https://file.aiquickdraw.com/mathbot/file/1726201166601yhq9ss4k.jpg

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)

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

If you still have difficulties entering the answer correctly

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

Learn how to find a particle's position, velocity, and acceleration at t = 3.0 s using a velocity-time graph. Understand key concepts in kinematics and learn to calculate displacement and acceleration. Suitable for Grades 9-12.

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