Interpreting and converting an acceleration-time graph

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The discussion centers on interpreting an acceleration-time graph to determine a particle's speed at a specific time. The initial confusion arises from misunderstanding how to calculate speed from the area under the velocity graph. It is clarified that the particle does not slow down to 15 m/s but rather decreases its speed by 15 m/s. The participants review the graph to ensure accurate representation of the deceleration phase. Ultimately, the correct interpretation leads to a better understanding of the graph's implications on the particle's speed.
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Homework Statement



WL9HGAZ.png


Homework Equations



a = Δv/Δt
v = Δx/Δt
x = area under velocity graph

The Attempt at a Solution



According to my second attempt, the answer to "particle's speed at t = 20.0 s?" is not 15m/s either.
Working out picture.
I don't understand where I went wrong. I've always thought that the area under a velocity graph would give the distance.
If t=20s isn't v=15m/s, then maybe that's why? But according to the a-t graph, 5x-3=-15. :S
 
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Take another look at your graph for the deceleration phase. It slows down by 15m/s.
 
Ibix said:
Take another look at your graph for the deceleration phase. It slows down by 15m/s.

Oh... I see now. It doesn't slow down to 15m/s but slows down 15m/s.
So like this, right?
lL42HNY.png
 
ACSC said:
Oh... I see now. It doesn't slow down to 15m/s but slows down 15m/s.
So like this, right?
[ IMG]http://i.imgur.com/lL42HNY.png[/PLAIN]
That looks good !
 

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