.6\); \(h = 0.01\), \(v_{avg}=-16.16\). For negative \(h\) approaching 0 (i.e., \(k\) approaching 0, \(h=-k\)): when \(h=-0.1\), \(v_{avg}=-14.4\); \(h=-0.01\), \(v_{avg}=-15.84\). As \(h ightarrow0\), the formula \(v_{avg}=-16 - 16h\) approaches \(-16\) (since \(\lim_{h ightarrow0}(-16 - 16h)=-16\)).

Answer:

The instantaneous velocity at \(t = 3\) is \(\boxed{-16}\)