Questions: The following problem is similar to a problem from your textbook. The table below shows the position s of a car at time t. Round your answers to one decimal place. t(sec) 0 0.2 0.4 0.6 0.8 1 s(ft) 0 2.7 2.8 5.2 5.8 7.6 Estimate the velocity at t=0.2 by averaging the average rate of change over the left and right intervals.

Solution Steps

Given the positions at times $t_a=0$ and $t_b=0.8$ are $s_a=0$ and $s_b=5.8$, respectively, the average velocity over the interval $[0, 0.8]$ is calculated using the formula: \[ \text{Average Velocity} = \frac{s_b - s_a}{t_b - t_a} = \frac{5.8 - 0}{0.8 - 0} = 7.2 \text{ ft/s} \]

To estimate the velocity at time $t_c=0.2$, we find two time points $t_{c1}=0$ and $t_{c2}=0.4$ such that $t_{c1} < t_c < t_{c2}$. The positions at these times are $s_{c1}=0$ and $s_{c2}=2.8$, respectively. The estimated velocity at $t_c$ is calculated as: \[ \text{Estimated Velocity at } t_c = \frac{s_{c2} - s_{c1}}{t_{c2} - t_{c1}} = \frac{2.8 - 0}{0.4 - 0} = 7 \text{ ft/s} \]

Final Answer: