To solve part (b), we need to find the acceleration of the particle at $t = 3 \, \text{s}$.

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Step 1: Recall the relationship between acceleration and velocity

Acceleration is the derivative of velocity with respect to time: $a(t) = \frac{d v_x}{dt}$

The velocity expression is given as: $v_x = 5t^2 - t^3$

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Step 2: Differentiate $v_x$ to find $a(t)$

$a(t) = \frac{d}{dt} \left( 5t^2 - t^3 \right)$

$a(t) = 10t - 3t^2$

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Step 3: Evaluate $a(t)$ at $t = 3$

Substitute $t = 3$ into the acceleration equation: $a(3) = 10(3) - 3(3)^2$

$a(3) = 30 - 27$

$a(3) = 3 \, \text{m/s}^2$

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Final Answer:

The acceleration of the particle at $t = 3 \, \text{s}$ is: $\boxed{3 \, \text{m/s}^2}$

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Do you have any further questions? Would you like a breakdown of part (a) as well?

Here are 5 related questions to deepen your understanding:

1. What is the physical significance of the derivative of velocity with respect to time?
2. How does the acceleration depend on $t$ based on the equation $a(t) = 10t - 3t^2$?
3. What is the velocity of the particle at $t = 3 \, \text{s}$?
4. What is the instantaneous acceleration of the particle at $t = 1 \, \text{s}$?
5. How would you find the time $t$ when the acceleration is zero?

Tip: The instantaneous acceleration tells us how the velocity is changing at a specific moment in time!