To calculate the acceleration and $g$-force from the given function, we need to understand the provided information and apply physics concepts to the context of motion.

Given function: $y = -0.1(x - 20)^2 + 30$

Steps to Derive Acceleration

1. Velocity (First Derivative): The first derivative $y' = \frac{dy}{dx}$ represents the rate of change of $y$ with respect to $x$, which could correspond to velocity if $x$ represents time.

   $y' = -0.2(x - 20)$

   At $x = 30$: $y'(30) = -0.2(10) = -2$

2. Acceleration (Second Derivative): The second derivative $y'' = \frac{d^2y}{dx^2}$ gives the rate of change of velocity, or the acceleration.

   Since $y' = -0.2(x - 20)$, we differentiate again with respect to $x$: $y'' = -0.2$

   This is a constant value, so the acceleration $a = -0.2$ units per second squared.

3. Calculating $g$-Force: The $g$-force is calculated by comparing the acceleration to the standard acceleration due to gravity $g = 9.8 \, \text{m/s}^2$.

   $\text{g-force} = \frac{|a|}{9.8} = \frac{0.2}{9.8} \approx 0.0204 \, g$

Summary:

• Acceleration: $-0.2 \, \text{units/s}^2$
• g-force: $0.0204 \, g$

Would you like further details or have any questions about this calculation?

Related Questions

1. What does each parameter (like 0.1 and 20) represent in the function $y = -0.1(x - 20)^2 + 30$?
2. How would the acceleration change if the function was modified?
3. Why is the $g$-force calculated in relation to the acceleration due to gravity?
4. Can we interpret this function in terms of displacement in physics?
5. What would the result be if we calculated the third derivative, $y'''$?

Tip:

In physics problems, identifying what each variable represents (e.g., time, displacement) can help connect mathematical results to physical interpretations.