Interpreting and converting an acceleration-time graph

[ACSC]

Homework Statement

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

 

[Ibix]

Take another look at your graph for the deceleration phase. It slows down by 15m/s.

 

[ACSC]

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?

 

[SammyS]

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 !