To find the best approximation of the instantaneous velocity at $t = 1$ second, we can calculate the average rate of change of $s(t)$ between two points close to $t = 1$.

The formula for the average rate of change is:

$v = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$

From the table:

• For $t = 1.0$, $s(t) = 13.44$
• For $t = 1.1$, $s(t) = 11.9676$
• For $t = 0.9$, $s(t) = 14.9276$

We calculate:

1. Forward difference: $v = \frac{11.9676 - 13.44}{1.1 - 1.0} = \frac{-1.4724}{0.1} = -14.724$

2. Backward difference: $v = \frac{13.44 - 14.9276}{1.0 - 0.9} = \frac{-1.4876}{0.1} = -14.876$

3. Central difference: $v = \frac{11.9676 - 14.9276}{1.1 - 0.9} = \frac{-2.96}{0.2} = -14.8$

Best Approximation:

The central difference often gives the best approximation of instantaneous velocity. Thus, the velocity is approximately $-14.8$, which corresponds to option A.

Let me know if you need detailed explanations or have questions.
Here are some related questions to consider:

1. Why is the central difference method more accurate than forward or backward difference methods?
2. How does the instantaneous velocity relate to the derivative of $s(t)$?
3. Can we estimate the instantaneous velocity more precisely with smaller intervals?
4. What would the velocity be if the table were extended beyond $t = 2.0$?
5. How can the graph of $s(t)$ help in visualizing instantaneous velocity?

Tip: Always check both forward and backward differences to validate your approximations for instantaneous rates.