VCE Specialist 3/4 Question Thread!

[paulsterio]

They're on the formula sheet, so you won't have to memorise them, but in all honesty, you might as well

[dc302]

They're on the formula sheet, so you won't have to memorise them, but in all honesty, you might as well

If it's on the formula sheet then you should know it--is how I went by

[acinod]

Just wondering if we need to know Derivates of inverse circular functions?

Most people in my class remembered it as they worked through the year.
Even during the exam I still remember I had to look it up in the formula sheet. It's kind of annoying and you're better off learning it but don't worry if you don't.

[ashoni]

hey guys, i need help with a problem about complex numbers. I'll type the question out .. show working out if possible, or tips on how to solve this kind of equation.

Solve to find x and y in the following.
a. (x+1) + (y-1)i = 2+3i
c. (2x+i) + (3-2yi) = x+3i

thanks in advancev

[ashoni]

ohhh thanks! i get it now, haha
could you please do this one, im kind of stuck on it D:

(2x-3i)+(-3+2y)i = y-xi

[trinh]

2x-3i-3i+2yi=y-xi
2x-6i+2yi=y-xi
2x+2i(y-3)=y-xi

2x=y [1]
2(y-3)=-x [2]

solve simultaneously.

correct me if i'm wrong xD

[trinh]

2x=y [1]
2(y-3)=-x [2]

sub [1] into [2]

2(2x-3)=-x
4x-6=-x
5x=6
x=6/5

sub x=6/5 into [1]
y=2x
y=12/5

[ashoni]

yeep, thats the answer thanks heaps trinh

[Special At Specialist]

Please help me with this kinematics problem:
v(t) = 4(1 - e^(-0.5t))
A particle travels for 30 seconds starting at t = 0. What distance does it cover?

I know area = definite integral under curve, but I have the velocity curve not the position curve I was thinking of maybe integrating it twice. Once as an indefinite integral to find x(t) and then once to find the area under the curve from 0 to 30. Would that be right?
Also, since distance and displacement are not the same thing, do I have to find when the velocity gets less than 0?

[xZero]

Distance  is the total area under the curve so graph it and see if it goes to the negative, if it does split it into two integrals, if not then just integrate it.

[Special At Specialist]

Oh I get it. I was overcomplicating it. All I had to do was x(30) - x(0), or the definite integral from 0 to 30 of the velocity function.

Thank you

[Planck's constant]

Oh I get it. I was overcomplicating it. All I had to do was x(30) - x(0), or the definite integral from 0 to 30 of the velocity function.

Thank you

Yes you were overcomplicating it, and now you are oversimplifying it.
xZero also pointed out that you must check that v(t) has no horizontal axis intercept for t in [0,30] ... which happens to be the case here.
Only then can you state with certainty that,

distance travelled = x(30) - x(0)

[Special At Specialist]

Ugh I'm so confused! The question is:

"A particle of mass 2kg hangs from a fixed point, O, by a light inextensible string of length 2.5m. It is pulled aside a horizontal distance of 2m by a force P N inclined at an angle of 75 degrees with the downward vertical, and rests in equilibrium. Find P and the tension of the string."

I thought that the length of the string didn't matter and the only thing that mattered was the force of the string. I tried making a triangle of forces but I didn't have enough information to solve it. Someone please tell me how the length of the string affects the forces.

[xZero]

the length of the string gives you the angle of the tension force, there should be 3 forces in total and you need to use simultaneous equations to solve for tension and P force

[Special At Specialist]

I thought the forces gave you the forces? After all, it wouldn't matter if the string was 1cm long or 1km long, there would still be the same force applied to it, right?

None of the previous problems I did even mentioned how long the string was. They just said the force and the angle and it was a simple matter of resolving forces. But I don't know what to do when I have forces, angles and lengths. How are they related?