Author Topic: VCE Specialist 3/4 Question Thread! (Read 2549881 times) Tweet Share

b^3 · « **Reply #2250 on:** September 09, 2013, 09:58:15 pm »

I actually read it initially as a method question, and was thinking more along those lines (yeah I now realise this is a spesh thread since we have a vector in 3 dimensions). As Alwin has said the approach still works as it's motion along a straight line, and I probably should have pointed that out, since it's a multiple choice question it doesn't really matter which way you get it, although for SA or ER it's probably best to stick to the integral (and well there's only the straight line case where this other method works anyways).

This is why I shouldn't mix so much red bull, low sleep, cramming and procrastination together before mid-sems.

EDIT: I'll just point it out again, so that some people don't get the wrong idea. You should be using the integration method, but this does work as we are dealing with motion in a straight line, and so you can look at it as you would with motion along one dimension as you would in methods, but generalise it to three dimensions.

EDIT2: I should add, sorry for missing stating some of the assumptions in a few threads lately, I kinda overlook the really small things now as they seem trivial and common sense compared to what I've been doing lately.

lzxnl · « **Reply #2251 on:** September 09, 2013, 10:11:33 pm »

Yeah I must have missed that. I just saw the quadratic term and the vector, so I immediately thought that it was a vector function that varied with time. Of course it was, but evidently not quite the same way as I had imagined.
Well, this is what happens after you force yourself to finish the 2009 VCAA exam 2 in 62 minutes. Your brain becomes dead.

Alwin · « **Reply #2252 on:** September 09, 2013, 10:20:29 pm »

Quote from: nliu1995 on September 09, 2013, 10:11:33 pm

Well, this is what happens after you force yourself to finish the 2009 VCAA exam 2 in 62 minutes. Your brain becomes dead.

62 min, a bit slack mate? I could fit 2 vcaa chem / methods ones or 3 physics ones at a pinch. Not sure about physics no more, but I used to when I still remembered the stuff

Holmes · « **Reply #2253 on:** September 10, 2013, 07:12:03 pm »

So I got dN/dt = kNt, and then I integrated with respect to N (by flipping it), to get t = 1/k ln(N) + c, not sure how to take it from here.

Done:

$dn/dt=kN$
$t=1/k \int 1/N$
$t=1/k*ln(N) + c$
$e^k(t-c)=N,<br />N=e^kt*e^-ck$
,let A = e^-ck

Then at t+100, N=3/2N (info given)
get an expression in terms of k
At t=0, N=A, and k= 0.01ln(3/2)

and so N=A*e^[0.01 ln (3/2)t], as required.

barydos · « **Reply #2254 on:** September 10, 2013, 08:16:49 pm »

If possible, I'd like some help with:

mainly part d) and e)

SocialRhubarb · « **Reply #2255 on:** September 10, 2013, 08:33:27 pm »

d.)

Spoiler

There's two ways to do this: convert both to cartesian form or introduce a new variable and solve for two variables. I'm going to use the cartesian method.

$\text{For } r_A:$

$\text{Let x=2t and y=t}$

$x=2y$

$\text{For } r_B:$

$\text{Let } x=4-4\sin(\alpha t)\text{ and }y=4\cos(\alpha t)$

$16\sin^2(\alpha t)=(x-4)^2\ \ \ 16\cos^2(\alpha t)=y^2$

$16\sin^2(\alpha t)+16\cos^2(\alpha t)=16 = y^2+(x-4)^2$

Substitute $x=2y$ , and

$16=y^2+(2y-4)^2=y^2+4y^2-16y+16$

$y(5y-16)=0$

$y=0\text{ or }\frac{16}{5}$

$x=2y\implies x=0\text{ or }\frac{32}{5}$

Coordinates of collision: $(0,\ 0)$ and $(\frac{32}{5},\ \frac{16}{5})$

e.)

Spoiler

We worked out the points of collision in part d. Now we work out what time particle A comes to these points of collision, and alter alpha such that particle B is also at the point of collision at those times.

$t=0\text{ or }\frac{16}{5}$

Looking at the position equation for particle B, you notice that for whatever value of alpha we have, particle B is never at a collision point at t=0. So we look at the other time.

At $t=\frac{16}{5}$ , we want the particle to be at the point $(\frac{32}{5},\ \frac{16}{5})$ . If we sub this information into the equation for particle B:

$\text{Eqn 1: }4-4\sin(\frac{16}{5}\alpha)=\frac{32}{5}$

$\text{Eqn 2: }4\cos(\frac{16}{5}\alpha)=\frac{16}{5}$

$\text{From Eqn 1: }\sin(\frac{16}{5}\alpha)=-\frac{3}{5}$

$\text{From Eqn 2: }\cos(\frac{16}{5}\alpha)=\frac{4}{5}$

So we can tell from these results that, since sin is negative and cos is positive, that we are in the fourth quadrant. Thus, we want $\frac{3\pi}{2}\leq \frac{16}{5}\alpha \leq 2\pi$ , since we want the smallest possible value of alpha that is greater than 0. This is important because the sin inverse and cos inverse functions will spit out results which are not in this domain, and we have to make sure we're answering the question correctly.

$\text{From Eqn 1: }\frac{16}{5}\alpha=\sin^{-1}(-\frac{3}{5})+2\pi$ The $2\pi$ is necessary to make sure alpha is in the domain which we want.

$\alpha=\frac{5}{16}(2\pi-\sin^{-1}(\frac{3}{5}))\text{ (since }\sin^{-1}\text{ is an odd function, we can manipulate the negatives a bit)}$

$\alpha \approx 1.76^c$

barydos · « **Reply #2256 on:** September 10, 2013, 09:01:46 pm »

Quote from: SocialRhubarb on September 10, 2013, 08:33:27 pm

d.)
Spoiler
There's two ways to do this: convert both to cartesian form or introduce a new variable and solve for two variables. I'm going to use the cartesian method.

$\text{For } r_A:$

$\text{Let x=2t and y=t}$

$x=2y$

$\text{For } r_B:$

$\text{Let } x=4-4\sin(\alpha t)\text{ and }y=4\cos(\alpha t)$

$16\sin^2(\alpha t)=(x-4)^2\ \ \ 16\cos^2(\alpha t)=y^2$

$16\sin^2(\alpha t)+16\cos^2(\alpha t)=16 = y^2+(x-4)^2$

Substitute $x=2y$ , and

$16=y^2+(2y-4)^2=y^2+4y^2-16y+16$

$y(5y-16)=0$

$y=0\text{ or }\frac{16}{5}$

$x=2y\implies x=0\text{ or }\frac{32}{5}$

Coordinates of collision: $(0,\ 0)$ and $(\frac{32}{5},\ \frac{16}{5})$

e.)
Spoiler
We worked out the points of collision in part d. Now we work out what time particle A comes to these points of collision, and alter alpha such that particle B is also at the point of collision at those times.

$t=0\text{ or }\frac{16}{5}$

Looking at the position equation for particle B, you notice that for whatever value of alpha we have, particle B is never at a collision point at t=0. So we look at the other time.

At $t=\frac{16}{5}$ , we want the particle to be at the point $(\frac{32}{5},\ \frac{16}{5})$ . If we sub this information into the equation for particle B:

$\text{Eqn 1: }4-4\sin(\frac{16}{5}\alpha)=\frac{32}{5}$

$\text{Eqn 2: }4\cos(\frac{16}{5}\alpha)=\frac{16}{5}$

$\text{From Eqn 1: }\sin(\frac{16}{5}\alpha)=-\frac{3}{5}$

$\text{From Eqn 2: }\cos(\frac{16}{5}\alpha)=\frac{4}{5}$

So we can tell from these results that, since sin is negative and cos is positive, that we are in the fourth quadrant. Thus, we want $\frac{3\pi}{2}\leq \frac{16}{5}\alpha \leq 2\pi$ , since we want the smallest possible value of alpha that is greater than 0. This is important because the sin inverse and cos inverse functions will spit out results which are not in this domain, and we have to make sure we're answering the question correctly.

$\text{From Eqn 1: }\frac{16}{5}\alpha=\sin^{-1}(-\frac{3}{5})+2\pi$ The $2\pi$ is necessary to make sure alpha is in the domain which we want.

$\alpha=\frac{5}{16}(2\pi-\sin^{-1}(\frac{3}{5}))\text{ (since }\sin^{-1}\text{ is an odd function, we can manipulate the negatives a bit)}$

$\alpha \approx 1.76^c$

Wow thanks a lot for that quick response, you're a genius

barydos · « **Reply #2257 on:** September 10, 2013, 10:31:54 pm »

Last set of questions for tonight from me:

Answers:

Spoiler

Thanks in advance.

Phy124 · « **Reply #2258 on:** September 11, 2013, 12:45:52 am »

12. a) i. $\underset{\sim}{\dot{r}_2}(t) = (2t + a)\underset{\sim}{i} + (t+b)\underset{\sim}{j} , a,b \in \mathbb{R}$ (I integrated the acceleration vector function)

$\underset{\sim}{\dot{r}_2}(0)=a\underset{\sim}{i} +b\underset{\sim}j} = -4\underset{\sim}{i} + 0\underset{\sim}{j}$ (We were told the initial velocity was $-4\underset{\sim}{i}$ )

$\Rightarrow a=-4, b = 0$

$\Rightarrow \underset{\sim}{\dot{r}_2}(t) = (2t -4) \underset{\sim}{i}+ t\underset{\sim}{j}$

Try doing the same for part ii. and see if you can arrive at the given answer.

b) i. This occurs when both of the i and j components for each of the vectors are equal. So, equating the i components; $2t-4 = t \Rightarrow t = 4$

ii. We can now equate the j components using this t value $t = k -t \Leftrightarrow 4 = k - 4 \Rightarrow k =8$

iii. Sub the value for t back into either of the equations to find the velocity at this point; $\underset{\sim}{\dot{r}_2}(4) = (2(4)-4)\underset{\sim}{i} + (4)\underset{\sim}{j} = 4(\underset{\sim}{i}+\underset{\sim}{j})$

barydos · « **Reply #2259 on:** September 11, 2013, 08:20:35 am »

Quote from: Snow Red on September 11, 2013, 12:45:52 am

12. a) i. $\underset{\sim}{\dot{r}_2}(t) = (2t + a)\underset{\sim}{i} + (t+b)\underset{\sim}{j} , a,b \in \mathbb{R}$ (I integrated the acceleration vector function)

$\underset{\sim}{\dot{r}_2}(0)=a\underset{\sim}{i} +b\underset{\sim}j} = -4\underset{\sim}{i} + 0\underset{\sim}{j}$ (We were told the initial velocity was $-4\underset{\sim}{i}$ )

$\Rightarrow a=-4, b = 0$

$\Rightarrow \underset{\sim}{\dot{r}_2}(t) = (2t -4) \underset{\sim}{i}+ t\underset{\sim}{j}$

Try doing the same for part ii. and see if you can arrive at the given answer.

b) i. This occurs when both of the i and j components for each of the vectors are equal. So, equating the i components; $2t-4 = t \Rightarrow t = 4$

ii. We can now equate the j components using this t value $t = k -t \Leftrightarrow 4 = k - 4 \Rightarrow k =8$

iii. Sub the value for t back into either of the equations to find the velocity at this point; $\underset{\sim}{\dot{r}_2}(4) = (2(4)-4)\underset{\sim}{i} + (4)\underset{\sim}{j} = 4(\underset{\sim}{i}+\underset{\sim}{j})$

Thank you! I got the initial positions mixed up with the wrong velocity vectors woops.

And I figured out q14.
a) is just a velocity in direction of j so path is linear (right?)
b) integrating the velocity gives you $r(t) = 2 i + (\frac{1}{2} t^2 -3t + 1) j$
$x = 2$
$y = \frac{1}{2}t^2-3t+1$
for $t>=0$ the min. value for y is -3.5, so $x= 2, y>=-3.5$
c) $y= t-3$ , so $y=0$ when $t=3$ , subbing in the value of t into the position vector gives (2,-3.5)

BasicAcid · « **Reply #2260 on:** September 14, 2013, 11:26:16 am »

Hey there, I'm just wondering whether the following diagram could also be expressed as Re(z) + Im(z) as well? (I know C is incorrect).

It's from the 2001 VCAA exam by the way, cheers.

Stevensmay · « **Reply #2261 on:** September 14, 2013, 11:29:29 am »

Wouldn't that give a line with equation y=x?

BasicAcid · « **Reply #2262 on:** September 14, 2013, 11:34:10 am »

Quote from: Stevensmay on September 14, 2013, 11:29:29 am

Wouldn't that give a line with equation y=x?

I said I know C is incorrect, I'm asking whether Re(z) + Im(z) would be the same.

Stevensmay · « **Reply #2263 on:** September 14, 2013, 11:37:08 am »

Turns out I was graphing in the wrong mode sorry. Re(z) + Im(z) = 0 would be the same.

BasicAcid · « **Reply #2264 on:** September 14, 2013, 11:41:10 am »

Quote from: Stevensmay on September 14, 2013, 11:37:08 am

Turns out I was graphing in the wrong mode sorry. Re(z) + Im(z) = 0 would be the same.

Like exactly the same? Are there any differences at all?

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