91Ó°ÊÓ

In the realm of kinematics, a velocity function describes how an object's speed changes over time. A common form is given by a mathematical expression, like the one used in our exercise:

• \(v_x(t) = \alpha + \beta t^2\)

This function tells us that velocity depends on both a constant term \(\alpha\) and a term that grows with the square of time \(t^2\). Here, \(\alpha = 3.00\text{ m/s}\), representing the initial velocity of the car, and \(\beta = 0.100\text{ m/s}^3\), describing how the velocity changes over time.
This quadratic nature implies that the car will accelerate, changing the rate of velocity as time progresses. Understanding the velocity function helps determine crucial elements of motion, like position changes and acceleration.
The behavior of the object is visually captured by plugging different values of \(t\) into the function to calculate corresponding \(v_x(t)\) values.

Average acceleration gives us a simplified overview of how the speed of an object changes over a time interval. It calculates the change in velocity divided by the change in time:

• \( a_{avg} = \frac{v_x(t_1) - v_x(t_0)}{t_1 - t_0} \)

For the given problem, average acceleration is calculated for the time interval \(t = 0\) to \(t = 5.00\) seconds. With a starting velocity \(v_x(0) = 3.00\text{ m/s}\) and an ending velocity \(v_x(5) = 5.50\text{ m/s}\), substituting these values results in \( a_{avg} = 0.50\text{ m/s}^2\).
This means on average, the car's speed increased by 0.50 meters per second every second over the 5-second interval. Average acceleration is invaluable in understanding how the speed accumulates over time, providing insights into the overall motion in a simplified manner.

Instantaneous acceleration captures the change in velocity at a specific moment in time, rather than over an interval. It is represented as the derivative of the velocity function:

• \(a_x(t) = \frac{d}{dt}(v_x(t))\)
• \(a_x(t) = 2\beta t\)

In our exercise, instantaneous acceleration is calculated at \(t = 0\) and \(t = 5\) seconds. For these specific instances,

• At \(t = 0\), \(a_x(0) = 0.00\text{ m/s}^2\)
• At \(t = 5\), \(a_x(5) = 1.00\text{ m/s}^2\)

This means at the starting point, the car experiences no acceleration, while at 5 seconds, its rate of velocity changes instantly at 1 meter per second squared.
Instantaneous acceleration provides an exact measurement of how fast the velocity is changing at any given time, making it crucial for understanding the dynamics at specific points.

A velocity-time graph visualizes the variation of velocity over a period. For a function like \(v_x(t) = 3.00 + 0.100 t^2\), the graph forms a parabola, reflecting the quadratic relationship.

• At \(t = 0\), velocity is \(3.00\text{ m/s}\)
• At \(t = 5\), velocity reaches \(5.50\text{ m/s}\)

The graph starts at the initial velocity and curves upward, illustrating the increasing velocity due to the growing \(t^2\) component.
This type of graphical representation helps visualize how quickly an object is moving at any point in time. The slope of the graph increases over time, indicating acceleration. It effectively shows both the magnitude and direction of velocity changes.

The acceleration-time graph represents how the instantaneous acceleration changes with time. For a linearly varying function like \(a_x(t) = 2 \times 0.100 t\), the graph is a straight line:

• Starting at \(a_x(0) = 0.00\text{ m/s}^2\)
• Ending at \(a_x(5) = 1.00\text{ m/s}^2\)

This linear increase in acceleration indicates a constant rate of acceleration increase.
The graph helps visualize how fast the car's velocity is changing and is particularly useful when comparing periods of constant acceleration versus changing acceleration.
It clarifies how an object's acceleration varies over a time interval in a clear and precise manner.