VCE Specialist 3/4 Question Thread!

[sin0001]

When it asks for the displacement in the, let's say, 5th second, I don't understand why we find the change in position from the 4th second to the 5th... Isn't there an way to solve for the instantaneous displacement? o.O
Help would be much appreciated- just trying to get my head around the terminology

[Stevensmay]

When it asks for the displacement in the, let's say, 5th second, I don't understand why we find the change in position from the 4th second to the 5th... Isn't there an way to solve for the instantaneous displacement? o.O
Help would be much appreciated- just trying to get my head around the terminology

Not to sure what you mean in this context. Have an example question?

[b^3]

It's not asking for the displacement 'at' the 5th second, i.e. not asking for displacement when t=5, rather it's asking for the displacement that occurs during the 5th second, that is starting at t=4 s and finishing at t=5 s. This will be the same as the change in position from t=4 s to t=5 s.

[sin0001]

It's not asking for the displacement 'at' the 5th second, i.e. not asking for displacement when t=5, rather it's asking for the displacement that occurs during the 5th second, that is starting at t=4 s and finishing at t=5 s. This will be the same as the change in position from t=4 s to t=5 s.

Righhhht, if they ask for displacement *during* the 5th second, I guess it makes more sense!

[b^3]

Righhhht, if they ask for displacement *during* the 5th second, I guess it makes more sense!

It's kinda annoying how in VCE it sometimes comes down to wording like this, for 'at' or 'in' e.t.c Although I guess ambiguous decisions will come up in life as well...

[sin0001]

Hey, I was wondering why t=1 isn't a valid solution for part a? I got t=0,1 and 2 seconds, but the answers only list t=0 & 2 as the answer to part a) o.O

[SocialRhubarb]

I'm assuming you used the dot product, to find it, which goes something like this:

.

At t=1, the acceleration has a magnitude of 0, so even though the dot product is equal to 0, it is not perpendicular to the velocity.

[dim_sim]

Hi can someone please help with part c? 

[09Ti08]

Let T1 (N) be the tension due to the 3 kg object
And T2(N) be the tension due to the 6kg object
Now we have 3 varibles with 3 equations:
1/ T1-3g=3a
2/ T2-T1=12a
3/ 6g-T2=6a
I believe that a=1.4m/s^2 

[dim_sim]

Let T1 (N) be the tension due to the 3 kg object
And T2(N) be the tension due to the 6kg object
Now we have 3 varibles with 3 equations:
1/ T1-3g=3a
2/ T2-T1=12a
3/ 6g-T2=6a
I believe that a=1.4m/s^2 

I got the same answer! But the answer in the textbook is 2.45m/s^2 

[Homer]

question 22 multiple choice from 2009 Exam 2, why do they calculate the displacement?

is it because they ask for the "distance of the body from its starting point" 

[e^1]

Hi everyone

I have trouble understanding page example 28 in Essentials (page 88-89). All I can see from the working out is that it proves the medians of the triangle intersect 2/3's of its length. However, I can't see how this proof shows them to be concurrent.

question 22 multiple choice from 2009 Exam 2, why do they calculate the displacement?

is it because they ask for the "distance of the body from its starting point" 

Yeah I think so, if they asked for "the total distance traveled within 10 seconds" then it might have been different.

[Aelru]

Kay guys, got two fun questions.

1. A line through P(-4,3) has equation and intersects the circle at point A and B.

The tangents to the circle at A and B intersect at C.
So, find the equation of the circle through A,B,C.
(answer is , if needed for checking)



2. OABC is a parallelogram. and . Y is a point on such that for some .
is perpendicular to .
So, show that OY intersects AB at

If you do require previous parts of the question for any of these two, please don't hesitate to ask
I just figured these parts selected are basically what's all that's needed.

Thanks!

[Homer]

Kay guys, got two fun questions.

1. A line through P(-4,3) has equation and intersects the circle at point A and B.

The tangents to the circle at A and B intersect at C.
So, find the equation of the circle through A,B,C.
(answer is , if needed for checking)

whats point C?

I have a question too. What exactly is the dot product? I know how to calculate it and use the formulas etcetc but what do we find by calculating the dot product of two vectors?

[SocialRhubarb]

Kay guys, got two fun questions.

Question 1:

Spoiler

This doesn't seem too hard, but it has quite a few steps and quite a bit of algebra.

Our first step is just finding the points of intersection between the line and the circle.







Fast forward:



These values of y correspond to the points .

These two points are both on the positive half of the circle, so we can represent the equation of the circle using .

.

At x=-1,

At x = 2,

So the tangents have the gradients of half and -2.

Given that we know their gradients and know the points which they must go through, we can work out their equations, which come out to be:



The intersection of these two tangents is (1,3).

We now know the points A, B and C, which are  (2,1), (-1,2) and (1,3).

Now all we need is to find a circle which passes through all three of these points.

If we use the standard equation for a circle, , and sub in our points, we'll get a system of three equations which we can solve, probably with a calculator or by hand. We get , which you could sub in and rearrange to obtain your answer, .

Question 2:

I got something close to the answer, but with a few different negatives, so I think my diagram is wrong.