Consider a particle moving along a straight line with velocity v(t) = t^3 - 6 t^2 
 + 9t. Determine the time(s) when the particle changes direction. What is the acceleration of the particle at those times? Calculate the total distance traveled by the particle from t = 0 to the first time it changes direction.

### Step 1: Determine when the particle changes direction
A particle changes direction when its velocity \( v(t) \) changes sign, which happens when \( v(t) = 0 \).

Given the velocity function:
\[
v(t) = t^3 - 6t^2 + 9t
\]
We need to find the times when \( v(t) = 0 \).

\[
t^3 - 6t^2 + 9t = 0
\]
Factor the equation:
\[
t(t^2 - 6t + 9) = 0
\]
\[
t(t - 3)^2 = 0
\]
This gives \( t = 0 \) and \( t = 3 \). So, the particle changes direction at \( t = 3 \).

### Step 2: Find the acceleration at those times
The acceleration is the derivative of the velocity function \( v(t) \), which is:
\[
a(t) = \frac{dv(t)}{dt} = 3t^2 - 12t + 9
\]

Now, evaluate the acceleration at the times the particle changes direction:

- At \( t = 0 \):
  \[
  a(0) = 3(0)^2 - 12(0) + 9 = 9
  \]

- At \( t = 3 \):
  \[
  a(3) = 3(3)^2 - 12(3) + 9 = 3(9) - 36 + 9 = 27 - 36 + 9 = 0
  \]

So, the acceleration is 9 at \( t = 0 \) and 0 at \( t = 3 \).

### Step 3: Calculate the total distance traveled by the particle from \( t = 0 \) to \( t = 3 \)
To calculate the total distance traveled, we need to integrate the velocity function over the interval \( [0, 3] \), but since the particle doesn't change direction between \( t = 0 \) and \( t = 3 \) (velocity remains positive or negative consistently), we can directly compute the displacement.

The displacement is the integral of the velocity function:
\[
\text{Distance} = \int_0^3 v(t) \, dt
\]
First, integrate \( v(t) \):
\[
\int v(t) \, dt = \int (t^3 - 6t^2 + 9t) \, dt = \frac{t^4}{4} - 2t^3 + \frac{9t^2}{2} + C
\]
Now, compute the definite integral from \( t = 0 \) to \( t = 3 \):

At \( t = 3 \):
\[
\left( \frac{(3)^4}{4} - 2(3)^3 + \frac{9(3)^2}{2} \right) = \left( \frac{81}{4} - 54 + \frac{81}{2} \right)
\]
Convert to a common denominator:
\[
= \frac{81}{4} - \frac{216}{4} + \frac{162}{4} = \frac{81 - 216 + 162}{4} = \frac{27}{4} = 6.75
\]

At \( t = 0 \):
\[
\left( \frac{(0)^4}{4} - 2(0)^3 + \frac{9(0)^2}{2} \right) = 0
\]

So, the total distance traveled is:
\[
\text{Distance} = 6.75 \text{ units}
\]

### Final Answer Summary:
- The particle changes direction at \( t = 3 \).
- The acceleration at \( t = 0 \) is 9, and at \( t = 3 \) it is 0.
- The total distance traveled by the particle from \( t = 0 \) to \( t = 3 \) is 6.75 units.

Would you like further details on any of these steps?

Here are some related questions to explore:
1. How do we interpret velocity and acceleration graphs for motion analysis?
2. How would the problem change if the velocity had more roots?
3. Can acceleration be negative while the particle changes direction?
4. How is the concept of total distance related to displacement?
5. What are the implications if acceleration were constant instead of varying with time?

**Tip:** The integral of velocity over time gives the displacement, but you must check for changes in direction to compute the actual distance traveled.

Consider a particle moving along a straight line with velocity v(t) = t^3 - 6 t^2 + 9t. Determine the time(s) when the particle changes direction. What is the acceleration of the particle at those times? Calculate the total distance traveled by the particle from t = 0 to the first time it changes direction.

Explore how to determine when a particle changes direction, find the acceleration at those points, and calculate the total distance traveled. This calculus-based problem involves velocity and acceleration functions, derivatives, and integrals, ideal for students in Grades 11-12 or early college.

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