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The velocity function of an object describes how its position changes with respect to time. Essentially, it's the rate of change of the position function. In our exercise, we have a position function given as \(s(t) = 5t^2 + 3\). To find the velocity function \(v(t)\), we need to differentiate the position function with respect to time.First, we consider the difference quotient, which gives us an average rate of change over a small interval \(h\):

• Start with \(v(t) = \lim_{h \to 0} \frac{s(t+h) - s(t)}{h}\).
• Substitute the position function: \(s(t) = 5t^2 + 3\).
• Calculate \(s(t+h) = 5(t+h)^2 + 3\).

When you expand \(s(t+h)\), you get \(5(t^2 + 2th + h^2) + 3\), simplifying leads to \(5t^2 + 10th + 5h^2 + 3\). This expression is then used in the difference quotient:\[ v(t) = \lim_{h \to 0} \frac{5t^2 + 10th + 5h^2 + 3 - 5t^2 - 3}{h} \].After simplifying the expression by canceling the terms \(5t^2\) and \(3\), we get \(10th + 5h^2\). Factoring out an \(h\), then taking the limit as \(h\) approaches zero, we find \(v(t) = 10t\).This result, \(v(t) = 10t\), means that the velocity is directly proportional to time, with a slope of \(10\). This tells us how fast the particle's position is changing at any given time \(t\).

Acceleration is a measure of how quickly an object's velocity is changing. It’s the derivative of the velocity function with respect to time. In this case, we've already determined the velocity function: \(v(t) = 10t\).To find the acceleration function \(a(t)\), we differentiate the velocity function:

• Calculate \(a(t) = \frac{d}{dt}(v(t)) = \frac{d}{dt}(10t)\).

With simple differentiation of a linear term, the derivative of \(10t\) is \(10\).Thus, \(a(t) = 10\), which means the acceleration is constant at all times. This constant acceleration indicates that the object's velocity is increasing uniformly over time. There's no change in the rate of acceleration; it's steady, making it a straightforward example of linear motion in physics.

The position function describes a particle's location along a particular axis as a function of time. It forms the starting point for understanding both velocity and acceleration. In our scenario, the position function is defined as \(s(t) = 5t^2 + 3\).Position function tells us where the particle is at any given time. For example:

• When \(t = 0\), \(s(0) = 3\). This is the initial position of the particle.
• As time increases, \(t\) affects the term \(5t^2\), showing how far the particle moves from its initial position.

The term \(5t^2\) in the function indicates that its movement is influenced by a quadratic relationship; its position depends quadratically on time \(t\). The particle's distance from the origin accelerates as \(t\) increases, fundamentally affecting how velocity and acceleration are calculated in subsequent steps.