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To solve this problem, we need to determine the velocity of the particle at \( t = 5.0 \) seconds based on the information provided.

### Given Information:
1. At \( t = 0 \), the particle is located at \( x = 25 \, \text{m} \) with a velocity of \( v(0) = 15 \, \text{m/s} \).
2. The acceleration \( a_x(t) \) varies with time as shown in the graph. The acceleration decreases linearly from \( 6 \, \text{m/s}^2 \) at \( t = 0 \) to \( 0 \, \text{m/s}^2 \) at \( t = 5 \, \text{s} \).
   
### Key Concept:
The velocity at time \( t = 5.0 \, \text{s} \) can be found using the relationship between velocity, acceleration, and time:
\[
v(t) = v(0) + \int_0^t a_x(t) \, dt
\]
The integral of the acceleration function gives the change in velocity. So, we need to compute the area under the acceleration-time graph from \( t = 0 \) to \( t = 5.0 \, \text{s} \) to find the total change in velocity.

### Step-by-Step Solution:
1. **Area under the acceleration curve**: The acceleration graph shows a triangle with a base of \( 5 \, \text{s} \) and a height of \( 6 \, \text{m/s}^2 \). The area of this triangle represents the total change in velocity.

\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \, \text{s} \times 6 \, \text{m/s}^2 = 15 \, \text{m/s}
\]

2. **Initial velocity**: The initial velocity is \( v(0) = 15 \, \text{m/s} \).

3. **Final velocity**: The total change in velocity is equal to the area under the curve, so the final velocity at \( t = 5 \, \text{s} \) is:
\[
v(5) = v(0) + \Delta v = 15 \, \text{m/s} + 15 \, \text{m/s} = 30 \, \text{m/s}
\]

### Conclusion:
The velocity of the particle at \( t = 5.0 \, \text{s} \) is \( 30 \, \text{m/s} \), which corresponds to option **c) +30 m/s**.

Would you like further details on any of the steps? Here are some additional questions related to the problem:

1. How do we derive the velocity from the acceleration graph using integrals?
2. How would the result change if the acceleration graph had a different shape?
3. Can we calculate the displacement of the particle during the same time interval?
4. What happens to the particle’s velocity after \( t = 5.0 \, \text{s} \) if the acceleration becomes negative?
5. How would the particle's motion differ if the initial velocity were negative?

**Tip**: Remember, the area under the acceleration-time graph always represents the change in velocity!

At t = 0, a particle is located at x = 25 m and has a velocity of 15 m/s in the positive x direction. The acceleration of the particle varies with time as shown in the diagram. What is the velocity of the particle at t = 5.0 s?

Learn how to find the velocity of a particle at t = 5.0 s using its acceleration-time graph. This problem involves kinematic equations, calculus, and graph analysis to compute the change in velocity. Suitable for students in Grades 11-12.

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