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Differentiation is a fundamental concept in calculus that helps us understand how things change. In the context of particle motion, differentiation will allow us to find how the position of a particle changes over time. By differentiating the position of a particle with respect to time, we obtain the velocity function.
The position function given is \(x = 2t^3 - 9t^2 + 12t + 10\). By differentiating this function with respect to \(t\), we calculate the velocity \(v(t)\) since velocity is the derivative of position with respect to time.
Differentiation is represented by \(\frac{d}{dt}\) (or "\(f'(x)\)" when dealing with other variables). Each term in the position equation is differentiated separately. Constants become zero because they do not change, and the power of \(t\) in each term decreases by one. Thus, the term \(2t^3\) becomes \(6t^2\), \(-9t^2\) becomes \(-18t\), and \(12t\) becomes \(12\). As a result, the velocity function is \(v(t) = 6t^2 - 18t + 12\).

In physics, velocity is defined as the rate of change of position with respect to time. It tells us how fast an object is moving and the direction of its motion. When we calculated \(v(t) = 6t^2 - 18t + 12\), this gives us a formula in terms of time \(t\) for the velocity of the particle at any time instant.
To find when the particle stops moving — what we call the "zero velocity" — we set \(v(t) = 0\). This results in the equation \(6t^2 - 18t + 12 = 0\), which describes the specific times \(t\) when the particle's motion comes to a halt. Solving this equation helps us find these key moments in time where velocity transitions through zero, meaning the particle has stopped momentarily or changed direction.
This step is crucial to comprehending when a particle ceases to move or reverses course, providing insight into the dynamics of its path.

Acceleration is the rate of change of velocity with respect to time. It indicates how quickly an object speeds up or slows down. To find the acceleration function for a particle, we differentiate its velocity function with respect to time.
For our exercise, the velocity function is \(v(t) = 6t^2 - 18t + 12\). Differentiating this function gives us the acceleration \(a(t)\). By applying differentiation rules, we calculate \(a(t) = 12t - 18\).
Once we have the acceleration function, we can find out the particle's acceleration at any given time by simply plugging in the time value. For instance, substituting \(t=1\) into \(a(t)\) gives us \(-6\, \text{ft/s}^2\), suggesting that the particle is slowing down at this particular time. Conversely, for \(t=2\), the acceleration is \(6\, \text{ft/s}^2\), indicating that the particle is speeding up. These results help us comprehend whether the particle's speed is increasing or decreasing at specific instances.

A quadratic equation is a second-degree polynomial equation, generally expressed in the form \(ax^2 + bx + c = 0\). It can be solved either by factoring, using the quadratic formula, or completing the square.
In the given exercise, when we set the velocity function \(v(t) = 6t^2 - 18t + 12\) to zero, we derived the quadratic equation \(t^2 - 3t + 2 = 0\). This equation is central to determining specific times when the velocity is zero.
Solving our specific quadratic equation involves factoring it into \((t-1)(t-2) = 0\). This gives us the roots \(t = 1\) and \(t = 2\). These solutions are the moments when the particle stops moving. Such an approach to solving quadratic equations by factoring assumes the equation can be easily split into distinct factors, revealing solutions or roots. This process is key to solving problems involving motion as seen by how it determines critical times in the particle's travel.