impulse, chapter 9?

Question

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). Im sorry if this looks confusing, but it wouldnt 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

I WILL RATE!

for parts a, b, and d I got those same answers, but for part b I got a different answer. And when I tried to plug in 0.4 (your answer) it said it was wrong. My answer for part c is wrong to though

Answer 1

Let I be impulse

then
I=integral (F dt) integrated from t1 to t2.

Does that help?

In simple terms impulse is the area bounded by the given curve.

At= A1+A2+A3
A1= (2-0)x4/2=4
A2=(3-2)x4=4
A3= (5-3)=4 (you made a mistake it is (5,0) not (0,5))
At=I=4+4+4=12 N s

b) F=m dv/dt
V= integral (F dt) /m from t1=0 to t2= 5sec
or
V=I/m= 12 / 3.15=3.8 m/s (avg)

c) Vf=V0 + V= -3.4 + 3.8 = 0.4 m/s

d) F=I/t= 12/5=2.4N