The x- coordinate of a particle moving on x-axis is given by `x = 3 sin 100 t + 8 cos^(2) 50 t`, where x is in cm and t is time in seconds. Find the maximum distance the particle can move from the origin (in cm).

Explanation

To find the maximum distance the particle can move from the origin given the equation \( x = 3 \sin(100t) + 8 \cos^2(50t) \), we will follow these steps:

### Step 1: Rewrite the equation
The equation given is:
\[ x = 3 \sin(100t) + 8 \cos^2(50t) \]
We can use the identity \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \) to rewrite \( \cos^2(50t) \).

### Step 2: Apply the identity
Using the identity, we have:
\[ \cos^2(50t) = \frac{1 + \cos(100t)}{2} \]
Substituting this back into the equation for \( x \):
\[ x = 3 \sin(100t) + 8 \left(\frac{1 + \cos(100t)}{2}\right) \]
This simplifies to:
\[ x = 3 \sin(100t) + 4(1 + \cos(100t)) \]
\[ x = 3 \sin(100t) + 4 + 4 \cos(100t) \]

### Step 3: Combine terms
Now we can combine the sine and cosine terms:
\[ x = 4 + 3 \sin(100t) + 4 \cos(100t) \]

### Step 4: Express in a single sine function
We can express \( 3 \sin(100t) + 4 \cos(100t) \) in the form \( R \sin(100t + \phi) \) where:
- \( R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)
- \( \tan(\phi) = \frac{4}{3} \)

Thus, we have:
\[ x = 4 + 5 \sin(100t + \phi) \]

### Step 5: Find the maximum value of \( x \)
The maximum value of \( \sin(100t + \phi) \) is 1. Therefore, the maximum value of \( x \) is:
\[ x_{\text{max}} = 4 + 5 \cdot 1 = 4 + 5 = 9 \text{ cm} \]

### Conclusion
Thus, the maximum distance the particle can move from the origin is:
\[ \boxed{9 \text{ cm}} \]