A few Specialist Problems

[GerrySly]

Alright got a couple questions after doing a couple practice exams...

I am constantly getting implicit differentiation incorrect. It seems to be the correct equation but not in their form, how flexible are they with this? E.g.



I did this and came up with and yet there answer is . This has happened more often than not but for the rest of the question I get the correct gradient so it's the correct form. I am just not sure how I can go further than what I have.

Also in reading time what are allowed to use? Calculators, reference material?

Cheers

[d0minicz]

i got the same as you lol
used quotient rule instead of their one?

[TrueTears]

I remember a similar problem /0 brought up [actually I think it's the same]

They should accept both forms, but to get their form use the product rule.

[GerrySly]

I remember a similar problem /0 brought up [actually I think it's the same]

They should accept both forms, but to get their form use the product rule.

Heh righto, I used the product rule to get that ()

[Mao]

(Image removed from quote.)

I did this and came up with and yet there answer is

[GerrySly]

2007 VCAA EXAM 2 MC Q 19
A particle is initially travelling at 5 m/s. A constant force of 18 newtons is applied to the particle, in the direction of motion, for 4 seconds. The particle accelerates at .
The momentum of the particle after 4 seconds is...

2007 VCAA EXAM 2 MC Q 21
A 12kg mass moves in a straight line under the action of a variable force F, so that is velocity when is is x metres from the origin is given by .
The force F acting on the mass is given by...

2007 VCAA EXAM 2 SA Q 4c


Just did 2007 VCAA Exam 2 as you can see, those were the questions I couldn't solve even after looking at the assessor's report. I always seem to have trouble with the style of questions as the last one (landing and angles) so if you could explain it generally it would be helpful

Thanks a bunch

[TonyHem]

Question 19)
F = mA
18 = M(1.5)
M = 12 kg

P = mv
Find v at t =4
u = 5, v =? t = 4, a = 1.5
v = u + at
v = 5 + 6
v = 11
p = 12*11 = 132

21) v^2 = 3x^2-x^3+16
(1/2)v^2 = 1/2(3x^2-x^3+16)
Derive it, to get a = whatever
then F = MA
so times both sides by m = 12

as for 4C, I asked that question myself so I can't help :|

[GerrySly]

Ah right thanks for that TonyHem, I knew I had to do something with other forms of acceleration for 21 but didn't know how, thanks for clearing that up

[kamil9876]

I always seem to have trouble with the style of questions as the last one (landing and angles) so if you could explain it generally it would be helpful

There's a good discussion about it starting from this post, and goes till the next page too. If unsure by the details after reading this then feel free to ask specific things.

http://vcenotes.com/forum/index.php/topic,11962.msg174393.html#msg174393

[Mao]

An alternative solution to Q19)

(Newton's Second Law, see Essentials Textbook)

[Mao]

Q4c)

The angle of landing deals with the angle the vector path meets with the horizontal plane. The angle is the angle between the vector tangent to the path and the horizontal, i.e. between the velocity vector and the horizontal.

The horizontal vector must point in the same direction as velocity vector, hence

, the rest is fairly trivial.

[GerrySly]

Q4c)

The angle of landing deals with the angle the vector path meets with the horizontal plane. The angle is the angle between the vector tangent to the path and the horizontal, i.e. between the velocity vector and the horizontal.

The horizontal vector must point in the same direction as velocity vector, hence

, the rest is fairly trivial.

Ah right, it's the velocity vector. The vector, , you just used the velocity vector but removed the k component yeah? Can I do that for every question like these? (Velocity vector).(velocity vector minus k component) etc. ?

[kamil9876]

That's assuming the "ground" is the x-y plane. What if the ground was the x-y plane tilted say ?  :coolsmiley:

[GerrySly]

That's assuming the "ground" is the x-y plane. What if the ground was the x-y plane tilted say ?  :coolsmiley:

Interesting, I assume you could just right angle triangle that shit and go j component is [tex]40\cdot \sin{(30^o)}=40\cdot \frac{1}{2}=20/tex]?

[kamil9876]

Ahh just realised the solution may not be as simple as I thought. Kinda post without thinking :/

Another kool way of doing the original version is to find the angle from taking . This angle is the angle between velocity and vector. A little geometry tells you that the required angle is minus this. In the new example, the vector taken from tilting 30 degrees acts like the new . Yeah... as i said not that as simple as i thought it could be :/. But at least you get the picture as to why the dot product is taken and not simply the tangent of the magnitudes of the i and k component.