VCE Specialist 3/4 Question Thread!

[Lejn]

Besides, isn't it v^2sin(2 theta)/g ??

My question: http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths2.pdf
For question 1. f)
Would I still be awarded the mark if I subbed z1 into LHS, and simplified
and then subbed to RHS and simplified, so LHS=RHS, or is it required that I sub in the x-value of z1 into y = sqr(3)x+2 (cartesian equivalent of the relation) and show that it equals the y-value of z1?

I believe both are sufficient, you just need to show that it works and either way does that, I don't think it's a very strict question. I used the latter though.

[psyxwar]

This question confuses me. What is there to show exactly? :/

[Alwin]

(Image removed from quote.)
This question confuses me. What is there to show exactly? :/

Is this a textbook question?

I personally think it is a typo and could be asking you to prove any of the following identities:

Note the first bar is meant to go over the whole fraction.
Or

and latex isn't working for me, mean to be bar on top of (z / w bar)

but yeah.. typo I think

[ahat]

Besides, isn't it v^2sin(2 theta)/g ??

Going back to the original question. That formula ^ is for range and is given by: range = v²sin(2θ)/g

A cricket ball is hit from ground level at an angle of 45* to the horizontal with an initial velocity of 20ms. The time taken, in secs, for it to return to the ground is?

t = 2vsin(θ)/g
t = 2(20)sin(45^o)/9.8
t = 2.87 s

Total time is seconds.

Which matches Stevensmay's answer. So I don't see what's wrong with using these equations? They've always worked for me.

[psyxwar]

Is this a textbook question?

I personally think it is a typo and could be asking you to prove any of the following identities:

Note the first bar is meant to go over the whole fraction.
Or

and latex isn't working for me, mean to be bar on top of (z / w bar)

but yeah.. typo I think

Cheers. I'll give that a shot then

[Alwin]

Which matches Stevensmay's answer. So I don't see what's wrong with using these equations? They've always worked for me.

Hm, "problem" not really. As people have pointed out already, if the question was something along the lines of: an orange cat with massive eyes is launched from a canon on top of a cliff 100m high at speed 1m/s 20^o to the horizontal, find the time taken for the cat to land on the ground below on all fours (it's a cat, 9 lives rmb) then you can't use that formula because the starting and ending hight are different. I'm assuming you know this already ahat, and I'm just stressing it for everyone else reading this.

Secondly, I think that method-wise this formula is fine, so long as you write it out or equivalent substitution to demonstrate to the examiner you know what you are doing and a little birdy hasn't flown in through the window and told you the answer. It's just that the formula applies for a very limited case... and knowing VCAA I doubt they'll set such questions

And erm, unless I remember wrongly from physics last year, I'm pretty sure that it's

?

[ahat]

I hardly want to start a debate on this issue, but I'm just reiterating what I've learnt in class and is explicitly stated in my notes

then you can't use that formula because the starting and ending hight are different. I'm assuming you know this already ahat, and I'm just stressing it for everyone else reading this.

Good point and yes, I have prepared for such a situation. If the particle doesn't start at (0,0) then you can express the position of the particle (at any time) as r = vtcos(θ) i + (vtsin(θ) - 0.5gt + d) j where d is the j component of the height of release.

The equation you make reference to (this is according to my notes) refers to the range of the particle. i.e. It's displacement from its point of release (when release is at (0,0)) to where it lands. The time taken for this to occur (for this displacement) can be expressed as: 2vsin(θ)/g, which is the equation I used, applicable in this case as the particle originated from (0,0).

knowing VCAA I doubt they'll set such questions

Noted

[jono88]

A particle travelling in a straight line has velocity v m/s at time t seconds. Its acceleration is given by dv/dt=3/(v^2-9). The time taken, in seconds, for the velocity to decrease from 2m/s to 1m/s is?
So dt/dv=(v^2-9)/3, therefore t=integral(v^2-9)/3, what would the upper and lower limits of the integral be in this case?

[sin0001]

A particle travelling in a straight line has velocity v m/s at time t seconds. Its acceleration is given by dv/dt=3/(v^2-9). The time taken, in seconds, for the velocity to decrease from 2m/s to 1m/s is?
So dt/dv=(v^2-9)/3, therefore t=integral(v^2-9)/3, what would the upper and lower limits of the integral be in this case?

The upper limit will be 1 and the lower limit will be 2 m/s, this is because the time at which particle is travelling at 1 m/s will be greater than the time when the particle is travelling at 2 m/s- so it should end up as a positive answer.
And since you're integrating with respect to 'v', the integrands MUST be the velocities. Similiarly, if you were integrating with respect to 't' (because of the 'dt'), then your upper/lower limits have to be times in seconds

[Daenerys Targaryen]

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths2.pdf
VCAA2007 MC11 and 4(c)

EditL oops 2007*

[Stevensmay]

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths2.pdf
VCAA2009 MC11 and 4(c)

2009 or 2007?

[Daenerys Targaryen]

2007 soz

Also does r.v=0 imply minimum distance?

[Stevensmay]

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths2.pdf
VCAA2007 MC11 and 4(c)

EditL oops 2007*

For 4c

Re: Specialist Question Thread!
Re: Specialist Question Thread!
Re: Specialist Question Thread!

[Alwin]

Also does r.v=0 imply minimum distance?

yes, minimum distance from O of r(t) if that makes sense. If everything is relative to a tree but the question asks the min distance to china town, you have to find r(t) of china town to the particle. then r(t).v(t) = 0 to solve for t hence r(t value)=min distance

[ashoni]

Redoing some questions from the past VCAA atm and I've come across some problem with my Classpad 330.
The last question in the 2009 Exam 2 for spesh, I've put in the integral betwee 0 and T of (g/2)(1-e^(-2t)) dt = 1200 and my CAS came up with the solution of T = -3.104... saying that 'More solutions may exist'.
I've looked at the assessment reports and their working was exactly the same as mine, but it was just something to do with my calc.
Anyone have any solutions to this?