Impulse, I have found a, b and d

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The discussion focuses on calculating impulse and final velocities for a 3.15 kg particle subjected to a varying net force over time. The impulse of the force is determined to be 12.0 N-s, leading to a final velocity of 3.8 m/s when starting from rest. The participant struggles with calculating the final velocity when the initial velocity is -3.4 m/s, using the impulse-momentum theorem incorrectly. The average force exerted over the time interval of 0 to 5.00 seconds is calculated to be 2.4 N.

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The magnitude of the net force exerted in the x direction on a 3.15 kg particle varies in time as shown in the figure below.

There is a graph that is supposed to be here. The y-axis is F(N) and the x-axis is t(s)
It starts from the orgin then goes to point (2,4), then to (3,4), then ends at (0,5). I am sorry if this looks confusing, but it wouldn't let me post the graph.

(a) Find the impulse of the force. The answer to this one is 12.0 N-s

(b) Find the final velocity the particle attains if it is originally at rest.
i m/s

(c) Find its final velocity if its original velocity is -3.4 m/s .
i m/s

(d) Find the average force exerted on the particle for the time interval between 0 and 5.00 s.
N

so far I have done..

for part A I found the answer to be 12 (4+4+4)

for part B I did I=mv which came out to v=3.8m/s

I am stuck on part c. I thought I am supposed to use I=m(vf-vi) but I keep getting a wrong answer.

And for part d I used I=f(tf-ti) and found the answer to be 2.4N

So I need part c!
 
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F=ma; a=F/m v(f) = v(i) +at
Do that for each section.
 

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