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July 22, 2025, 09:06:57 am

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2547670 times)  Share 

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Phy124

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Re: Specialist 3/4 Question Thread!
« Reply #2955 on: February 14, 2014, 01:07:16 am »
+1
Looks alright to me. Although, just as a pointer I think in future it would be easier to multiply both the numerator and denominator of by . i.e.







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alchemy

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Re: Specialist 3/4 Question Thread!
« Reply #2956 on: February 14, 2014, 09:02:08 pm »
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A car travels at 60km/h for the first half of a journey. It travels at x km/h for the second half of the journey. Find an expression for its average speed throughout the journey.

m.Chemia

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Re: Specialist 3/4 Question Thread!
« Reply #2957 on: February 14, 2014, 09:19:18 pm »
+2
A car travels at 60km/h for the first half of a journey. It travels at x km/h for the second half of the journey. Find an expression for its average speed throughout the journey.

Let be the total distance.
Then the distance of the first half of the journey is , the distance of the second half of the journey is also .

As , the time for the first half would be , the time for the second half would be .

The total time would be
.

The average speed is then
.

clıppy

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Re: Specialist 3/4 Question Thread!
« Reply #2958 on: February 15, 2014, 04:38:37 pm »
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Would someone please be able to explain vector resolutes to me? I just can't seem to wrap my head around them or understand what they are or how they work.
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Putting this here so I don't forget about it: http://www.codecogs.com/latex/eqneditor.php

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Ancora_Imparo

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Re: Specialist 3/4 Question Thread!
« Reply #2959 on: February 15, 2014, 11:09:04 pm »
+3
This is much easier to understand when drawn out, so please forgive me if the following explanation is confusing.

Say we have two vectors, and and we are trying to determine the vector resolute of in the direction of . What this means is that we are trying to determine how much of vector lies in the direction of vector .

Let's take a simple example to illustrate. Let and
Notice that is in the direction of the x-axis. So if I wanted to find the vector resolute of in the direction of , what I want to actually find is 'how much of lies on the x-axis'. Since vector has already been nicely split into and components for us, we know that the answer to this is just , which was the x-axis component of .

What makes things tricky is that usually neither vector lies on the x-axis or y-axis. This means that their direction will be something totally random and we can't easily answer the question 'how much of lies in the direction of ?'. How do we overcome this?

Join the two vectors such that their tails both start at the same point. Now, draw a line starting from the head of vector towards vector such that it is perpendicular to vector . If this line doesn't actually pass through vector , extend vector so that it does. Now, you should have a right-angled triangle with vector as the hypotenuse.

What we have done here is resolve vector in two directions. One is parallel to vector (it lies on vector in our diagram) and one is perpendicular to vector . The side of the right-angled triangle that lies on vector is our vector resolute.

Let be the angle between our two vectors. The length of this side is thus given by . This length is known as the scalar resolute.
Since the magnitude of is 1, this expression can be rewritten as:

Hence, the scalar resolute of in the direction of is .

Since the vector resolute is in the same direction as , and we know that has a magnitude of 1, all we need to do is multiply the length of the side by .
Ie: The vector resolute is given by the scalar resolute multiplied by the unit vector of .
 
As a formula, this is: Vector resolute of in the direction of
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BLACKCATT

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Re: Specialist 3/4 Question Thread!
« Reply #2960 on: February 16, 2014, 03:01:04 pm »
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Can someone explain to my why the answer to j is 5pi/6 and not -pi/6?

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drake

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Re: Specialist 3/4 Question Thread!
« Reply #2961 on: February 16, 2014, 03:55:59 pm »
+1
arccos(sin(-pi/3)) = arccos(-sqrt(3)/2) = 5pi/3... think about the range of arccos(x). the range is [0,pi]... so -pi/6 is not in the range whilst 5pi/6 is.
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Sanguinne

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Re: Specialist 3/4 Question Thread!
« Reply #2962 on: February 16, 2014, 07:02:06 pm »
0
cambridge textbook chapter 4 review multiple choice, question 2.

2) The complex number z shown in the diagram is best represented by:
I don't understand why C is correct when the only difference between B and C is that B is in degrees and C is in radians.



Thanks
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m.Chemia

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Re: Specialist 3/4 Question Thread!
« Reply #2963 on: February 16, 2014, 07:26:34 pm »
+1
cambridge textbook chapter 4 review multiple choice, question 2.

2) The complex number z shown in the diagram is best represented by:
I don't understand why C is correct when the only difference between B and C is that B is in degrees and C is in radians.



Thanks

For polar form of a complex number, the argument (angle) is default to be in radians if the degree sign is not explicitly shown. So option B might look like it has an argument in degrees, but it is actually still in radians, and that's why C is correct while B is not.

Sanguinne

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Re: Specialist 3/4 Question Thread!
« Reply #2964 on: February 19, 2014, 05:55:07 pm »
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find all real values of r for which ri is a solution of the equation
z^4-2z^3+11z^2-18z+18=0
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RKTR

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Re: Specialist 3/4 Question Thread!
« Reply #2965 on: February 19, 2014, 07:14:23 pm »
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find all real values of r for which ri is a solution of the equation
z^4-2z^3+11z^2-18z+18=0
sub ri into the equation

(ri)^4-2(ri)^3+11(ri)^2-18(ri)+18=0
r^4-2r^3(-i)-11r^2-18ri+18=0
r^4+2ir^3-11r^2-18ri+18=0
r^4-11r^2+18=18ri-2r^3i
r must be real value
so (18r-2r^3)i=0
     18r-2r^3=0
      2r^3-18r=0
     2r(r^2-9)=0
   r=0,r=-3,r=+3
r=0 cant be accepted because if you sub into equation you get 18=0
therefore r=-3 or r=3
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Sanguinne

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Re: Specialist 3/4 Question Thread!
« Reply #2966 on: February 19, 2014, 07:18:02 pm »
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sub ri into the equation

(ri)^4-2(ri)^3+11(ri)^2-18(ri)+18=0
r^4-2r^3(-i)-11r^2-18ri+18=0
r^4+2ir^3-11r^2-18ri+18=0
r^4-11r^2+18=18ri-2r^3i
r must be real value
so (18r-2r^3)i=0
     18r-2r^3=0
      2r^3-18r=0
     2r(r^2-9)=0
   r=0,r=-3,r=+3
r=0 cant be accepted because if you sub into equation you get 18=0
therefore r=-3 or r=3
just to clarify, your equation (18r-2r^3)i=0 is based on the fact that r must be real, hence the complex = 0?? hence you can solve for r?
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RKTR

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Re: Specialist 3/4 Question Thread!
« Reply #2967 on: February 19, 2014, 07:33:59 pm »
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Yup
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eagles

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Re: Specialist 3/4 Question Thread!
« Reply #2968 on: February 19, 2014, 08:44:32 pm »
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How to do this question without calc?

sin (cos-1(1/3)) = ?

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RKTR

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Re: Specialist 3/4 Question Thread!
« Reply #2969 on: February 19, 2014, 08:53:41 pm »
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How to do this question without calc?

sin (cos-1(1/3)) = ?

Thanks.
let cos-1 (1/3)=x
    cos(x)=1/3
   draw a triangle and use pythagoras theorem
   sin(cos-1(1/3) ) = (2root2)/3?
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