91Ó°ÊÓ

The velocity function, represented as \( v_x(t) = \alpha - \beta t^2 \), describes how the velocity of an object changes over time. In this exercise, \( \alpha \) is a constant with a value of 4.00 m/s, which represents the initial velocity of the object when time \( t = 0 \). In contrast, \( \beta \) is 2.00 m/s³, introducing a quadratic term, \( \beta t^2 \), that affects how velocity decreases as time progresses.
This velocity function signifies that the object starts with a positive velocity and decelerates as time goes by due to the negative \( \beta t^2 \) term.
When the velocity becomes zero, the object changes direction or stops momentarily.
Understanding this function helps us predict how an object is moving at any given moment by simply plugging in a time value \( t \).

Acceleration is the rate of change of velocity with respect to time. It is derived from the velocity function by taking its derivative. In this problem, the acceleration function is determined by differentiating the given velocity function \( v_x(t) = \alpha - \beta t^2 \).
Differentiating gives us \( a(t) = -2\beta t \). When we substitute \( \beta = 2.00 \, \text{m/s}^3 \), it simplifies to \( a(t) = -4.00 t \, \text{m/s}^2 \).
This indicates that the object's acceleration is directly proportional to time, meaning it increases in magnitude as time increases, but in the negative direction.
The negative sign tells us that the object is decelerating or accelerating in the opposite direction of its initial motion.

Displacement refers to the object's overall change in position, from its starting point to where it is at a given time. It is important to note that displacement is different from distance, as displacement considers only the change in position.
To find the object's displacement over time, we integrate the velocity function. By integrating \( v_x(t) = \alpha - \beta t^2 \), we find the position function \( x(t) \). The integration process yields:
\[ x(t) = \int (\alpha - \beta t^2) \, dt = \alpha t - \frac{1}{3} \beta t^3 + C \]
Given that the initial position \( x(0) = 0 \), the constant \( C \) is determined to be zero, yielding the displacement function:
\[ x(t) = 4.00 t - \frac{2.00}{3} t^3 \]
This function helps us ascertain the object's displacement at any given moment by substituting the time \( t \).

The position of an object as a function of time tells us the exact location of the object relative to its starting point, considering its velocity changes over time.
From the velocity function, we've integrated to obtain the position function \( x(t) = 4.00 t - \frac{2.00}{3} t^3 \).
This function captures both the linear and quadratic aspects of motion - the linear term \( 4.00 t \) suggests a steady progression in position affected linearly by time, and the quadratic term \(-\frac{2.00}{3} t^3 \) introduces the variation due to changing velocity.
To find the maximum positive displacement from the origin, we look for when \( v_x(t) = 0 \), indicating the point where the object stops or changes direction. Calculating for \( t \) gives approximately 1.41 s. Substituting this into the position function provides the maximum position:

• \( x(1.41) \approx 3.77 \, \text{m} \)

This is when the object reaches its furthest point from the starting position and is key in solving trajectory-related problems.