Let's go through the problem step by step. Here's a breakdown of the situation based on the image you provided:

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.