91Ó°ÊÓ

In physics, a displacement function describes how the position of an object changes over time. In this exercise, the displacement function is given by \( x = 3t^2 \) meters. This tells us that the displacement or position of the object is directly proportional to the square of the time \( t \).

What makes this function particularly interesting is the quadratic term, \( t^2 \), which implies that the object's position changes rapidly as time progresses. For example, when \( t = 0 \), the displacement is 0 meters, indicating that the object starts from a reference point at the origin. As time increases, the displacement increases quadratically, not linearly, meaning that the object moves faster as time goes by.

Functions like these are often used to model accelerated motion, where velocity changes over time. Understanding how displacement relates to time is crucial for calculating other motion-related concepts, like velocity.

Time intervals are essential for analyzing motion because they define the specific periods over which changes in position are measured. In our problem, we need to calculate the average velocity over two distinct time intervals: from \( 0 \) to \( 2.0 \) seconds, and from \( 2.0 \) to \( 4.0 \) seconds.

A time interval is simply the difference between the final time and the initial time. In mathematical terms, the interval \( \Delta t \) is expressed as \( \Delta t = t_{final} - t_{initial} \).

In this exercise:

• For the first interval \(( \Delta t = 2.0 \, \text{s} - 0 = 2.0 \, \text{s})\), we are observing the motion from the start point \(t = 0\) to \(t = 2.0\) seconds.
• The second interval \(( \Delta t = 4.0 \, \text{s} - 2.0 \, \text{s})\) focuses on the motion from \(t = 2.0\) to \(t = 4.0\) seconds.

These intervals allow us to determine the average velocity during specific periods of time, showing how an object's speed varies over these durations.

Velocity calculation involves determining how fast an object is moving and in which direction. We calculate velocity using the change in displacement over a specific time interval. In the given exercise, we focus on calculating the average velocity during each time interval.

To find the displacement change (\( \Delta x \)) in each interval, we subtract the initial displacement from the final displacement:

• In interval (a) \((0 - 2.0 \, \text{s})\), the initial displacement is 0 meters \((x(0) = 0)\) and the final displacement at \(t = 2.0\) seconds is 12 meters \((x(2) = 12)\).
• For interval (b) \((2.0 - 4.0 \, \text{s})\), the initial displacement is 12 meters \((x(2) = 12)\) and the final displacement at \(t = 4.0\) seconds is 48 meters \((x(4) = 48)\).

This step-by-step calculation illustrates the changes in position, enabling us to determine the velocity by considering the distance traveled in each time frame.

The average velocity is a measure of how an object's position changes over a specific time interval. It is calculated using the formula:

\[ v_{avg} = \frac{\Delta x}{\Delta t} \]

where \( \Delta x \) is the change in displacement, and \( \Delta t \) is the change in time.

In this exercise:

• For time interval (a) from \( 0 \) to \( 2.0 \, \text{s} \), \( \Delta x = 12 \, \text{m} \) and \( \Delta t = 2.0 \, \text{s} \) yields an average velocity of \( \frac{12 \, \text{m}}{2.0 \, \text{s}} = 6 \, \text{m/s} \).
• For time interval (b) from \( 2.0 \) to \( 4.0 \, \text{s} \), \( \Delta x = 48 \, \text{m} - 12 \, \text{m} = 36 \, \text{m} \) and \( \Delta t = 2.0 \, \text{s} \) gives \( \frac{36 \, \text{m}}{2.0 \, \text{s}} = 18 \, \text{m/s} \).

This formula is fundamental because it reflects the object's overall speed between any two points in time, providing valuable information about its motion.