what does negative time in velocity-time graph mean?

[PG1995]

Hi

I'm really confused about negative time axis on **broken link removed**? What does negative time mean in this context? I know it's all about interpretation but in this case I can't come up any interpretation.

Let's say if a car is moving toward north then it's velocity is positive and if it's direction of travel is south then the velocity is negative. When we have negative time and negative velocity the displacement will be positive. This again confuses me because I can't think of any real world interpretation.

I think I should try here to illustrate what I really mean by 'interpretation' above. Suppose, every week I go to an amusement park. For this every week my parents give me $50. I can denote this amount of money as positive money. When I'm there I spend all of the money and that is negative money, i.e. -50. Then, when I have exams for two weeks I don't go there. I get curious to calculate the money it has saved me; money saved = (money it cost me enjoy the park) x (number of times I go there). When I go there I take it positive and when I decide not to go there I take it negative. Therefore, money saved = (-50)x(-2)=$100.

Please help me with this. Thank you.

Regards
PG

 

[steveB]

The point t=0 is usually a particular point we like to label as time zero for convenience. It rarely has any significance other than providing a label for a typical starting time. If an event starts at t=-12 and ends at t=-2, it is no different than an event starting at t=0 and ending at t=10. Both things happen with time flowing in the positive direction and over the same time difference.

Sometimes people will talk about time flowing backwards and this would be moving in the negative direction, but again, the place you start is usually not very relevant. It simply provides a label or reference point. Essentially, time difference is typically the important thing to worry about.

This question seems related to the question you asked about integration and you thought that integrating on the negative side of the x-axis resulted in negative area. In both cases you should think more about the direction you are traveling (whether distance or time) and not on which side of the axis you are on. When you go to the right you are increasing in value and when you go to the left you are decreasing in value. For both time and distance, the difference of values between two points matters more than the actual values at those points.

Of course, I'm kind of mixing physics into the math here and your question is really as much of a math question, but I do this deliberately to help you visualize the right way to look at it.

 

[PG1995]

Thank you.

This question seems related to the question you asked about integration and you thought that integrating on the negative side of the x-axis resulted in negative area.

You are correct. That question really made me think about it. But then I started thinking about multiplication of two negative numbers in general and then came the idea of velocity-time graph.

But how would you explain negative velocity into negative time producing positive displacement? Likewise, how would you explain positive displacement resulting from multiplication of negative velocity and positive time? Could you please let me know? Thanks.

Regards
PG

 

[steveB]

... how would you explain positive displacement resulting from multiplication of negative velocity and positive time?

Before I can try to explain how positive displacement results from negative velocity over a positive time interval, you would first need to convince me it can happen.

When an object moves with negative velocity, it is traveling in the negative x direction (let's assume we are dealing with one dimension only). This means that as time moves forward, displacement is negative between any two time intervals.

Maybe you just worded this question incorrectly?

... how would you explain negative velocity into negative time producing positive displacement?

Well since negative velocity over a positive time gives a negative displacement, it seems reasonable that if you change the flow of time from positive to negative, you will then get a positive displacement. In reality time does not flow backward, but we are free to imagine time flowing backwards. The math still works out fine.

 

[KeepItSimpleStupid]

Who cares?

Whatever the function of s(t) is? t=0 is arbitrary. It's noting more than an initial condition.

For fun take look at a rocket launch. t=0 is liftoff. t-(whatever) is just another unit of time.

Quit bunching up your underwear for something silly.

Another example:

Backing into a parking space and pulling out. T=0 is parked. So there, you got your negative velocity (ds/dt) too depending on your reference.

 

[PG1995]

Thank you, Steve, KISS.

I will return to this thread some time later because right now I need to do some work which can give me good grades!

Best wishes
PG