CSSA 04 trial questions (1 Viewer)

[ToO LaZy ^*]

if anyone could give me the solutions to these questions that would be really great..
these were the ones i couldn't get.., not full marks anyway..

 

[withoutaface]

1.a. I think you end up with -n^2(x-1) because you remove the '1' through differentiation.
b. find the endpoints by looking at where v=0 and find their range
c. basically just find the point where the particle starts, and that is your distance. sub in x=0 to the x=.... equation to find the time
2. use pytagoras' and remember that all radii of the same circle are the same length
3. can't remember
4. use the velocities of the particles with the values you obtained in the last part and use pythagoras' theorem to show that one is 45' above the horizontal and one is 45' below at the time of collision

sorry if that was a bit vague, hope it helped.

 

[CrashOveride]

Q3 was a rip-off of an Excel test question in the back ^_^

I'll give you the remainder when i get back

 

[Estel]

1.
a) x = 4cos^2t-2sin^2t
cos2t = 2cos^2t - 1 = 1 - 2sin^2t
so x = (4/2)(1+cos2t) - (2/2)(1-cos2t)
= 1 + 3cos2t
x' = -6sin2t
x"= -12cos2t
= -4((1+3cos2t)-1)
=-4(x-1)
b) -2<=x<=4
Period = 2pi/n = pi
c) to reach origin, particle moves 4m,
solve 0 = 1 + 3cos2t
cos2t = -1/3
t = 0.955
ie 1.0 seconds

2
(x+8)^2 + (x+12)^2 = 20^2
and expand

3.
dx/dt = sinxcosx
dt/dx = 1/sinxcosx
t = ln(tanx) + C
e^(t-c) = tanx
x = tan(-1)e^(t)
as t --> infinity, x -->pi/2
sketch urself

4.
bah can't be bothered doin... too many sections

If you really need it, I'll do it later... kind of in a rush rite now.
If i got any wrong, it's cuz I suck + in rush.

Cheers

 

[ToO LaZy ^*]

4. use the velocities of the particles with the values you obtained in the last part and use pythagoras' theorem to show that one is 45' above the horizontal and one is 45' below at the time of collision

sorry if that was a bit vague, hope it helped.

hmmm..i don't know where you're coming from, but what i've tried to do is find their point of contact (which is done before) then and finding the tangents of both particles which should hopefully equal -1 for them to be perpendicular....but it doesn't for some reason...

if you have time, could you show me your way.?...



ESTEL: you rock dude!

 

[withoutaface]

You wouldn't believe this, but I have my answers to every other question but q7 sitting on my desk -__-.

Well i think from part ii) you can find a value for theta, and you know that time=1sec.

Sub these values into x' and y' for particle O

then draw a right angled triangle with these velocities as the components, then find the angle @ above the horizontal


|\
|..\
|....\
|__@\

yes that diagram is shit, but you get the idea.

And you do the same thing for particle P (angle *). You then subtract * from @ and show that this gives a value of 90'.

If I can find my answers I might be able to give you a more detailed solution, but this will have to do for now.

 

[ToO LaZy ^*]

ohkk..i'll give it a go later, but for now, sleep is more important.

 

[withoutaface]

ohkk..i'll give it a go later, but for now, sleep is more important.

2+2=11.


Prove me wrong.

 

[ToO LaZy ^*]

let's see, what do we have here...2+2..
phhwhoaa.!
what do you think i am?...a calculator!?

how bout you prove yourself wrong..:uhhuh:

 

[withoutaface]

let's see, what do we have here...2+2..
phhwhoaa.!
what do you think i am?...a calculator!?

how bout you prove yourself wrong..:uhhuh:

I am using base 3.


:. I am right

 

[Xayma]

But I am a bit worried about your notation. Normally you would write:
2₃+2₃=11₃

Anyway who uses base 3, Binary, Octal, Decimal and Hexadecimal are the main ones used.

 

[withoutaface]

But they are useless to disprove his custom status.

 

[ToO LaZy ^*]

curse you faceboy.!..leave my custom status alone! barney and friends hater.

 

[withoutaface]

curse you faceboy.!..leave my custom status alone! barney and friends hater.

Well I can see why this is so popular!

 

[ToO LaZy ^*]

i take back what i said...