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Mathematics 22 Online
OpenStudy (aveline):

I need help with a velocity graph

OpenStudy (aveline):

http://i65.tinypic.com/2v0okgp.jpg

OpenStudy (518nad):

positive velocity means she was moving upward

OpenStudy (518nad):

remember that velocity is displacement/second so if displacement is 0, it means 0 displacement/second, it was stationary

OpenStudy (518nad):

therefore upward is when positive velocity downward is when negative velocity rest is at zero velocity

OpenStudy (518nad):

can u identify the intervals now?

OpenStudy (aveline):

Let me see...

OpenStudy (aveline):

Well, I thought that from (0,2) and (2,3) it was moving downward, but my teacher's work says that it was (2, 22/5). How does that work?

OpenStudy (aveline):

I mean (0,1) and (2,3)

OpenStudy (518nad):

not 2,3

OpenStudy (518nad):

its not the slope that matters, its all the portion below 0

OpenStudy (aveline):

Oh....

OpenStudy (518nad):

after t=3 the velocity is still negative going to 0, that just means she was slowing down as she was walking backwards

OpenStudy (aveline):

So the particle is at rest from 1≤t≤2 ?

OpenStudy (518nad):

yes

OpenStudy (aveline):

Ok cool, I think I've got it. Thanks for clearing that up for me!

OpenStudy (518nad):

sure, one sneaky thing, u can ask ur teacher about it

OpenStudy (518nad):

at the point t=22/5, it is technically at rest, its an infintessimally small time frame, but i dont know how ur teacher wants to deal with that, its just logistics.

OpenStudy (aveline):

We're only on chapter 2, so I don't think we'll get that specific until some time in the future...

OpenStudy (518nad):

technically its not really an interval, so i think u can ignore that

OpenStudy (518nad):

ok, just wanted to throw that out there

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