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The problem you've shared provides the equation of motion for a particle moving in a straight line:

\[
x = t^3 - 6t^2 + 9t + 4
\]

### Part (a): Find the position and acceleration when the velocity \(v = 0\).

1. **Position \(x(t)\)** is already given by the equation.
2. **Velocity \(v(t)\)** is the first derivative of \(x(t)\) with respect to time \(t\):

\[
v(t) = \frac{dx}{dt} = 3t^2 - 12t + 9
\]

Set \(v(t) = 0\) and solve for \(t\):

\[
3t^2 - 12t + 9 = 0
\]

Dividing the equation by 3:

\[
t^2 - 4t + 3 = 0
\]

Factoring the quadratic:

\[
(t - 3)(t - 1) = 0
\]

Thus, \(t = 3\) or \(t = 1\).

Now, we calculate the **position** \(x\) at these times:

- For \(t = 1\):

\[
x(1) = 1^3 - 6(1)^2 + 9(1) + 4 = 1 - 6 + 9 + 4 = 8
\]

- For \(t = 3\):

\[
x(3) = 3^3 - 6(3)^2 + 9(3) + 4 = 27 - 54 + 27 + 4 = 4
\]

So, when \(v = 0\), the positions are \(x = 8\) at \(t = 1\) and \(x = 4\) at \(t = 3\).

Next, we calculate the **acceleration** \(a(t)\), which is the derivative of the velocity function:

\[
a(t) = \frac{dv}{dt} = 6t - 12
\]

- At \(t = 1\):

\[
a(1) = 6(1) - 12 = -6
\]

- At \(t = 3\):

\[
a(3) = 6(3) - 12 = 6
\]

Thus, the accelerations are:
- At \(t = 1\), \(a = -6\)
- At \(t = 3\), \(a = 6\)

### Part (b): Find the position and velocity when \(a = 0\).

We already know that the acceleration is given by:

\[
a(t) = 6t - 12
\]

Set \(a(t) = 0\) to solve for \(t\):

\[
6t - 12 = 0 \quad \Rightarrow \quad t = 2
\]

Now, we find the **position** \(x(t)\) and **velocity** \(v(t)\) at \(t = 2\):

- Position \(x(2)\):

\[
x(2) = 2^3 - 6(2)^2 + 9(2) + 4 = 8 - 24 + 18 + 4 = 6
\]

- Velocity \(v(2)\):

\[
v(2) = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = -3
\]

Thus, when \(a = 0\), the position is \(x = 6\) and the velocity is \(v = -3\).

---

Would you like any further clarification or additional details?

Here are 5 related questions to deepen understanding:
1. What is the significance of velocity being zero in motion?
2. How do changes in acceleration affect the motion of a particle?
3. What is the relationship between the second derivative of position and acceleration?
4. Can there be motion when both velocity and acceleration are zero?
5. How would the particle’s motion change if the equation of motion had different coefficients?

**Tip:** Always check the units of time, distance, and velocity to ensure they align with the real-world scenario when solving problems related to motion.

The path of a particle moving in a straight line is given by x = t^3 - 6t^2 + 9t + 4. a) Find the position and acceleration of the body when v = 0. b) Find the position and velocity of the body when a = 0.

Learn how to find the position, velocity, and acceleration of a particle moving along a straight path. The motion equation x = t^3 - 6t^2 + 9t + 4 is analyzed step by step to find when v = 0 and a = 0. Suitable for students in Grades 10-12.

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