We are given the position of a particle as a function of time $t$, defined by the equation:

$x(t) = 6t^4 - 2t^3 - 12t^2 + 3t + 3$

To find the time, position, and velocity when the acceleration $a = 0$, we will follow these steps:

1. Find acceleration $a(t)$:

Acceleration $a(t)$ is the second derivative of position $x(t)$ with respect to time $t$. First, let's find the velocity $v(t)$, which is the first derivative of position.

$v(t) = \frac{dx}{dt} = \frac{d}{dt}(6t^4 - 2t^3 - 12t^2 + 3t + 3)$ Differentiate to get: $v(t) = 24t^3 - 6t^2 - 24t + 3$

Now, to get acceleration, differentiate the velocity $v(t)$ with respect to $t$:

$a(t) = \frac{dv}{dt} = \frac{d}{dt}(24t^3 - 6t^2 - 24t + 3)$ Differentiate to get: $a(t) = 72t^2 - 12t - 24$

2. Set $a(t) = 0$ and solve for $t$:

Now, set the acceleration equal to zero to find the time when the acceleration is zero.

$72t^2 - 12t - 24 = 0$ Divide the whole equation by 12 to simplify: $6t^2 - t - 2 = 0$

Solve this quadratic equation using the quadratic formula:

$t = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(6)(-2)}}{2(6)}$

$t = \frac{1 \pm \sqrt{1 + 48}}{12} = \frac{1 \pm \sqrt{49}}{12} = \frac{1 \pm 7}{12}$

Thus, the two solutions are: $t = \frac{1 + 7}{12} = \frac{8}{12} = \frac{2}{3} \quad \text{and} \quad t = \frac{1 - 7}{12} = \frac{-6}{12} = -\frac{1}{2}$

Since time cannot be negative, the only valid solution is $t = \frac{2}{3} \, \text{seconds}$.

3. Find the position $x(t)$ when $t = \frac{2}{3}$:

Substitute $t = \frac{2}{3}$ into the position equation:

$x\left(\frac{2}{3}\right) = 6\left(\frac{2}{3}\right)^4 - 2\left(\frac{2}{3}\right)^3 - 12\left(\frac{2}{3}\right)^2 + 3\left(\frac{2}{3}\right) + 3$

Let's calculate this step by step.

4. Find the velocity $v(t)$ when $t = \frac{2}{3}$:

Now substitute $t = \frac{2}{3}$ into the velocity equation:

$v\left(\frac{2}{3}\right) = 24\left(\frac{2}{3}\right)^3 - 6\left(\frac{2}{3}\right)^2 - 24\left(\frac{2}{3}\right) + 3$

I'll now compute these values to get the position and velocity at $t = \frac{2}{3}$.At $t = \frac{2}{3} \, \text{seconds}$:

• The position of the particle is approximately $x = 0.26 \, \text{meters}$.
• The velocity of the particle is approximately $v = -8.56 \, \text{m/s}$.

Would you like any further details or explanations?

Here are 5 related questions to explore this concept further:

1. How would the particle's motion change if the initial equation for $x(t)$ had different coefficients?
2. What is the significance of the velocity being negative at $t = \frac{2}{3}$?
3. Can you derive the maximum and minimum positions the particle reaches?
4. How would you graph the position, velocity, and acceleration functions for better understanding?
5. What if acceleration was given as a constant? How would the analysis change?

Tip: The second derivative of a function helps identify concavity and points of inflection in addition to acceleration in motion problems.