1. **State the problem:** We have a position function $y = 28t - 10t^2$ describing the position of an object over time $t$ in seconds. We want to find the velocity at specific small time intervals and then estimate the instantaneous velocity at $t=2$ seconds. 2. **Recall the formula:** Velocity is the rate of change of position with respect to time. The average velocity over a small interval $\Delta t$ is given by: $$v_{avg} = \frac{y(t + \Delta t) - y(t)}{\Delta t}$$ The instantaneous velocity at time $t$ is the limit of the average velocity as $\Delta t \to 0$, which is the derivative: $$v(t) = \frac{dy}{dt}$$ 3. **Calculate average velocities:** - At $t=2$, calculate $v_{avg}$ for $\Delta t = 0.01, 0.005, 0.002, 0.001$ seconds. Calculate $y(2)$: $$y(2) = 28(2) - 10(2)^2 = 56 - 40 = 16$$ Calculate $y(2 + \Delta t)$ for each $\Delta t$: - For $\Delta t=0.01$: $$y(2.01) = 28(2.01) - 10(2.01)^2 = 56.28 - 10(4.0401) = 56.28 - 40.401 = 15.879$$ Average velocity: $$v_{avg} = \frac{15.879 - 16}{0.01} = \frac{-0.121}{0.01} = -12.1$$ - For $\Delta t=0.005$: $$y(2.005) = 28(2.005) - 10(2.005)^2 = 56.14 - 10(4.020025) = 56.14 - 40.20025 = 15.93975$$ $$v_{avg} = \frac{15.93975 - 16}{0.005} = \frac{-0.06025}{0.005} = -12.05$$ - For $\Delta t=0.002$: $$y(2.002) = 28(2.002) - 10(2.002)^2 = 56.056 - 10(4.008004) = 56.056 - 40.08004 = 15.97596$$ $$v_{avg} = \frac{15.97596 - 16}{0.002} = \frac{-0.02404}{0.002} = -12.02$$ - For $\Delta t=0.001$: $$y(2.001) = 28(2.001) - 10(2.001)^2 = 56.028 - 10(4.004001) = 56.028 - 40.04001 = 15.98799$$ $$v_{avg} = \frac{15.98799 - 16}{0.001} = \frac{-0.01201}{0.001} = -12.01$$ 4. **Find instantaneous velocity by differentiation:** $$v(t) = \frac{dy}{dt} = 28 - 20t$$ At $t=2$: $$v(2) = 28 - 20(2) = 28 - 40 = -12$$ 5. **Interpretation:** The average velocities approach $-12$ as $\Delta t$ gets smaller, confirming the instantaneous velocity at $t=2$ is $-12$ ft/sec. **Final answers:** - At 0.01 seconds: $-12.1$ ft/sec - At 0.005 seconds: $-12.05$ ft/sec - At 0.002 seconds: $-12.02$ ft/sec - At 0.001 seconds: $-12.01$ ft/sec - Instantaneous velocity at $t=2$: $-12$ ft/sec