91Ó°ÊÓ

Displacement in kinematics describes an object's change in position over time. It's essential to know that displacement isn't the same as distance; displacement considers the direction as well. In this exercise, we are asked to calculate the displacement of an object between two time intervals, from \(t=4\) seconds to \(t=8\) seconds.
To find the displacement, we use the position function provided: \(x(t)=12-6t+3.2t^2\). We find the position at these two time points and take the difference:

• First, calculate \(x(4)\), substituting \(t = 4\) into the equation.
• Next, calculate \(x(8)\) by substituting \(t = 8\).
• The displacement is then \(x(8) - x(4)\).

This approach helps us evaluate how much the position of the object has changed within the given time frame.

The velocity function is a cornerstone in understanding motion, as it provides information about the rate of change of position over time. Velocity, in essence, tells us how fast something is moving and in what direction it moves. In mathematics, velocity is derived as the first derivative of the position function.
To find the velocity function \(v(t)\), we differentiate the position function \(x(t) = 12 - 6t + 3.2t^2\) with respect to time \(t\):

• The derivative of \(-6t\) is \(-6\).
• The derivative of \(3.2t^2\) is \(6.4t\).

Thus, the velocity function is \(v(t) = -6 + 6.4t\).
Additionally, we evaluate this function at a specific time \(t=3\) seconds by substituting \(3\) into the velocity function. This calculation provides us with the object's velocity at that moment in time.

Acceleration provides insights into how an object's velocity changes over time. It is crucial in kinematics to understand how quickly an object speeds up or slows down. To comprehend this concept, one new derivative step is required.
The acceleration function is the derivative of the velocity function. From our earlier calculation of the velocity function \(v(t) = -6 + 6.4t\), we take its derivative with respect to time \(t\):

• Since the constant term \(-6\) disappears upon differentiation, only \(6.4t\) contributes to the acceleration.
• Therefore, the derivative is \(6.4\), giving us the acceleration function \(a(t) = 6.4\).

This constant value of acceleration indicates that the object's velocity changes steadily over time, making it an example of uniform acceleration.

Derivation, in the realm of kinematics, is a powerful mathematical tool used to find rates of change. Every time we shift from one motion property—be it position, velocity, or acceleration—to another, we perform a derivation.
In this exercise, we started with the given position function \(x(t)=12-6t+3.2t^2\).

• First, finding the velocity involved taking the first derivative of the position function, resulting in \(v(t) = -6 + 6.4t\).
• Next, calculating acceleration required taking the derivative of the velocity function, yielding \(a(t) = 6.4\).

Each derivation step clarifies how one aspect of motion is linked to another. This sequence of derivations helps break down the complex relationships between these kinematic quantities into simpler, understandable forms.