To calculate the acceleration and the g-force for the function given in the image, we need to understand the context in which "acceleration" is being applied here. In mathematical terms, acceleration is often the second derivative of a position function $f(x)$ with respect to time. Here, $f(x)$ appears to be given in terms of $x$, so without additional context (such as interpreting $x$ as time), I'll assume we're looking for the second derivative of $f(x)$, which would represent "acceleration" if $x$ were time.

Step-by-Step Solution

The function $f(x)$ given is: $f(x) = e^{(0.5x - 5)}$

1. First Derivative $f'(x)$: The first derivative, $f'(x)$, has already been calculated in the image: $f'(x) = e^{(0.5x - 5)} \cdot 0.5 = \frac{e^{(0.5x - 5)}}{2}$

2. Second Derivative $f''(x)$ (for "acceleration"): To find the acceleration, we take the derivative of $f'(x)$. $f''(x) = \frac{d}{dx} \left( \frac{e^{(0.5x - 5)}}{2} \right)$ Using the chain rule, we get: $f''(x) = \frac{1}{2} \cdot e^{(0.5x - 5)} \cdot 0.5 = \frac{e^{(0.5x - 5)}}{4}$

So the "acceleration" $f''(x)$ is: $f''(x) = \frac{e^{(0.5x - 5)}}{4}$

3. Calculating $f''(16)$: To find the specific acceleration at $x = 16$: $f''(16) = \frac{e^{(0.5 \cdot 16 - 5)}}{4} = \frac{e^{8 - 5}}{4} = \frac{e^3}{4}$

   So, the acceleration at $x = 16$ is $\frac{e^3}{4}$.

4. Calculating the g-force: The g-force is typically calculated as $g = \frac{a}{9.8}$, where $a$ is the acceleration in units of $\text{m/s}^2$ and $9.8 \, \text{m/s}^2$ is the acceleration due to gravity.

   If $f(x)$ is measured in meters (which we assume unless otherwise specified), the g-force at $x = 16$ would be: $\text{g-force} = \frac{\frac{e^3}{4}}{9.8} = \frac{e^3}{39.2}$

Final Answer

• Acceleration at $x = 16$: $f''(16) = \frac{e^3}{4}$
• G-force at $x = 16$: $\frac{e^3}{39.2}$

Would you like a more detailed explanation of any part?

Here are 5 related questions that may deepen your understanding:

1. How do you interpret acceleration when dealing with exponential functions?
2. Why do we use the second derivative to find acceleration?
3. What would happen if the exponent of $e$ were a more complex function?
4. How do g-forces vary with changing acceleration values?
5. What are practical applications of calculating g-forces in physics?

Tip: Remember that exponential functions grow very quickly, so even a small change in $x$ can lead to a significant change in the function value and its derivatives.