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Let's solve each part step-by-step.

### Part (a)
To estimate \( v'(8) \) using the data table, we need to approximate the derivative at \( t = 8 \). Since \( v'(t) \) represents the rate of change of \( v(t) \), we can use the values of \( v(t) \) at \( t = 6 \) and \( t = 10 \) to estimate \( v'(8) \) with the average rate of change formula:

\[
v'(8) \approx \frac{v(10) - v(6)}{10 - 6} = \frac{600 - 500}{10 - 6} = \frac{100}{4} = 25 \, \text{meters per minute squared}
\]

So, \( v'(8) \approx 25 \, \text{m/min}^2 \).

### Part (b)
To find John's average acceleration over the interval \( 6 \leq t \leq 12 \), we calculate the average rate of change of \( v(t) \) between \( t = 6 \) and \( t = 12 \):

\[
\text{Average acceleration} = \frac{v(12) - v(6)}{12 - 6} = \frac{-430 - 500}{12 - 6} = \frac{-930}{6} = -155 \, \text{m/min}^2
\]

So, the average acceleration over this interval is \( -155 \, \text{m/min}^2 \).

### Part (c)
Sam’s velocity is given by \( v(t) = 0.5t^3 - 2t^2 + 100 \). To find her acceleration at \( t = 3 \), we need to find \( v'(t) \), the derivative of \( v(t) \), and then evaluate it at \( t = 3 \).

1. Differentiate \( v(t) \):
   \[
   v'(t) = \frac{d}{dt} (0.5t^3 - 2t^2 + 100) = 1.5t^2 - 4t
   \]

2. Substitute \( t = 3 \):
   \[
   v'(3) = 1.5(3)^2 - 4(3) = 1.5 \cdot 9 - 12 = 13.5 - 12 = 1.5 \, \text{m/min}^2
   \]

So, Sam’s acceleration at \( t = 3 \) is \( 1.5 \, \text{m/min}^2 \).

### Part (d)
At \( t = 3 \), John is at \( x_J = 200 \) meters, and Sam is at \( x_S = 300 \) meters. We are asked to find the rate at which the distance between John and Sam is changing at \( t = 3 \).

1. The distance between John and Sam at any time \( t \) is \( D(t) = |x_S(t) - x_J(t)| \).
2. Since we are given only their velocities, we will interpret \( v_J(3) \) and \( v_S(3) \) as the rates of change of their positions with respect to time.

Using \( v_J(3) = 500 \) m/min (from the table) and \( v_S(3) = v(3) = 0.5(3)^3 - 2(3)^2 + 100 = 94.5 \) m/min:

\[
\frac{dD}{dt} = v_S(3) - v_J(3) = 94.5 - 500 = -405.5 \, \text{m/min}
\]

Thus, the distance between John and Sam is decreasing at a rate of \( 405.5 \, \text{m/min} \) at \( t = 3 \).

Would you like further explanations on any part? Here are related questions for practice:

1. How would you estimate \( v'(t) \) if more data points were available around \( t = 8 \)?
2. What would be John's position at \( t = 10 \) if his initial position was 0?
3. How can we determine Sam's position function if we know her velocity function?
4. What is the significance of a negative acceleration in the context of John's motion?
5. How would the distance between John and Sam change if Sam had a different initial velocity?

**Tip:** For average rates of change, remember to focus on initial and final values within the interval!

John is riding a bike along a straight path. For 0 ≤ t ≤ 20, John's velocity is given by a differentiable function v. Selected values of v(t), where t is measured in minutes and v(t) is measured in meters per minute, are given in the table. Solve the following:

(a) Use the data table to estimate the value of v'(8).
(b) Find John's average acceleration over the interval 6 ≤ t ≤ 12.
(c) Sam is jogging along the same path. For 0 ≤ t ≤ 5, Sam's velocity is modeled by v(t) = 0.5t^3 - 2t^2 + 100. Find her acceleration at time t = 3.
(d) At time t = 3, John is at a position xJ = 200 meters and Sam is at a position xS = 300 meters. Find the rate, in meters per minute, at which the distance between John and Sam is changing at time t = 3.

This problem explores how to estimate derivatives, calculate average acceleration, and determine rate of distance change using velocity data. Key concepts include differentiation, average rate of change, and instantaneous acceleration, suitable for high school students in Grades 11-12.

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