Calculating Impulse from a Graph: How to Solve the Problem

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Nitrate
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Homework Statement


[The problem is in the attached image]


Homework Equations





The Attempt at a Solution


My best guess is to break up the graph into two triangles one positive, one negative and then calculate and add the impulse of both of these triangles... Is that what you're supposed to do?
 
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  • Line graph titled Net Force vs Time for a Football Player, Net Force (N) from -1000 to 600 against Time (s) from 0 to 1.4, showing a peak near 400 N at 0.1 s, a dip to about -800 N at 0.8 s, and a return to 0 at 1.2 s.
    notsurehow.png
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Impulse is defined as the change in momentum, or Δp=Δ(mv)=mΔv. This quantity is related to the force F by the following logic.

F = dp/dt (you should memorize this equation; it's very useful.)

= d(mv)/dt
= m(dv)/dt
= m(dv/dt)
= ma, which is the definition of force.

Given this simple relationship,

∫dp = Δp = ∫F dt

What is an integral? An integral is the signed area between the x-axis (F=0) and the curve. So yes, calculate them separately, and then glue them together. Don't forget that everything below F=0 counts as negative impulse.