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The problem describes the motion of a particle under varying acceleration. The acceleration graph indicates a linear decrease in acceleration, starting at \( a_x = 6 \, \text{m/s}^2 \) at \( t = 0 \) and reaching \( a_x = 0 \, \text{m/s}^2 \) at \( t = 6 \, \text{s} \).

### Step-by-step analysis:

1. **Acceleration Function:**
   The acceleration is given as \( a_x = -t + 6 \). This equation is derived from the slope of the graph. It's a line with a negative slope of \(-1\) and a y-intercept of 6.

2. **Velocity Function:**
   To find velocity, integrate the acceleration function:
   \[
   v(t) = \int (-t + 6) \, dt = -\frac{t^2}{2} + 6t + C
   \]
   We know that at \( t = 0 \), the velocity is 15 m/s, so:
   \[
   v(0) = 15 = -\frac{0^2}{2} + 6(0) + C \quad \Rightarrow \quad C = 15
   \]
   Thus, the velocity function becomes:
   \[
   v(t) = -\frac{t^2}{2} + 6t + 15
   \]

3. **Position Function:**
   Now, integrate the velocity function to find the position function:
   \[
   x(t) = \int \left( -\frac{t^2}{2} + 6t + 15 \right) dt = -\frac{t^3}{6} + 3t^2 + 15t + C'
   \]
   Given that at \( t = 0 \), \( x(0) = 25 \) m, we find:
   \[
   x(0) = 25 = -\frac{0^3}{6} + 3(0^2) + 15(0) + C' \quad \Rightarrow \quad C' = 25
   \]
   Therefore, the position function is:
   \[
   x(t) = -\frac{t^3}{6} + 3t^2 + 15t + 25
   \]

4. **Finding the position at \( t = 5 \, \text{s} \):**
   Substitute \( t = 5 \) into the position function:
   \[
   x(5) = -\frac{5^3}{6} + 3(5^2) + 15(5) + 25
   \]
   \[
   x(5) = -\frac{125}{6} + 3(25) + 75 + 25
   \]
   \[
   x(5) = -20.83 + 75 + 75 + 25 = 154.17 \, \text{m}
   \]
   So, the particle's position at \( t = 5 \, \text{s} \) is approximately \( 154 \, \text{m} \).

The correct answer is **154 m**, which corresponds to option **(d)**.

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Do you have any questions or would you like further details on any step? Here are some related questions you could explore:

1. How does the position function change if the initial velocity is different?
2. How would the acceleration equation change if the slope of the graph was steeper?
3. What would happen if the acceleration was constant instead of linear?
4. How can we use the velocity function to find when the particle stops moving?
5. What is the displacement of the particle between \( t = 0 \) and \( t = 5 \)?

**Tip:** Always integrate acceleration to get velocity and velocity to get position. Use initial conditions to solve for constants of integration.

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 position of the particle at t = 5.0 s?

Solve a kinematics problem involving a particle moving under varying acceleration. Learn how to calculate the velocity and position by integrating the acceleration function. The initial velocity is 15 m/s and initial position is 25 m. Find the position at t = 5.0 seconds.

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