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At  t
  = 3.0  s
 , what is the particle's position?
Express your answer with the appropriate units.
View Available Hint(s)for Part A
Hint 1for Part A. How to approach the problem
  
  
  
 
Hint 2for Part A. Simplify: equation for displacement
  
  
  
  
  
 
 
 
 
 
Hint 3for Part A. Simplify: find displacement

  
 

 
Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value type

x
  =
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Enter your answer using units of distance.
Part B
At  t
  = 3.0  s
 , what is the particle's velocity?
Express your answer with the appropriate units.
View Available Hint(s)for Part B
Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value type

vx
  =
nothing
nothing
Part C
At  t
  = 3.0  s
 , what is the particle's acceleration?
Express your answer with the appropriate units.
View Available Hint(s)

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.

At t = 3.0 s, what is the particle's position? Express your answer with the appropriate units.

Learn how to determine the position and acceleration of a particle at t = 3.0 seconds using the velocity-time graph. This problem covers concepts like displacement as the area under the curve and constant acceleration, making it ideal for high school physics students.

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