Understanding Negative Velocity and Displacement on a Velocity-Time Graph

[grzz]

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.

 

[malawi_glenn]

Depends if you mean area or signed area (integral)

 

[grzz]

Thanks

 

[PeroK]

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.

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.

The integral, however, assigns a negative sign to areas below the $x$ axis. The integral may be positive or negative or zero.

In terms of physics, the integral of a velocity time graph gives the displacement. Whereas, the total area under a velocity time graph gives the total distance.

 

[grzz]

Hence such a question makes sense only to students who are familiar with integration. Am I correct to say this?

 

[PeroK]

Hence such a question makes sense only to students who are familiar with integration. Am I correct to say this?

Not necessarily. You can take an area below the $x$ axis to be negative, without relying on integration.

 

[grzz]

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?

 

[PeroK]

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?

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.