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In calculus, the velocity function represents the rate of change of the position with respect to time. It's essentially the derivative of the position function. If you know the position function, finding the velocity involves taking its derivative. For this exercise, the position function is given by \( s(t) = t^3 - 4.5t^2 - 7t \).
To find the velocity function, differentiate the position function with respect to time \( t \). The derivative \( v(t) = \frac{d}{dt}(s(t)) = 3t^2 - 9t - 7 \).
This velocity function \( v(t) = 3t^2 - 9t - 7 \) tells us how fast the position changes at any time \( t \). When you need to find when the particle reaches a specific velocity, like 5 m/s, set \( v(t) \) equal to that value and solve for \( t \). This helps in pinpointing precise moments in time that have particular speeds.

The position function in calculus describes where a particle is located over time. It gives a direct relationship between the position and time, essentially illustrating the path of motion. For this problem, the position function is represented by \( s(t) = t^3 - 4.5t^2 - 7t \).
By having a precise formula, you can determine exact points on the particle's trajectory at any given \( t \). The function consists of polynomial terms where each influences how the graph behaves:

• \( t^3 \) adds cubic growth, causing a steeper change in position as time increases.
• \( -4.5t^2 \) encourages curves that dip downwards.
• \( -7t \) adds a linear element pulling it in a downward slope.

Knowing the position gives insight into how to derive velocity (as seen in the velocity function) and further understand the particle's entire motion behavior.

The quadratic formula offers a reliable way to solve for the roots of quadratic equations. It's immensely helpful when you set your velocity function equal to a specific number, convert it to a quadratic form, and need to find \( t \). Its form is \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
For the velocity function \( 3t^2 - 9t - 12 = 0 \), identify:

• \( a = 3 \)
• \( b = -9 \)
• \( c = -12 \)

Plug these coefficients into the quadratic formula.
This process provides possible \( t \) values where the velocity reaches the desired speed of 5 m/s. You may get 2 solutions due to the \( \pm \) sign, requiring careful consideration for valid physical context, like discarding negative time values for practical scenarios.

Acceleration is another key concept in motion analysis. The acceleration function derives from the velocity function and indicates the rate of change of velocity. It essentially describes how the velocity is increasing or decreasing over time.
Consider the velocity function from earlier, \( v(t) = 3t^2 - 9t - 7 \). Differentiating it with respect to \( t \) gives us the acceleration function:\[ a(t) = \frac{d}{dt}(3t^2 - 9t - 7) = 6t - 9 \]This function, \( a(t) = 6t - 9 \), enables us to determine when acceleration is precisely zero by setting \( a(t) = 0 \). Solving \( 6t - 9 = 0 \) gives \( t = 1.5 \). At this moment, the particle switches from increasing to decreasing acceleration or vice-versa. This is significant as it suggests the particle's velocity is temporarily constant, providing a glimpse into its dynamic motion behavior.