My Suggested Solutions to SP2008 Exam 2 [Confirmed]

[xers]

For e.i, how the hell is that a 1 mark question, i hate vcaa.

[pastiche]

I am ... doubly confused, but hey, is that written out w/ a tablet-top? I want one.

[Nelle91]

I think actually Tsintheta = 979 N my bad

[kurrymuncher]

Hey, was it harder than last years?

[csq7]

Synesthetic:
i think you forgot to square root the y to get the cartesian equation..

[Synesthetic]

I think actually Tsintheta = 979 N my bad

Okay thanks for confirmation, and bad luck.

[jamestaR`]

As if your train one is right...you went from y^2 = whatever to say that y = the same thing?

and speed = 0? dun think so

[Synesthetic]

Synesthetic:
i think you forgot to square root the y to get the cartesian equation..

Do you mean this? a)ii) y=ħsqrt[x^2(1-x^2)], by substituting the parametric equation for x into y
I included 'sqrt'.

As if your train one is right...you went from y^2 = whatever to say that y = the same thing?

and speed = 0? dun think so

I didn't have speed=0 ...

Hey, was it harder than last years?

YES.

[jamestaR`]

is that a joke? u said v= 0i = 0j therfore the exact speed is 0m/s

edit sorry meant for the guy who hand wrote his train solution

[darlok]

Im sorry but alot of the solutions that neobeo posted are wrong. =/.

[csq7]

soz I didnt get speed zero too..

[xers]

Lol, how can u be not moving and passing through a point at the same time?

[Ahmad]

When you throw a ball in the air you get an inverted parabola and at the top of its flight it has ZERO velocity (the gradient of the parabola is 0 at the maximum). This situation is analogous, if you inspect the graph it is evident that there is a stationary point and so the velocity MUST be zero.

I have also verified Neobeo's other solutions and they are indeed correct.

[Synesthetic]

Regarding Neobeo's solution here, I think I spotted the error:



The fourth line of working for a)ii) :

y^2 = sin^2t/3cos^2t/3, not y !!!!

So does that make the rest of his solutions all incorrect? Is that why we have different answers to [my solutions:] t= 6pi, v=sqrt(2)/3, and s=6.1m?

[xers]

Wait. Why can't you just substitute 3pi into the velocity vector then find the modulus of it?