Finding Velocity with Varying Force and Friction

[CGI]

Homework Statement

The 3kg collar is initially at rest and is acted upon a force Q which varies by the the graph shown. Knowing that the coefficient of kinetic friction is 0.25, determine the velocity of the collar at t = 1 s and t = 2s.

Homework Equations

FΔt = mΔv
F_friction = .25 x N

The Attempt at a Solution

I know that the answer for b = 3.43 m/s, but I'm trying to think of how I would get there.
I thought about the impulse equation so I tried plugging it in. Since at t = 2, F = 5, and Δt = 2, I said

10 = 3(V_final - V_Initial

where V initial would be 0. I get that the velocity at t = 2 is 3.33, which is close, but no cigar. I then thought about the the force of friction that would be acting in the opposite direction of Q, so I should subtract the two forces. I found that the force of friction is 7.3575 N. Once i subtract the two, I get a negative value, which I don't think is right. Any help would be much appreciated!

 

[Simon Bridge]

The impulse is the area under the force-time graph.

 

[haruspex]

the graph shown

I don't understand the scale on the graph. The y=20 line seems to be only 3 times the height of the y=5 line.

I found that the force of friction is 7.3575 N. Once i subtract the two, I get a negative value, which I don't think is right.

When the calculated force of friction exceeds Q, there are two possibilities for the acceleration. What are they? (Note that you make a certain assumption when you calculate kinetic friction.)