How Do F-t and F-d Graphs Relate to Momentum and Work Done?

[semc]

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?

Any help would be appreciated :!) :!)

 

[radou]

Right, the area of the F-t diagram gives the change of momentum of the object, since $$\vec{F}=\frac{d(m\vec{v})}{dt} \Rightarrow \int_{0}^{t_{1}} \vec{F}dt = m\vec{v_{1}}-m\vec{v_{0}}$$. Considering the work, it equals $$W=\int_{1}^{2} \vec{F}d\vec{s}$$, 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..

 

[semc]

hmm...so i suppose when the meaning of integrating or differentiating a graph follows their basic definitions?

 

[radou]

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.