VCE Specialist 3/4 Question Thread!

[Hancock]

6a)

a = v * (dv/dx) = (dx/dt)(dv/dx) = dv/dt = a

dv/dx = 12(x-5)^2

a = (4(x-5)^3)*(12(x-5)^2)) = 48(x-5)^5 m/s


7a)

a = 9 - 4x

0.5v^2 = int (9-4x) dx
0.5v^2 = 9x - 2x^2 + c
v^2 = 18x - 4x^2 + c

v = 1, x = 2

1 = 36 - 4(4) + c
1 = 36 - 16 + c
1 = 20 + c
c = -19

v^2 = 18x - 4x^2 - 19
v = (18x - 4x^2 - 19)^0.5

Taken positive root because when x = 2, v = +1

and is defined where 18x - 4x^2 - 19 > 0

-4x^2 + 18x - 19 = 0
4x^2 - 18x + 19 = 0

x = (18 +/- (18^2 - 4(4)(19))^0.5) / 8
x = (1/4)(9 +/- 5^0.5)

Therefore, velocity is defined between (1/4)(9-root5) =< x =< (1/4)(9+root5)

[Bhootnike]

i got the same ans. as you for q6a,
but the answer is

LOL.


and for 7, nope
its


EDIT. What Hancock got
thanks guys.
ill try and digest it now!

[Bhootnike]

for q7
i just realised...
i accidentally wrote down 9-4x^2 rather than just 9-4x
ffs... no wonder i was getting it wrong

for q6
that's awesome! i didnt even think about that...
your told v, and you can find out dv/dx
so you multiply them together to find a.
genius!

[Bhootnike]

alright
im so bad at this i couldnt do the next one either...

this time i employed the use of v. dv/dx and did it
and i got x = 2log|v| + 3v^2/2  + c
which is not right cos once you find c, and solve for v, it doesnt give the right answer..

[Hancock]

a = (1/6)(3v^2 + 2)
v(dv/dx) = (1/6)(3v^2 + 2)
dv/dx = (1/6)(3v + 2/v)

dv/dx = (1/6v)(3v^2 + 2)
FLIP
You didn't flip here and integrated with respect to v, which screwed it up

dx/dv = 6v / (3v^2 + 2)

x = int 6v/(3v^2 + 2) .dv
x = ln |3v^2 + 2| + c
x + c = ln |3v^2 + 2|
e^(x+c) = |3v^2 + 2|
e^(x)*e^(c) = |3v^2 + 2|

Let e^c = +/- A
Ae^x = 3v^2 + 2

3v^2 = Ae^(x) - 2
v = 3 when x = 0

3(9) = Ae^0 - 2
27 = A - 2
A = 29

3v^2 = 29e^(x) - 2
v = ([29e^(x) - 2]/3)^0.5

[Bhootnike]

this is what i did:

a= (1/6)(3v^2 + 2)
v(dv/dx) = (1/6)(3v^2 + 2)
dv/dx = (1/6)(3v + 2/v)

dv/dx = (1/6v)(3v^2 + 2)

so from here,
i simplified the fraction to get
i then FLIPPED
to get
when i integrated that w.r.t v,
i got

i dont understand how simplifying it to get two fractions, then flipping, gave me a different answer to not simplifying it and flipping initially ?

[Hancock]

this is what i did:

a= (1/6)(3v^2 + 2)
v(dv/dx) = (1/6)(3v^2 + 2)
dv/dx = (1/6)(3v + 2/v)

dv/dx = (1/6v)(3v^2 + 2)

so from here,
i simplified the fraction to get
i then FLIPPED
to get
when i integrated that w.r.t v,
i got

i dont understand how simplifying it to get two fractions, then flipping, gave me a different answer to not simplifying it and flipping initially ?

They problem is that you have to flip after you simplify into one fraction, not flip both.

Example:

2 = 1/2 + 3/2

Does flipping both sides now work?

1/2 =/= 2 + 2/3

However,

2 = 4/2 (simplified RHS fractions into one term)
1/2 = 2/4

[Bhootnike]

ahhh dammit'
you cant divide additive expressions ,.....
*facepalm*

thanks man!

[Mr. Study]

Not too sure if this is a noob question.

The solutions state that . Why exactly can they say this?

Thanks.

[b^3]

The m's are representing gradients.

Now we know that where is the angle the line makes with the positive direction of the x-axis. Here that angle is
So

(originally I looked at this wrongly as if the m's were representing lengths, and tried to overcomplicate it..., might have been what you were thinking too )

[Mr. Study]

Haha, yeah ... I was thinking of lengths. 

Thanks .

[soccerboi]

In a lift that is accelerating downwards at 1m/s², a spring balance shows the apparent weight of the body to be 2.5 kg wt. What would be the reading if the lift was:
a) at rest?
b) accelerating upwards at 2 m/s²

I can do part a) and got 2.784 kg wt, but i don't know how they got 3.35 kg wt for part b)
Can someone show me part b) ?
Thanks

[brightsky]

accelerating upwards at 2 ms^(-2) means there is a force of magnitude 2m acting downwards in addition to the weight force of the obect mg. so anwer is 2m + mg N (plug in the value of m you obtained for part 1).

[stephanieteddy]

Hello, just a quick question:

What exactly are we doing when we resolve vector a in the direction of vector b?

I know how to do it, just not sure why!

[monkeywantsabanana]

^ the same way i know how to use a^ but not know what i'm actually doing. I've got so many whys in spesh.