To find the average velocity and estimate the velocity at $t = 0.2$, we will proceed step by step:

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Step 1: Average Velocity Over $0 \leq t \leq 0.2$

The formula for average velocity over an interval $[t_1, t_2]$ is:
$v_{\text{avg}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$

For $0 \leq t \leq 0.2$, the values from the table are:
$t_1 = 0$, $s(t_1) = 0$
$t_2 = 0.2$, $s(t_2) = 0.5$

Substitute these values: $v_{\text{avg}} = \frac{s(0.2) - s(0)}{0.2 - 0} = \frac{0.5 - 0}{0.2} = 2.5 \, \text{ft/s}$

Thus, the average velocity is: $v_{\text{avg}} = 2.5 \, \text{ft/s}$

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Step 2: Estimate the Velocity at $t = 0.2$

To estimate the instantaneous velocity at $t = 0.2$, we calculate the slope of the secant lines surrounding $t = 0.2$. These slopes approximate the velocity.

1. Using the interval $[0, 0.2]$: We already calculated: $v = \frac{s(0.2) - s(0)}{0.2 - 0} = 2.5 \, \text{ft/s}$

2. Using the interval $[0.2, 0.4]$:
   From the table: $s(0.2) = 0.5$, $s(0.4) = 1.5$

   Slope: $v = \frac{s(0.4) - s(0.2)}{0.4 - 0.2} = \frac{1.5 - 0.5}{0.2} = 5 \, \text{ft/s}$

3. Average the two slopes to approximate $v(0.2)$:
   $v(0.2) \approx \frac{2.5 + 5}{2} = 3.75 \, \text{ft/s}$

Thus, the estimated velocity at $t = 0.2$ is: $v(0.2) \approx 3.75 \, \text{ft/s}$

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Final Results:

1. Average velocity over $0 \leq t \leq 0.2$:
   $v_{\text{avg}} = 2.5 \, \text{ft/s}$

2. Estimated velocity at $t = 0.2$:
   $v(0.2) \approx 3.75 \, \text{ft/s}$

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Would you like a graph or further explanation on secant slopes?
Here are some follow-up questions:

1. How does instantaneous velocity differ from average velocity?
2. Why does taking smaller intervals improve the estimate for instantaneous velocity?
3. What is the relationship between velocity and the derivative of the position function?
4. How could you improve the velocity estimate at $t = 0.2$?
5. What tools can be used to visualize motion problems like this one?

Tip: For a better approximation of instantaneous velocity, consider taking smaller intervals around the point of interest!