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### Problem Overview:
The motion of a particle is defined by a law of motion \( s(t) = t^3 - 15t^2 + 72t \), where \( t \) is in seconds, and \( s \) is in feet. The problem involves finding the velocity, acceleration, rest intervals, motion direction, and plotting graphs.

### (a) Velocity at time \( t \):
The velocity function \( v(t) \) is the derivative of the position function \( s(t) \). So,

\[
v(t) = \frac{d}{dt}(t^3 - 15t^2 + 72t) = 3t^2 - 30t + 72
\]

### (b) Velocity after 5 seconds:
Substitute \( t = 5 \) into the velocity equation \( v(t) = 3t^2 - 30t + 72 \):

\[
v(5) = 3(5)^2 - 30(5) + 72 = 3(25) - 150 + 72 = 75 - 150 + 72 = -3 \, \text{ft/s}
\]

The velocity after 5 seconds is \( v(5) = -3 \, \text{ft/s} \).

### (c) When is the particle at rest?
The particle is at rest when the velocity is zero, i.e., \( v(t) = 0 \):

\[
3t^2 - 30t + 72 = 0
\]

Divide by 3 to simplify:

\[
t^2 - 10t + 24 = 0
\]

Now solve this quadratic equation using the quadratic formula:

\[
t = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(24)}}{2(1)} = \frac{10 \pm \sqrt{100 - 96}}{2} = \frac{10 \pm \sqrt{4}}{2} = \frac{10 \pm 2}{2}
\]

So, \( t = \frac{10 + 2}{2} = 6 \) and \( t = \frac{10 - 2}{2} = 4 \).

The particle is at rest at \( t = 4 \, \text{s} \) and \( t = 6 \, \text{s} \).

### (d) When is the particle moving in the positive direction?
The particle moves in the positive direction when \( v(t) > 0 \). From the velocity equation \( 3t^2 - 30t + 72 \), solve for when the velocity is positive:

The roots of \( v(t) = 0 \) are \( t = 4 \) and \( t = 6 \). Using a sign chart for \( v(t) \), the velocity is positive when \( t \in [0, 4) \cup (6, \infty) \).

### (e) Total distance traveled in the first 8 seconds:
The total distance traveled requires evaluating the position changes and taking into account direction changes at rest points \( t = 4 \) and \( t = 6 \). We find the position at these critical points:

\[
s(0) = 0, \quad s(4), \quad s(6), \quad s(8)
\]

This part requires specific calculations for \( s(t) \) at these values to find the total distance.

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Let me know if you would like further details for any part, and here are some related questions to expand your understanding:

1. How do you derive the acceleration function from the velocity function?
2. Why is it important to consider direction changes in calculating total distance traveled?
3. How can the quadratic formula be used in physical problems like this one?
4. Can you explain the significance of the velocity being negative at certain points in time?
5. How do the graphs of position, velocity, and acceleration relate to each other?

### Tip:
Always check for intervals where the particle changes direction (e.g., where velocity changes sign) when calculating total distance, since distance is the sum of absolute displacements, not just the net change in position.

A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s is in feet. The function given is s(t) = t^3 - 15t^2 + 72t. Various aspects of the motion are analyzed, including velocity, acceleration, rest intervals, and graphs.

Analyze the motion of a particle defined by s(t) = t^3 - 15t^2 + 72t. This detailed solution covers velocity, acceleration, and graphical interpretations, including when the particle is at rest and its total distance traveled in the first 8 seconds. Suitable for high school students learning calculus.

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