Let's break down the graph and solve the problem step by step.

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

The graph shows the velocity $v_x$ as a function of time $t$. The slope of the velocity-time graph indicates acceleration, and the area under the velocity-time curve represents displacement.

• The initial velocity $v_x(0) = 6 \, \text{m/s}$.
• At $t = 3.0 \, \text{s}$, $v_x(3) = 0 \, \text{m/s}$.

The graph is a straight line, so the velocity decreases linearly, meaning the acceleration is constant. To find the position at $t = 3.0 \, \text{s}$, we use the area under the velocity curve from $t = 0$ to $t = 3.0 \, \text{s}$, which is the displacement:

The shape under the curve is a right triangle. The area of this triangle gives the displacement.

• The base of the triangle is $3.0 \, \text{s}$ (the time interval).
• The height of the triangle is $6.0 \, \text{m/s}$ (the initial velocity).

The area of the triangle is: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3.0 \, \text{s} \times 6.0 \, \text{m/s} = 9.0 \, \text{m}$

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, we can directly see that the velocity at $t = 3.0 \, \text{s}$ is: $v_x = 0 \, \text{m/s}$ This is because the velocity linearly decreases to zero at $t = 3.0 \, \text{s}$.

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

The acceleration is the slope of the velocity-time graph. Since the velocity decreases uniformly, the slope (acceleration) is constant.

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

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 $-2.0 \, \text{m/s}^2$.

Would you like more details or clarifications?

5 Related Questions:

1. How do you calculate displacement when velocity varies non-linearly?
2. How does the area under a curve relate to motion in other types of graphs?
3. What would happen to the particle’s motion if acceleration were positive?
4. How can you calculate average velocity from a velocity-time graph?
5. How does changing initial velocity affect total displacement?

Tip:

Always remember, for a velocity-time graph, the area under the curve represents displacement, and the slope gives acceleration.