91Ó°ÊÓ

Displacement and time share a profound relationship in physics, especially when we're looking into how objects move through space. In this context, the displacement is not just a line connecting two points. It's more of a storyteller of how a point moves through time. In our problem, the displacement as a function of time is expressed as \(x^3 = t^3 + 1\). This involvement of cubic terms indicates a non-linear relationship between displacement \(x\) and time \(t\).
Understanding this relationship is critical because it forms the starting point for analyzing the motion further. The equation, \(x^3 = t^3 + 1\), can be interpreted as a path that takes both the shape of a curve and the substance of time's passage. Each moment in time yields a unique position for the moving point, illustrating how displacement is indeed time-dependent. Using this equation is like having a map that guides us through the motion's development by letting us calculate where the object will be at each given moment.

To delve deeper into motion, we need to understand how quickly displacement changes over time, which is essentially velocity. Calculating velocity involves taking the derivative of displacement with respect to time.
Given our displacement equation \(x^3 = t^3 + 1\), differentiating both sides concerning \(t\) reveals how to take the story of displacement and convert it into velocity. By applying differentiation, we have:

• The equation becomes \(3x^2 \frac{dx}{dt} = 3t^2\).
• Solving for \(\frac{dx}{dt}\) gives us \(\frac{dx}{dt} = \frac{t^2}{x^2}\).

Velocity, \(v\), thus defined by the formula \(\frac{dx}{dt}\), informs us of the rate at which the position of the point is changing. It's not a mere speed; it's a vector quantity that accounts for direction as well. By understanding this derivative, we gain insights into how the object accelerates or decelerates as time progresses.

Acceleration is another step further in the story of motion. While velocity tells us how fast an object moves, acceleration explains how velocity itself changes over time. Evaluating acceleration requires another layer of differentiation. We begin by differentiating the velocity function.
Starting from the derived velocity \(\frac{dx}{dt} = \frac{t^2}{x^2}\), further differentiation using the product and chain rules provides the acceleration \(\frac{d^2x}{dt^2}\). Doing this results in:

• The derivative becomes \(\frac{d^2x}{dt^2} = \frac{2t}{x^2} - \frac{2t^3}{x^5}\).
• Simplifying this expression leads to \(\frac{2t}{x^5}(x^3 - t^3)\).

Utilizing the original displacement relation \(x^3 = t^3 + 1\) further refines the acceleration to \(\frac{2t}{x^5}\).
This final form represents how quickly the velocity of a moving point changes. Different from uniform motion, it reflects the reality of varied moving environments, hinting at a dynamic interplay between forces impacting an object. It underlines how deeply interconnected position and velocity are with time, illustrating the intricate ballet of motion dynamics.