The problem in the images provides a function for the position of a particle moving along a line, $s = f(t)$, where $t \geq 0$ is time in seconds and $s$ is the position in feet. The position function given is:

$f(t) = t^3 - 12t^2 + 36t$

Let's work through the questions:

(f) Finding the acceleration function $a(t)$:

To find the acceleration, we need the second derivative of the position function with respect to time.

1. First Derivative (Velocity Function $v(t)$): $v(t) = \frac{d}{dt}[t^3 - 12t^2 + 36t] = 3t^2 - 24t + 36$

2. Second Derivative (Acceleration Function $a(t)$): Now, differentiate the velocity function to get the acceleration: $a(t) = \frac{d}{dt}[3t^2 - 24t + 36] = 6t - 24$ Thus, the acceleration function is: $a(t) = 6t - 24 \, \text{(in ft/s}^2\text{)}$

(f) Finding the acceleration at $t = 5$:

To find $a(5)$, substitute $t = 5$ into the acceleration function:

$a(5) = 6(5) - 24 = 30 - 24 = 6 \, \text{ft/s}^2$

So, the acceleration after 5 seconds is $6 \, \text{ft/s}^2$.

(g) Graphing the position, velocity, and acceleration functions for the first 8 seconds:

To graph these functions, we use the following equations:

1. Position function: $f(t) = t^3 - 12t^2 + 36t$
2. Velocity function: $v(t) = 3t^2 - 24t + 36$
3. Acceleration function: $a(t) = 6t - 24$

These functions can be plotted for $t$ in the range $0 \leq t \leq 8$ seconds.

Would you like further assistance with these graphs or any other details?

5 Related Questions:

1. How is the velocity related to the position function?
2. What is the physical meaning of acceleration being positive or negative?
3. Can the particle ever stop moving? If so, at what times does this occur?
4. What is the maximum or minimum position of the particle in the given time interval?
5. How would the acceleration function change if the position function was different?

Tip:

When analyzing motion, always remember that the velocity is the derivative of position and the acceleration is the derivative of velocity.