Let us solve each part step by step.

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(a) Find the average velocity in the time interval $t = 1.00 \, \text{s}$ to $t = 3.50 \, \text{s}$:

The formula for average velocity is: $v_{\text{avg}} = \frac{\Delta x}{\Delta t}$ Here:

• At $t = 1.00 \, \text{s}$, $x = 11 \, \text{m}$ (as given in the problem).
• At $t = 3.50 \, \text{s}$, $x = 0 \, \text{m}$ (from the line equation).

So: $\Delta x = x_{\text{final}} - x_{\text{initial}} = 0 - 11 = -11 \, \text{m}$ $\Delta t = t_{\text{final}} - t_{\text{initial}} = 3.50 - 1.00 = 2.50 \, \text{s}$

Substitute these values: $v_{\text{avg}} = \frac{-11}{2.50} = -4.4 \, \text{m/s}$

Answer (a): $v_{\text{avg}} = -4.4 \, \text{m/s}$

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(b) Determine the instantaneous velocity at $t = 2.00 \, \text{s}$:

The instantaneous velocity at $t = 2.00 \, \text{s}$ is the slope of the tangent line at this point.

From the graph:

• The tangent line at $t = 2.00 \, \text{s}$ touches the curve at approximately $(2.00, 6.00)$.
• Use the tangent slope formula: $\text{slope} = \frac{\Delta x}{\Delta t}$ From the tangent line, approximate the coordinates of two points:
• Point 1: $(1.50, 9.0)$
• Point 2: $(2.50, 3.0)$

So: $\Delta x = 3.0 - 9.0 = -6.0, \quad \Delta t = 2.50 - 1.50 = 1.00$ $\text{slope} = \frac{-6.0}{1.00} = -6.0 \, \text{m/s}$

Answer (b): $v_{\text{inst}} = -6.0 \, \text{m/s}$

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(c) At what value of $t$ is the velocity zero?

Velocity is zero where the slope of the curve is horizontal. From the graph:

• The slope becomes zero at approximately $t = 4.00 \, \text{s}$.

Answer (c): $t = 4.00 \, \text{s}$

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Summary of Answers:

1. (a) $v_{\text{avg}} = -4.4 \, \text{m/s}$
2. (b) $v_{\text{inst}} = -6.0 \, \text{m/s}$
3. (c) $t = 4.00 \, \text{s}$

Let me know if you'd like additional clarifications or details! 😊

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Related Questions:

1. How do you determine the average velocity from a graph when the motion is non-linear?
2. What are the key differences between average and instantaneous velocity?
3. How does the slope of a position-time graph relate to motion direction?
4. Can you have a positive velocity and negative acceleration simultaneously?
5. How can a particle’s velocity change direction while maintaining constant speed?

Tip: Always check units and graph scales carefully while interpreting motion data!