91Ó°ÊÓ

In kinematics, velocity is a fundamental concept that describes the rate of change of an object's position with respect to time. It is a vector quantity, which means it has both a magnitude and a direction. For the given exercise, the velocity of the object at time zero is provided as 20.0 m/s along the x-axis. This is known as the initial velocity, denoted as \( v_{0x} \).

The change in velocity over time can be determined by integrating the acceleration function. In this problem, the acceleration is a time-dependent function, \( a_x = -Ct \), where \( C \) is a constant. To find the velocity as a function of time, we need to integrate the acceleration:

• Velocity function: \( v_x(t) = \int a_x \, dt = \int (-Ct) \, dt = -\frac{1}{2}Ct^2 + v_{0x} \)

In this case, the velocity function becomes \( v_x(t) = -\frac{1}{2}Ct^2 + 20.0 \), showing us how the velocity changes with time due to the acceleration being \( -Ct \). The negative sign indicates that the velocity is decreasing over time.

Acceleration is a key concept in kinematics that describes how an object's velocity changes over time. It is also a vector quantity, meaning it has both magnitude and direction. In this exercise, the object's acceleration is given by the equation \( a_x = -Ct \), where \( C \) is a constant expressed in units of \( \text{m/s}^3 \).

This equation tells us that the acceleration is not constant but changes linearly with time. The negative sign indicates that the object is slowing down over time, as its acceleration is acting in the opposite direction of the initial velocity.

• Acceleration equation: \( a_x = -Ct \)

Considering this function, it was necessary to determine the value of \( C \) such that the object would come to a stop at \( t = 8.00 \) seconds. By solving the velocity equation with the condition \( v_x(8) = 0 \), the value of \( C \) was found to be 0.625 \( \text{m/s}^3 \). This value shows that at exactly 8 seconds, the deceleration caused by the time-dependent acceleration has reduced the velocity to zero.

Integration is an essential mathematical technique used in kinematics to determine various physical quantities from their rates of change. In this context, integration allows us to derive velocity from acceleration and position from velocity. This is particularly useful when dealing with non-constant accelerations.

In the given exercise, the integration process is used twice:

• First, to determine the velocity as a function of time from the acceleration \( a_x = -Ct \).
• Second, to find the position as a function of time from the velocity \( v_x(t) = -\frac{1}{2}Ct^2 + 20.0 \).

By integrating the velocity function, the position equation is obtained:
\( x(t) = \int v_x(t) \, dt = \int \left(-\frac{1}{2}Ct^2 + 20.0\right) \, dt = -\frac{1}{6}Ct^3 + 20t + x_0 \)
Assuming the object starts at an initial position \( x_0 = 0 \), the position at \( t = 8 \) seconds is calculated, resulting in the object having traveled 106.67 meters during this time period. This demonstrates how integration provides a powerful way to determine the displacement based on changing velocity.