To determine the acceleration of the particle at $x = 1 \, \text{m}$, we need to find the derivative of the velocity function with respect to time, $a = \frac{dv}{dt}$.

Given the velocity function $v = \sqrt{100 + 4x - x^2}$, we use the chain rule to express $\frac{dv}{dt}$ in terms of $x$:

$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v$

First, we find $\frac{dv}{dx}$:

$v = (100 + 4x - x^2)^{1/2}$

Using the chain rule for differentiation:

$\frac{dv}{dx} = \frac{d}{dx} \left( (100 + 4x - x^2)^{1/2} \right)$

Applying the chain rule:

$\frac{dv}{dx} = \frac{1}{2} (100 + 4x - x^2)^{-1/2} \cdot (4 - 2x)$ $\frac{dv}{dx} = \frac{4 - 2x}{2\sqrt{100 + 4x - x^2}}$

Simplifying:

$\frac{dv}{dx} = \frac{4 - 2x}{2 \sqrt{100 + 4x - x^2}} = \frac{2 - x}{\sqrt{100 + 4x - x^2}}$

Next, we use the given value of $x = 1$ to find $\frac{dv}{dx}$ at that point:

$v \text{ at } x = 1 = \sqrt{100 + 4(1) - (1)^2} = \sqrt{100 + 4 - 1} = \sqrt{103}$

$\frac{dv}{dx} \text{ at } x = 1 = \frac{2 - 1}{\sqrt{103}} = \frac{1}{\sqrt{103}}$

Now we can find the acceleration:

$a = \frac{dv}{dx} \cdot v$ $a = \frac{1}{\sqrt{103}} \cdot \sqrt{103} = 1 \, \text{m/s}^2$

Thus, the acceleration of the particle at $x = 1 \, \text{m}$ is $1 \, \text{m/s}^2$.

Would you like more details or have any questions?

Here are some other questions you might find interesting:

1. How to find the velocity of a particle at a given time?
2. How does one derive the equation of motion for a particle under constant acceleration?
3. What is the relationship between velocity and acceleration?
4. How do you integrate acceleration to find velocity?
5. What are the applications of derivatives in physics?

Tip: When dealing with kinematics problems, always check if you need to apply the chain rule for differentiation, especially when velocity is given as a function of position.