Equations of motion

[Admin]

If you have any problems with acceleration-time graphs, velocity-time graphs, displacement-time graphs and the equations of motion you can post here.

[jamie]

sir stuck with the h/w the past paper Q .

[Admin]

Jamie,
The graph is not for equations of motion problem but internal resistance.
Remember Y = mx + c is V = -rI + E
so where it cuts the y-axis is your emf, the gradient is -r and the x-axis is your current.
The following will help
1. Extend the graph to cut the y-axis, this will give you E.
2. Then find the gradient this will give you -r. Multiply gradient by -1 you get r.
3. If you take a point on the graph this will give (I,V) i.e. (x,y). So at that point you can finr the correspnding value of R remember that R = V/I i.e. (y/x).
If you are still struggling please give me more specific information from the question.

[jamie]

lol ano its got nothing 2 do with equations of motion but am not sure how 2 start my own topic so a just put it in here but cheers anyway a think a get it now

[Admin]

Excellent!