In each of the following,  s (in ft) is the directed distance of a moving body or particle from the origin at time t (in sec) on a coordinate line. Describe and diagram the motion for t is greater than equal to 0. 
s=2t^3-15t^2+36t

We are given the position function of a moving particle on a coordinate line as:

\[
s(t) = 2t^3 - 15t^2 + 36t
\]

where \( s(t) \) is the directed distance from the origin at time \( t \geq 0 \).

### Step 1: Velocity Function
The velocity of the particle is the derivative of the position function:

\[
v(t) = \frac{ds}{dt} = 6t^2 - 30t + 36
\]

### Step 2: Critical Points of Velocity
To find when the particle is at rest or changing direction, set the velocity equal to zero:

\[
6t^2 - 30t + 36 = 0
\]

Dividing through by 6:

\[
t^2 - 5t + 6 = 0
\]

Factoring:

\[
(t - 2)(t - 3) = 0
\]

Thus, \( t = 2 \) and \( t = 3 \) are critical points.

### Step 3: Acceleration
The acceleration is the derivative of the velocity:

\[
a(t) = \frac{dv}{dt} = 12t - 30
\]

This tells us how the velocity is changing over time.

### Step 4: Analyzing Motion
Now, we analyze the motion in intervals based on the critical points \( t = 2 \) and \( t = 3 \).

- For \( t < 2 \): Choose \( t = 1 \), we substitute into the velocity function:
  \[
  v(1) = 6(1)^2 - 30(1) + 36 = 6 - 30 + 36 = 12
  \]
  So the particle is moving forward.
  
- For \( 2 < t < 3 \): Choose \( t = 2.5 \), we substitute into the velocity function:
  \[
  v(2.5) = 6(2.5)^2 - 30(2.5) + 36 = 37.5 - 75 + 36 = -1.5
  \]
  So the particle is moving backward.
  
- For \( t > 3 \): Choose \( t = 4 \), we substitute into the velocity function:
  \[
  v(4) = 6(4)^2 - 30(4) + 36 = 96 - 120 + 36 = 12
  \]
  So the particle is moving forward again.

### Step 5: Diagram of the Motion
- The particle starts moving forward when \( t < 2 \).
- It stops momentarily at \( t = 2 \), then moves backward between \( 2 < t < 3 \).
- The particle stops again at \( t = 3 \), and then continues to move forward for \( t > 3 \).

### Diagram:

- Between \( t = 0 \) and \( t = 2 \), the velocity is positive, so the particle is moving away from the origin.
- Between \( t = 2 \) and \( t = 3 \), the velocity is negative, so the particle moves toward the origin.
- After \( t = 3 \), the particle moves away from the origin again.

The turning points at \( t = 2 \) and \( t = 3 \) represent moments when the particle changes direction.

---

Would you like further details, or do you have any specific questions?

### Related Questions:
1. What happens to the particle's motion if the initial velocity is changed?
2. How do acceleration and velocity behave over time for this function?
3. What is the total distance traveled by the particle for \( t \geq 0 \)?
4. How would the motion differ if the particle starts at a different initial position?
5. How do inflection points affect the motion in problems like this?

### Tip: 
Always check the sign of velocity to determine the direction of the motion: positive means moving forward, and negative means moving backward.

In each of the following, s (in ft) is the directed distance of a moving body or particle from the origin at time t (in sec) on a coordinate line. Describe and diagram the motion for t is greater than equal to 0. s = 2t^3 - 15t^2 + 36t

Explore particle motion for s = 2t^3 - 15t^2 + 36t on a coordinate line, with critical points at t = 2 and t = 3. Learn how to compute velocity and acceleration, analyze direction changes, and create a motion diagram. Suitable for Grades 11-12.

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