To find the particle's acceleration $a(t)$, we start by recalling the relationship between position, velocity, and acceleration. Acceleration is the second derivative of the position function $x(t)$ with respect to time $t$.

Step 1: Differentiate $x(t)$ to find velocity $v(t)$

Given: $x(t) = t^3 - 3t^2 + 7t - 6$ First derivative (velocity): $v(t) = \frac{dx(t)}{dt} = 3t^2 - 6t + 7$

Step 2: Differentiate $v(t)$ to find acceleration $a(t)$

Second derivative (acceleration): $a(t) = \frac{dv(t)}{dt} = 6t - 6$

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

$a(3) = 6(3) - 6 = 18 - 6 = 12$

Final Answer:

The particle's acceleration at $t = 3$ is: $\boxed{12 \, \text{units/time}^2}$

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Related Questions:

1. How do you compute velocity from a position function?
2. What is the physical meaning of the second derivative in motion?
3. If $x(t)$ represents the position, what does $v(t)$ at $t = 3$ represent?
4. How would the acceleration change for $x(t) = t^4$ instead?
5. Can acceleration ever be zero? What would that signify?

Tip: When working with motion problems, always check if derivatives are consistent with the physical interpretation (e.g., units, direction, and sign).