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Need help with F-t graph

  1. Sep 15, 2006 #1
    The area under the F-t graph gives the momentum of the object right?But why when you intergate the graph you get rate of change of momentum with respect to time?Why is it that when u integrate F-d graph you get the work done?:confused:

    Any help would be appreciated :!!) :!!)
  2. jcsd
  3. Sep 15, 2006 #2


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    Right, the area of the F-t diagram gives the change of momentum of the object, since [tex]\vec{F}=\frac{d(m\vec{v})}{dt} \Rightarrow \int_{0}^{t_{1}} \vec{F}dt = m\vec{v_{1}}-m\vec{v_{0}}[/tex]. Considering the work, it equals [tex]W=\int_{1}^{2} \vec{F}d\vec{s}[/tex], where 1 and 2 are the two points on the trajectory. This all folows from basic definitions, so you should be more specific if you still don't understand..
  4. Sep 15, 2006 #3
    hmm...so i suppose when the meaning of integrating or differentiating a graph follows their basic definitions?:uhh:
  5. Sep 15, 2006 #4


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    You can use this shorthand to understand it: integrating a graph means finding the area beneath the graph between some two points ; differentiating a graph means finding the tangent on the graph in some point.
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