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!