Spesh Exam Questions

[Special At Specialist]

So now we can integrate both sides with respect to v.

I don't understand how you knew to integrate from v0 to v1. When you integrate both sides, aren't you supposed to integrate indefinitely?

I did it this way:
a = f(v)
dv/dt = f(v)
dt/dv = 1 / f(v)
dt = dv / f(v)
∫dt = ∫dv / f(v)
t = ∫dv / f(v)
Let g(v) = 1 / f(v)
t = ∫g(v) dv
t = G(v) + C

when t = t0, v = v0
t0 = G(v0) + C
C = t0 - G(v0)
t = G(v) + t0 - G(v0)

Since we know that the definite integral from v0 to v of g(v) = G(v) - G(v0)
t = ∫(v0, v) of g(v) dv + t0
t = ∫(v0, v) of (1 / f(v)) dv + t0
when t = t1, v = v1
t1 = ∫(v0, v1) of (1 / f(v)) dv + t0

Option B of multiple choice.

[b^3]

Well yeh you can do it that way. I probably should of had . But yeh, I could integrate from v0 to v1 because I was integrating from the points that matched up with it for the rhs, the t0 to t1.

But yeh, if you do it the indefinite way then you just have to make sure you get the C right later.

[TheRajinator]

More questions: I need help with Q27 on 2003 exam 1, wasn't really clear why it's D. Thanks 

[b^3]

Since the blocks are stationary, the reaction force of block M on block m, will have a reaction force of the same magnitude opposing it. That is there is a force acting down (this is basically the weight force of block m) on the top surface of Block M. There is the reaction force from the ground acting on block M upwards, that is and there is the weight force of block M, that is .

So that leaves us with which is option D.

[BigAl]

VCAA 2008 exam 2 Q22 I am really not sure how to approach this question. I also don't understand what the kilbaha solutions have done. Can someone please explain this question? Thanks.

think of this way. when you antififf you get a c value right? A constant that the function passes through. so whatever your function is (t,y etc) after integrating the function you plug in initial value t0 y0 whatever...

[TheRajinator]

Since the blocks are stationary, the reaction force of block M on block m, will have a reaction force of the same magnitude opposing it. That is there is a force acting down (this is basically the weight force of block m) on the top surface of Block M. There is the reaction force from the ground acting on block M upwards, that is and there is the weight force of block M, that is .

So that leaves us with which is option D.

Thanks man, It's actually a pretty simple question. I guess I just interperted it wrong.

[TheRajinator]

How do we approach vector proof questions like this:

[BubbleWrapMan]

Find OM in terms of a and c, then you can find AP and PC in terms of them as well, so you can compare the two