Question

for the position function ( s(t) = -16t^2 + 104t ), complete the following table with the appropriate average velocities. then make a conjecture about the value of the instantaneous velocity at ( t = 1 ). complete the following table (type exact answers. type integers or decimals.)
time interval | average velocity
--- | ---
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 \(s(t)=-16t^{2}+104t\).

Step 1: Interval \([1, 2]\)

First, find \(s(1)\) and \(s(2)\):
\(s(1)=-16(1)^{2}+104(1)=-16 + 104 = 88\)
\(s(2)=-16(2)^{2}+104(2)=-64 + 208 = 144\)
Then, average velocity \(=\frac{s(2)-s(1)}{2 - 1}=\frac{144 - 88}{1}=56\)

Step 2: Interval \([1, 1.5]\)

Find \(s(1.5)\):
\(s(1.5)=-16(1.5)^{2}+104(1.5)=-36 + 156 = 120\)
Average velocity \(=\frac{s(1.5)-s(1)}{1.5 - 1}=\frac{120 - 88}{0.5}=\frac{32}{0.5}=64\)

Step 3: Interval \([1, 1.1]\)

Find \(s(1.1)\):
\(s(1.1)=-16(1.1)^{2}+104(1.1)=-16(1.21)+114.4=-19.36 + 114.4 = 95.04\)
Average velocity \(=\frac{s(1.1)-s(1)}{1.1 - 1}=\frac{95.04 - 88}{0.1}=\frac{7.04}{0.1}=70.4\)

Step 4: Interval \([1, 1.01]\)

Find \(s(1.01)\):
\(s(1.01)=-16(1.01)^{2}+104(1.01)=-16(1.0201)+105.04=-16.3216 + 105.04 = 88.7184\)
Average velocity \(=\frac{s(1.01)-s(1)}{1.01 - 1}=\frac{88.7184 - 88}{0.01}=\frac{0.7184}{0.01}=71.84\)

Step 5: Interval \([1, 1.001]\)

Find \(s(1.001)\):
\(s(1.001)=-16(1.001)^{2}+104(1.001)=-16(1.002001)+104.104=-16.032016 + 104.104 = 88.071984\)
Average velocity \(=\frac{s(1.001)-s(1)}{1.001 - 1}=\frac{88.071984 - 88}{0.001}=\frac{0.071984}{0.001}=71.984\)

Answer:

s for each interval:

• \([1, 2]\): \(56\)
• \([1, 1.5]\): \(64\)
• \([1, 1.1]\): \(70.4\)
• \([1, 1.01]\): \(71.84\)
• \([1, 1.001]\): \(71.984\)

Conjecture: As the interval length approaches \(0\) (i.e., the end of the interval approaches \(t = 1\)), the average velocity approaches \(72\) (the instantaneous velocity at \(t = 1\), which can be found by taking the derivative \(s^\prime(t)=-32t + 104\), so \(s^\prime(1)=-32 + 104 = 72\)).