The new image you've uploaded is identical to the previous graph, showing the velocity $v_x$ as a function of time $t$. The approach to solving Parts A, B, and C will be the same as before:

Part A: Find the particle's position at $t = 3.0 \, \text{s}$

From the velocity-time graph, we know that the area under the curve represents displacement. The shape under the curve from $t = 0 \, \text{s}$ to $t = 3.0 \, \text{s}$ is a right triangle.

• Base of the triangle: $3.0 \, \text{s}$
• Height of the triangle: $6.0 \, \text{m/s}$

The area (displacement) is calculated as: $\text{Displacement} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3.0 \, \text{s} \times 6.0 \, \text{m/s} = 9.0 \, \text{m}$

Thus, the particle's position at $t = 3.0 \, \text{s}$ is $9.0 \, \text{m}$.

Part B: Find the particle's velocity at $t = 3.0 \, \text{s}$

From the graph, the velocity at $t = 3.0 \, \text{s}$ is: $v_x = 0 \, \text{m/s}$

Part C: Find the particle's acceleration at $t = 3.0 \, \text{s}$

The slope of the velocity-time graph gives the acceleration. The slope is calculated as: $a = \frac{\Delta v_x}{\Delta t} = \frac{0 \, \text{m/s} - 6 \, \text{m/s}}{3.0 \, \text{s}} = -2.0 \, \text{m/s}^2$

Final Answers:

• Part A: The particle's position at $t = 3.0 \, \text{s}$ is $9.0 \, \text{m}$.
• Part B: The particle's velocity at $t = 3.0 \, \text{s}$ is $0 \, \text{m/s}$.
• Part C: The particle's acceleration at $t = 3.0 \, \text{s}$ is $-2.0 \, \text{m/s}^2$.

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