Average Velocity 4Ee088

1. **State the problem:** We have the position function $$s(t) = -16 t^2 + 104 t$$ and need to find the average velocity over several time intervals starting at $t=1$. 2. **Recall the formula for average velocity:** $$\text{Average velocity} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$$ where $t_1$ and $t_2$ are the endpoints of the time interval. 3. **Calculate $s(t)$ at the relevant points:** - $s(1) = -16(1)^2 + 104(1) = -16 + 104 = 88$ - $s(2) = -16(2)^2 + 104(2) = -16(4) + 208 = -64 + 208 = 144$ - $s(1.5) = -16(1.5)^2 + 104(1.5) = -16(2.25) + 156 = -36 + 156 = 120$ - $s(1.1) = -16(1.1)^2 + 104(1.1) = -16(1.21) + 114.4 = -19.36 + 114.4 = 95.04$ - $s(1.01) = -16(1.01)^2 + 104(1.01) = -16(1.0201) + 105.04 = -16.3216 + 105.04 = 88.7184$ - $s(1.001) = -16(1.001)^2 + 104(1.001) = -16(1.002001) + 104.104 = -16.032016 + 104.104 = 88.071984$ 4. **Calculate average velocities for each interval:** - For $[1, 2]$: $$\frac{s(2) - s(1)}{2 - 1} = \frac{144 - 88}{1} = 56$$ - For $[1, 1.5]$: $$\frac{s(1.5) - s(1)}{1.5 - 1} = \frac{120 - 88}{0.5} = \frac{32}{0.5} = 64$$ - For $[1, 1.1]$: $$\frac{s(1.1) - s(1)}{1.1 - 1} = \frac{95.04 - 88}{0.1} = \frac{7.04}{0.1} = 70.4$$ - For $[1, 1.01]$: $$\frac{s(1.01) - s(1)}{1.01 - 1} = \frac{88.7184 - 88}{0.01} = \frac{0.7184}{0.01} = 71.84$$ - For $[1, 1.001]$: $$\frac{s(1.001) - s(1)}{1.001 - 1} = \frac{88.071984 - 88}{0.001} = \frac{0.071984}{0.001} = 71.984$$ 5. **Conjecture about instantaneous velocity at $t=1$:** As the interval shrinks towards $t=1$, the average velocity approaches approximately $72$. Therefore, the instantaneous velocity at $t=1$ is likely $$72$$. **Final answers:** | Time Interval | Average Velocity | |---------------|------------------| | $[1, 2]$ | 56 | | $[1, 1.5]$ | 64 | | $[1, 1.1]$ | 70.4 | | $[1, 1.01]$ | 71.84 | | $[1, 1.001]$ | 71.984 |