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March 19, 2024, 04:05:23 pm

Author Topic: Specialist 1/2 Question Thread!  (Read 119702 times)  Share 

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Sine

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Re: Specialist 1/2 Question Thread!
« Reply #150 on: October 15, 2017, 09:12:47 pm »
+1
just did the question got same answer as VanillaRice - so there is no answer given ahah
you can also do this question (or check answer by) converting the vectors to lines in the cartesian plane to find the min distance.

KiNSKi01

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Re: Specialist 1/2 Question Thread!
« Reply #151 on: October 17, 2017, 06:49:35 pm »
0
How do you answer a question like this? I know once I have values for cos u and sin u I just need to use one of the double angle formulas for cosine. But I have no idea how to find out what these values are
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LifeisaConstantStruggle

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Re: Specialist 1/2 Question Thread!
« Reply #152 on: October 17, 2017, 07:29:29 pm »
+3
I don't know if you've learnt the double angle formulas in spec 1/2, but the formula for this would be:

cos(2x)=1-2sin2(x)

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/specmaths-formula-w.pdf

Here are all the trig identity formulas.
« Last Edit: October 17, 2017, 07:33:49 pm by LifeisaConstantStruggle »
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A TART

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Re: Specialist 1/2 Question Thread!
« Reply #153 on: October 17, 2017, 07:43:18 pm »
+4
How do you answer a question like this? I know once I have values for cos u and sin u I just need to use one of the double angle formulas for cosine. But I have no idea how to find out what these values are

Hi!

If you look at your forumale sheet, the double angle forumula for cosine has 3 variations (couldn't think of a better word).

You are not limited to cos(2x)=(cos x)^2 - (sin x)^2 (see attached images)

Working:
[url=https://i.imgur.com/vNqbeiR.jpg]
[url=https://i.imgur.com/M7InnYt.jpg]
(I have a history of changing a 1 into a 2 and 2 into a 12 in an exam so take this with a grain of salt)
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KiNSKi01

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Re: Specialist 1/2 Question Thread!
« Reply #154 on: October 17, 2017, 07:53:43 pm »
0
Yeah I know the 3 different cosine double angle formulas but I was asking about what values you set sin/cos equal to. For this example, if you used a double angle formula with cos in it, would the value for cos u or cos v (Depending) be 4/5 because of pythag theorem?
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VanillaRice

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Re: Specialist 1/2 Question Thread!
« Reply #155 on: October 17, 2017, 08:23:04 pm »
+1
Yeah I know the 3 different cosine double angle formulas but I was asking about what values you set sin/cos equal to. For this example, if you used a double angle formula with cos in it, would the value for cos u or cos v (Depending) be 4/5 because of pythag theorem?
You can absolutely use the Pythagorean theorem to find the value of cos, however be sure to consider the domain of alpha and beta  (i.e. is it 4/5 or -4/5?).

Hope this helps :)
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Shadowxo

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Re: Specialist 1/2 Question Thread!
« Reply #156 on: October 18, 2017, 07:16:56 pm »
+2
Yeah I know the 3 different cosine double angle formulas but I was asking about what values you set sin/cos equal to. For this example, if you used a double angle formula with cos in it, would the value for cos u or cos v (Depending) be 4/5 because of pythag theorem?
You can find cos(alpha) or cos(beta) using pythag but you wouldn't need to for this question.

If they wanted you to find sin(2*alpha) however, you'd have to find out sin(alpha) as well, as sin(2*alpha)=2sin(alpha)cos(alpha)
Note that if you'd already found cos(2*alpha) you could use pythag / trig to find sin(2*alpha)
Hope this answers your question :)
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KiNSKi01

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Re: Specialist 1/2 Question Thread!
« Reply #157 on: October 19, 2017, 09:14:17 pm »
0
Thanks for the help!  ;D Circular functions are the worst   :-\
« Last Edit: October 19, 2017, 09:26:02 pm by KiNSKi01 »
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Re: Specialist 1/2 Question Thread!
« Reply #158 on: October 20, 2017, 07:28:14 pm »
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Hi guys, first post here  :o,

Just a kinematics question that I want to make sure I've gotten right:

"Two particles, travelling in the same direction, pass a given point together at 8 m/s. Thereafter, one particle begins to slow down with a constant retardation of 2/7 m/s^2. The other continues for 6 seconds at 8 m/s and then slows down with a retardation of 1/9(t-6) m/s^2 where t is the time after the two particles were together. "

"Find how far each of the particles gave travelled when they come to rest."

The first particle was pretty simple; I got 112m. For the second one, I got an answer of 97m, which is the one I'm not quite sure about. Let me know! Cheers
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Guideme

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Re: Specialist 1/2 Question Thread!
« Reply #159 on: October 22, 2017, 06:30:51 pm »
0


Question 6 pls thank you
:0 :)

Dais_Deorum

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Re: Specialist 1/2 Question Thread!
« Reply #160 on: October 22, 2017, 06:53:47 pm »
+2
Hey Guideme,

If the object being acted on by the force is in equilibrium (not moving at all), that force is balanced by an equal one acting in the opposite direction to it. So that force of 20kgwt acting down the plane happens to be a component of the gravity force, and is the result of mg * sin40.
So we know that:

mg * sin40 = 20

Therefore:

Mass = 20/sin40

In this case, the mass equals 31.1145kg

*edit*

To find the pressure this weight exerts on the surface, you must find the normal force, or the component of the force acting perpendicular to the plane. As the object is in equilibrium, the pressure acting on the surface will be the same as the normal force.  To find this we use the parallel force (20kgwt) and tan.

tan(theta) = opposite/adjacent

tan40 = 20/N

N = 20/tan40

N (pressure  on surface) = 26.8416kgwt
« Last Edit: October 22, 2017, 08:19:42 pm by Dais_Deorum »
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Shadowxo

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Re: Specialist 1/2 Question Thread!
« Reply #161 on: October 22, 2017, 09:40:12 pm »
+3
Hi guys, first post here  :o,

Just a kinematics question that I want to make sure I've gotten right:

"Two particles, travelling in the same direction, pass a given point together at 8 m/s. Thereafter, one particle begins to slow down with a constant retardation of 2/7 m/s^2. The other continues for 6 seconds at 8 m/s and then slows down with a retardation of 1/9(t-6) m/s^2 where t is the time after the two particles were together. "

"Find how far each of the particles gave travelled when they come to rest."

The first particle was pretty simple; I got 112m. For the second one, I got an answer of 97m, which is the one I'm not quite sure about. Let me know! Cheers
Hey :)
So the kinematics formulas only work for constant acceleration, which the second one isn't, so we have to use integration.
The distance travelled during the first section (no acceleration) is v*t = 6*8=48m
The velocity in the second section (with acceleration) is the integral of acceleration with respect to time


Edit: Also note there are a couple ways to do this. One other way would be to find x as a function of t for t > 6 and sub in the value where v=0 (ie t=18), avoiding definite integrals
« Last Edit: October 22, 2017, 09:49:37 pm by Shadowxo »
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HighSchoolerRS

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Re: Specialist 1/2 Question Thread!
« Reply #162 on: October 24, 2017, 06:22:03 pm »
0
Hi guys.
I can't seem to figure out how to prove this and was hoping somebody could run me through it?
I am pretty sure that the question was the first one but there is a slight possibility that it is the second one.




Thank you!
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VanillaRice

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Re: Specialist 1/2 Question Thread!
« Reply #163 on: October 24, 2017, 06:50:52 pm »
+3
Hi guys.
I can't seem to figure out how to prove this and was hoping somebody could run me through it?
I am pretty sure that the question was the first one but there is a slight possibility that it is the second one.




Thank you!
It's the second one.
Beginning from the RHS and using the identity 1 + tan2(x) = sec2(x)
Spoiler


I'll let you finish it off :) Please post if you're having any issues.

Hope this helps :)
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HighSchoolerRS

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Re: Specialist 1/2 Question Thread!
« Reply #164 on: October 28, 2017, 04:49:21 pm »
+1
It's the second one.
Beginning from the RHS and using the identity 1 + tan2(x) = sec2(x)
Spoiler


I'll let you finish it off :) Please post if you're having any issues.

Hope this helps :)

Thank you! I actually got the answer now haha
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