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The position function for the particle is given by:
\[
s(t) = t^3 - 6t^2 + 3t + 5
\]
where \( s(t) \) is the position in meters and \( t \) is time in seconds.

### Part (a) - Instantaneous Velocity
To find the velocity, we differentiate the position function \( s(t) \) with respect to \( t \). The velocity function \( v(t) \) is the first derivative of \( s(t) \):

\[
v(t) = \frac{d}{dt}[t^3 - 6t^2 + 3t + 5] = 3t^2 - 12t + 3
\]
So, the instantaneous velocity is:
\[
v(t) = 3t^2 - 12t + 3
\]

### Part (b) - Acceleration
The acceleration is the derivative of the velocity function \( v(t) \), which gives us the acceleration function \( a(t) \):

\[
a(t) = \frac{d}{dt}[3t^2 - 12t + 3] = 6t - 12
\]
Thus, the acceleration at any time \( t \) is:
\[
a(t) = 6t - 12
\]

### Part (c) - When is the Particle at Rest?
The particle is at rest when the velocity is zero, i.e., when \( v(t) = 0 \). We solve:

\[
3t^2 - 12t + 3 = 0
\]
Dividing through by 3:
\[
t^2 - 4t + 1 = 0
\]
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -4 \), and \( c = 1 \):

\[
t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}
\]
Approximating \( \sqrt{3} \approx 1.732 \), we get:
\[
t = 2 + 1.732 \approx 3.732 \quad \text{and} \quad t = 2 - 1.732 \approx 0.268
\]
So, the particle is at rest at \( t \approx 3.732 \) and \( t \approx 0.268 \).

### Part (d) - Motion of the Particle
The particle changes direction when the velocity changes sign, which occurs at the times when \( v(t) = 0 \), i.e., at \( t \approx 0.268 \) and \( t \approx 3.732 \).

Now, analyzing the direction of motion:
- For \( t < 0.268 \), \( v(t) > 0 \), so the particle moves in the positive direction.
- For \( 0.268 < t < 3.732 \), \( v(t) < 0 \), so the particle moves in the negative direction.
- For \( t > 3.732 \), \( v(t) > 0 \), so the particle moves in the positive direction again.

Let me know if you need further details or clarification! Here are some follow-up questions to explore:

1. Can you explain the physical significance of velocity and acceleration at \( t = 0 \)?
2. What are the units of velocity and acceleration in this problem?
3. How can you interpret the particle’s position at \( t = 0 \)?
4. What happens to the acceleration at the instants when the particle changes direction?
5. What does the position function tell us about the behavior of the particle over time?

**Tip:** When solving motion problems involving position, velocity, and acceleration, always differentiate the position function to find velocity and then differentiate the velocity to find acceleration.

A particle moves along a line so that its position at any time t ≥ 0 is given by the function s(t) = t^3 - 6t^2 + 3t + 5 where s is measured in meters and t is measured in seconds.
(a) Find the instantaneous velocity at any time t.
(b) Find the acceleration of the particle at any time t.
(c) When is the particle at rest?
(d) Describe the motion of the particle. At what values of t does the particle change directions?

Explore a calculus problem involving a particle’s motion with position function s(t) = t^3 - 6t^2 + 3t + 5. This problem covers how to compute instantaneous velocity, acceleration, and determine when the particle is at rest and changes direction. Suitable for high school students studying calculus.

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