The discussion clarifies that the area under a velocity-time graph represents displacement, while the total area accounts for distance. It emphasizes that the integral of the graph assigns a negative value to areas below the x-axis, indicating negative velocity and displacement. Understanding these concepts requires familiarity with integration, as it differentiates between signed and unsigned areas. The conversation highlights that students unfamiliar with integration may struggle with the concept of negative areas.
PREREQUISITESStudents studying physics, educators teaching calculus and physics concepts, and anyone interested in understanding the mathematical interpretation of motion through velocity-time graphs.
The area under a curve is always, by definition, a positive number. If a curve crosses the ##x## axis, then the total area between the curve and ##x## axis is the sum of all the separate areas.grzz said:Homework Statement:: Is area under a velocity-time graph a distance or displacement?
Relevant Equations:: Velocity = rate of change of displacement with time.
I think the area required is a displacement.
Not necessarily. You can take an area below the ##x## axis to be negative, without relying on integration.grzz said:Hence such a question makes sense only to students who are familiar with integration. Am I correct to say this?
If we are talking about a velocity-time graph, then velocity is negative below the time axis. And displacement is negative. That is nothing to do with integration.grzz said:A student who knows about integration will not ask why the area can be negative while one who does not know about integration will ask why the area below the x axis is negative. What answer can I give him then?