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.