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Let's break down the problem and solve it step by step based on the provided information.

The position function of the particle is given by:

\[
s(t) = 2t^3 - 30t^2 + 126t
\]
where \( s(t) \) is the position in feet, and \( t \) is the time in seconds.

### 1. Finding the velocity at \( t = 0 \):

The velocity function is the derivative of the position function \( s(t) \). So, we need to find \( v(t) = s'(t) \).

\[
v(t) = \frac{d}{dt}(2t^3 - 30t^2 + 126t) = 6t^2 - 60t + 126
\]

Now, substitute \( t = 0 \) into \( v(t) \):

\[
v(0) = 6(0)^2 - 60(0) + 126 = 126
\]

So, the velocity at \( t = 0 \) is **126 ft/sec**.

### 2. Finding when the particle stops moving:

The particle stops moving when its velocity is zero. So, we need to solve \( v(t) = 0 \):

\[
6t^2 - 60t + 126 = 0
\]

Simplify by dividing the equation by 6:

\[
t^2 - 10t + 21 = 0
\]

Now, solve the quadratic equation using the quadratic formula:

\[
t = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(21)}}{2(1)}
\]
\[
t = \frac{10 \pm \sqrt{100 - 84}}{2}
\]
\[
t = \frac{10 \pm \sqrt{16}}{2}
\]
\[
t = \frac{10 \pm 4}{2}
\]

So the two solutions are:

\[
t = \frac{10 + 4}{2} = 7 \quad \text{and} \quad t = \frac{10 - 4}{2} = 3
\]

Thus, the particle stops moving at \( t = 3 \) seconds and \( t = 7 \) seconds.

### 3. Finding the position of the particle at \( t = 20 \):

Substitute \( t = 20 \) into the position function \( s(t) \):

\[
s(20) = 2(20)^3 - 30(20)^2 + 126(20)
\]
\[
s(20) = 2(8000) - 30(400) + 126(20)
\]
\[
s(20) = 16000 - 12000 + 2520 = 6520 \, \text{feet}
\]

So, the position of the particle at \( t = 20 \) is **6520 feet**.

### 4. Finding the total distance traveled between \( t = 0 \) and \( t = 20 \):

To find the total distance, we need to consider where the velocity changes sign. The particle stops at \( t = 3 \) and \( t = 7 \), so we need to calculate the positions at these times and sum the absolute values of the distances.

- Position at \( t = 0 \):
  \[
  s(0) = 2(0)^3 - 30(0)^2 + 126(0) = 0
  \]

- Position at \( t = 3 \):
  \[
  s(3) = 2(3)^3 - 30(3)^2 + 126(3)
  \]
  \[
  s(3) = 2(27) - 30(9) + 126(3) = 54 - 270 + 378 = 162
  \]

- Position at \( t = 7 \):
  \[
  s(7) = 2(7)^3 - 30(7)^2 + 126(7)
  \]
  \[
  s(7) = 2(343) - 30(49) + 126(7) = 686 - 1470 + 882 = 98
  \]

- Position at \( t = 20 \) (already calculated):
  \[
  s(20) = 6520
  \]

Now, calculate the total distance traveled:

- From \( t = 0 \) to \( t = 3 \), the particle travels \( |0 - 162| = 162 \) feet.
- From \( t = 3 \) to \( t = 7 \), the particle travels \( |162 - 98| = 64 \) feet.
- From \( t = 7 \) to \( t = 20 \), the particle travels \( |98 - 6520| = 6422 \) feet.

So, the total distance traveled is:

\[
162 + 64 + 6422 = 6648 \, \text{feet}
\]

Thus, the total distance traveled is **6648 feet**.

---

Would you like further details or have any questions?

Here are 5 related questions:
1. How does the velocity change over time between \( t = 0 \) and \( t = 20 \)?
2. What is the acceleration of the particle at \( t = 5 \)?
3. At what time is the particle moving fastest?
4. Can we calculate the average velocity of the particle over the interval \( [0, 20] \)?
5. How does the total distance compare to the displacement of the particle?

**Tip:** When dealing with motion, always distinguish between *total distance traveled* and *displacement*. Total distance accounts for all changes in direction, while displacement only considers the initial and final positions.

A particle moves along a straight line and its position at time t is given by s(t) = 2t^3 - 30t^2 + 126t. Find the velocity at t = 0, when the particle stops moving, the position at t = 20, and the total distance traveled between t = 0 and t = 20.

Learn how to calculate the velocity of a particle moving along a straight line, the times when it stops, the position at t = 20 seconds, and the total distance traveled. This involves differentiation and solving quadratic equations, suitable for Grades 10-12.

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