Mechanics Help (1 Viewer)

[someth1ng]

Hey guys, I was just looking at this: http://www.boredofstudies.org/courses/maths/4u/1228483617_2008_Mathematics_Extension_2_Notes.pdf On page 6, it says substituting x=0, v=0 then: [the equation here] I don't see how they got to that equation...

 

[barbernator]

The equation was the integral. As there was a v on the top and a v^2 on the bottom, it is a log function, so they had to manipulate the constants to get the derivative on top of the fraction and then integrate i can latex it up if u want.

 

[deswa1]

They skipped a lot of lines of working but see the line before where they have the integral of v/(g-kv^2). If you integrate that you'll end up with the next line and then that's what they sub into

 

[someth1ng]

I understand how they integrated that but I don't understand how they got: v^2=g/k ... From subbing in x=0, v=0 What happened to the C?

 

[deswa1]

They evaluated the C I'm guessing in that line as well- that's why you sub in the boundary conditions (x=0, v=0) so you can work out what C is.

 

[barbernator]

<a href="http://www.codecogs.com/eqnedit.php?latex=x=\frac{-ln(g-kv^{2})}{2k}@plus;C\\ \therefore C=\frac{ln(g)}{2k}\\\\ \therefore x=\frac{-ln(\frac{g-kv^{2}}{g})}{2k}\\\\ -2kx=ln(\frac{g-kv^{2}}{g})\\\\ e^{-2kx}=\frac{g-kv^{2}}{g}\\\\ ge^{-2kx}=g-kv^{2}\\\\ kv^{2}=g-ge^{-2kx}\\\\ v^{2}=\frac{g}{k}(1-e^{-2kx})~as~required" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x=\frac{-ln(g-kv^{2})}{2k}+C\\ \therefore C=\frac{ln(g)}{2k}\\\\ \therefore x=\frac{-ln(\frac{g-kv^{2}}{g})}{2k}\\\\ -2kx=ln(\frac{g-kv^{2}}{g})\\\\ e^{-2kx}=\frac{g-kv^{2}}{g}\\\\ ge^{-2kx}=g-kv^{2}\\\\ kv^{2}=g-ge^{-2kx}\\\\ v^{2}=\frac{g}{k}(1-e^{-2kx})~as~required" title="x=\frac{-ln(g-kv^{2})}{2k}+C\\ \therefore C=\frac{ln(g)}{2k}\\\\ \therefore x=\frac{-ln(\frac{g-kv^{2}}{g})}{2k}\\\\ -2kx=ln(\frac{g-kv^{2}}{g})\\\\ e^{-2kx}=\frac{g-kv^{2}}{g}\\\\ ge^{-2kx}=g-kv^{2}\\\\ kv^{2}=g-ge^{-2kx}\\\\ v^{2}=\frac{g}{k}(1-e^{-2kx})~as~required" /></a>

 

[someth1ng]

Okay, that makes more sense - it's just normal integration...same as 3U except with kv^2, I guess.