VCE Specialist 3/4 Question Thread!

[hel256]

Hi, I have some questions, any help would be appreciated

It's from the Exercise 4.3 of Maths Quest 12 Specialist, if that helps, but the questions are here below.

11 From a hot air balloon rising vertically upward with a speed of 8 m/s, a sandbag is
dropped which hits the ground in 4 seconds. Determine the height of the balloon
when the sandbag was dropped.

12 A missile is projected vertically upward with a speed of 73.5 m/s, and 3 seconds
later a second missile is projected vertically upward from the same point with the
same speed. Find when and where the two missiles collide.

13 A flare, A, is fired vertically upwards with a velocity of 35 m/s from a boat. Four
seconds later, another flare, B, is fired vertically upwards from the same point with
a velocity of 75 m/s. Find when and where the flares collide.

Thanks guys!!!

I haven't done any maths in over two months now so if my solutions are incorrect someone please correct me.
For all of these questions I would use the suvat equations with a=9.8(m/s). In fact, for all of these questions I believe only is needed.

11. In this question you are simply solving for height, or s
The question tells you the initial velocity (u) is 8m/s and the time taken to traverse the distance between the position of the hot air balloon when the sandbag is dropped and the ground is 4 seconds, so you now have your u and t values respectively.
Remember that the hot air balloon is rising vertically, and thus you must assign opposite values to u and a (I personally would assign the positive value to the upwards direction).
So your equation would go from:

to

Hence s=-46.4m, and thus the height of the hot air balloon when the sandbag was dropped was 46.4m (in this case the negative sign is indicating that the bag fell in the direction of gravity, ie. downwards).

12. This time you have two objects and are trying to find the point at which they collide, thus I would recommend equating their s values.
If we let t be the time passed (in seconds) since the first missile was launched, then the time passed since the second missile was launched would be t-3.
Both missiles have the same initial velocity of 73.5m/s, and both have gravity as their constant acceleration.
Therefore we can equate the values of s for each missile to create the equation

where u=73.5, a=-9.8 and we are solving for t.
Therefore t=9 (seconds) and by substituting that into the original s=ut+1/2at^2 equation we can find s=264.6m
Hence the missiles collide 264.6m above the ground 9 seconds after the launch of the first missile.

13. Though the flares don't start with the same initial velocity, we can apply the same methodology in this question as the last. The only difference here would be that the values of u on each side of the equation would differ (we'll use b and c to denote these different initial velocities).
Again, by letting t be the time since the launch of the first flare, the time since the launch of the second flare would be t-4
So the equation would go from

to

Thus t=43/9 secs, and by substituting that in s=44849/810m or 55.37m to 2 d.p.
Therefore the flares collide 43/9 (or 4.78 to 2 d.p.) seconds after the launch of the first flare 55.37m above the ground.

Again it's late rn and I haven't done anything remotely mathematical for two months now so if these are incorrect please correct me and I apologise in advance.

[LE-0130]

Hi! Just hoping to get some help for this question here
I think I have an idea as to what I did wrong.
The answers say that the mean of the 20 batteries is 140 and the standard deviation of the 20 batteries is 2.236.
If you use the method of...
E(20X) = 20E(X) and Var(20X) = 400Var(X0, then you'll get the wrong answer.
If you use this method of adding X 20 times instead, you'll get
E(X + X + X...+X) = 20E(X) and Var(X + X + X+....+X) = 20Var(X).
When do I know which method to use?
Any help or advice would be appreciated! Thanks

[Harrycc3000]

Hi Guys,
Just asking about this q (16e) from the 3/4 Cambridge textbook (chapt 2F has diagram)
16 The points A, B, C, D and E shown in the diagram have
position vectors
a = i + 11 j b = 2i + 8 j c = −i + 7 j
d = −2i + 8 j e = −4i + 6 j
respectively. The lines AB and DC intersect at F as shown.
a Show that E lies on the lines DA and BC.
b Find −−→AB and −−→DC.
c Find the position vector of the point F.
d Show that FD is perpendicular to EA and that EB is perpendicular to AF.
e Find the position vector of the centre of the circle through E, D, B and F

The solution said that because angle EDF and angle EBF were 90 degree angles it implied EF was the diameter (One of the circle theorems) and that you just needed to find the midpoint to get the vector for the centre of the circle.
I was wondering that if I just didn't realise this (didn't catch it) in the exam would subbing in x and y values for (x-h)^2 + (y-k)^2 = r^2 for each of the position vectors E, D, B and F and then simultaneously solving each of these equations work? This was how I tried to solve the equation but it didn't work but I'm wondering whether this is an arithmetic error because this method has worked for a past non-vector q that asked for the eqn of a circle from 3 known points so I want to see if it could be applied to a q like this. Also idk if I want to go through an hour of pain looking for whether I did an arithmetic error so I'm just gonna ask the q here lol.

Any responses are much appreciated!!!!!

[Harrycc3000]

Hi again guys lol
I've spent an extremely long time on this question (19 b) and i still can't quite get it. 
Would appreciate any help!
Thanks

[kinslayer]

Hi! Just hoping to get some help for this question here
I think I have an idea as to what I did wrong.
The answers say that the mean of the 20 batteries is 140 and the standard deviation of the 20 batteries is 2.236.
If you use the method of...
E(20X) = 20E(X) and Var(20X) = 400Var(X0, then you'll get the wrong answer.
If you use this method of adding X 20 times instead, you'll get
E(X + X + X...+X) = 20E(X) and Var(X + X + X+....+X) = 20Var(X).
When do I know which method to use?
Any help or advice would be appreciated! Thanks

If Y is the sum of the lifetimes of the batteries, then saying that , and by extension is saying that the actual (ie. deterministic) sum of the lifetimes is equal to 20 times the lifetime of the first battery (or any single battery), which is obviously wrong and I believe the reason for your confusion. In other words,

Given the information that the X_i are independent and identically distributed (so they all have the same mean and variance and are uncorrelated), you'll get the correct variance by putting:

Var(Y) = Var(X₁ + X₂ + ... + X₂₀)
    = Var(X₁) + Var(X₂) + ... + Var(X₂₀)
    = 20*Var(X₁) = 5

This gives the correct standard deviation and you can then use the normal approximation to solve the problem. The key is that X₁ is not equal to X₂, but Var(X₁) = Var(X₂). Same for expectation, but it doesn't matter in this case. See below.

PS. For the expectation, both "ways" give 140 since expectation is linear whereas variance is bilinear. The fact that E[Y] = E[20*X₁] is simply coincidence in this case.

[fun_jirachi]

--snip--

Hey there

If possible, please try and include the diagram - really helps people skimming the forum and especially people that want to answer your questions, not everyone has the textbook in question! Just something to note for next time.

The method you suggest should work - we can look at your arithmetic for you if you still can't spot an error. However, this method is often redundant when there's a quicker method like the solution has.

Given distinct points \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) all lie on the same circle, you can set up two equations

then solve for h and k, which basically gives you the position vector for the centre.

Hi again guys lol
I've spent an extremely long time on this question (19 b) and i still can't quite get it. 
Would appreciate any help!
Thanks

Only going to give the full working if you really really can't do it. That being said, here are a few hints that you might want to consider (they do go progressively, so look at them in order).

What do you notice about \(\triangle EFX\) and \(\triangle CDX\)?
Note that you are given that BE = AF = BC - is there a convenient way to link this to the fact that \(|\vec{AB}| = k|\vec{BC}|\)?
Can you set up an equation that uses one of EC and EX or FX and FD in terms of k?

[Harrycc3000]

Hey there

If possible, please try and include the diagram - really helps people skimming the forum and especially people that want to answer your questions, not everyone has the textbook in question! Just something to note for next time.

The method you suggest should work - we can look at your arithmetic for you if you still can't spot an error. However, this method is often redundant when there's a quicker method like the solution has.

Given distinct points \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) all lie on the same circle, you can set up two equations

then solve for h and k, which basically gives you the position vector for the centre.

Only going to give the full working if you really really can't do it. That being said, here are a few hints that you might want to consider (they do go progressively, so look at them in order).

What do you notice about \(\triangle EFX\) and \(\triangle CDX\)?
Note that you are given that BE = AF = BC - is there a convenient way to link this to the fact that \(|\vec{AB}| = k|\vec{BC}|\)?
Can you set up an equation that uses one of EC and EX or FX and FD in terms of k?

Hi!
I managed to get a solution and will attach here, just wondering if it was the same method as yours?
Thanks for the tips! Really helped

[fun_jirachi]

It's similar but not the same - you don't have to worry about solutions being 'bang on' as long as they make logical sense and result in the correct answer. I'm glad you worked it out for yourself!

[jasmine24]

Hi, would anyone be able to explain where the sec = 5/4 came from?
Thank you!

[fun_jirachi]

Since we are calculating the rate of change at h = 30, \(\tan (\alpha) = \frac{h}{40} = \frac{3}{4}\). Drawing up a right angled triangle with sides of length 3, 4, and 5 with corresponding angle \(\alpha\) shows that \(\sec (\alpha) = \frac{5}{4}\)

[a weaponized ikea chair]

Hello

Please see attached, Question C.

The answer says 8454.02, but I got 8454.01.

What did I do wrong?

[S_R_K]

Hello

Please see attached, Question C.

The answer says 8454.02, but I got 8454.01.

What did I do wrong?

I think you meant "8854.01", but anyway:

I got 8854.02 by rounding the balance of the investment to 2 decimal places after each $400 deposit is made.

(I wouldn't say you've done anything wrong, but that accounts for the difference between the two answers).

[Jinju-san]

Heyo!

I was stuck on this particular question (attached below)…
I am mainly confused about how the gradient of the asymptotes should be calculated… I calculated them to be plus or minus 7/6, but the answer section of my textbook says it should be plus or minus 6/7…

Thanks! 

[fun_jirachi]

Hey!

When drawing hyperbolae, you have to be a bit more careful with asymptotes if they're vertical hyperbolae like in part c), as opposed to horizontal like in part a). It's not just as simple as 'the asymptotes are \(y = \pm \frac{b}{a}x\)', though this is true for horizontal hyperbolae - for vertical hyperbolae the asymptotes are \(y = \pm \frac{a}{b}x\) - see if you can figure out why!

Hope this helps

[jkfleur]

hi, need help with the following question my teacher set on an assignment... i can't seem to get it

given that sin(x) + cos(x) = sqrt(2)/3, where x is in Q4, find:
i) the exact value of sin(2x)
ii) the exact value of sin(x) - cos(x)
iii) the exact value of tan(x)
thanks for the help in advance!