VCE Specialist 3/4 Question Thread!

[507]

it's uniform acceleration so first find the acceleration:

let P be the initial position, Q be the final position:
v^2 = u^2 + 2as
(40)^2 = (20)^2 + 2as
a = ((40)^2-(20)^2)/(2s)
a = 600/s

now, let the distance be s/2 (half the original distance), take this point as the final position, use v^2 = u^2 + 2as again
v^2 = (20)^2 + 2a(s/2)
v^2 = (20)^2 + 2(600/s)(s/2)
v^2 = (20)^2 + 600
v^2 = 1000
v = sqrt(1000) - positive direction as P to Q
v= 10sqrt(10) ms^-1

Thanks

[duhherro]

ahh another question guys!

Q6 of 1D from essentials . How in the ... ":/

[pi]

Fyi, that stuff isn't on the spesh course so I wouldn't worry

[Will T]

I'm calling the essentials out on Question 21 of the short answer chapter review chapter 1.

Can someone please confirm or deny that the answer is a typo on the part of the editor?

[Will T]

Also calling for a double check of essestnials chapter 1 extended response question 1d.

Pretty sure it's 11.93 and not 11.43km.

[BubbleWrapMan]

I'm calling the essentials out on Question 21 of the short answer chapter review chapter 1.

Can someone please confirm or deny that the answer is a typo on the part of the editor?

I believe that is a typo

Also calling for a double check of essestnials chapter 1 extended response question 1d.

Pretty sure it's 11.93 and not 11.43km.

That's what I get as well, I vaguely recall that it might have been a typo...

[Will T]

Thanks very much.

[Will T]

Essentials chapter review chapter 1 extended response 3d and 3e I guess.

I can only manage to find one of the circle's, namely: (x-2)^2 + (y-2)^2 = 4.

[Will T]

No worries guys, just got it then.

[b^3]

You would have gotten to
Now we know that the point lies on that curve, that is


That is we have


Typed it out, so posting anyways, it might help someone else.

[zvezda]

essentials text q6a in Ex1H, the answers have:

Im thinking typo, but I just want to make sure that i'm not wrong

[polar]

if you're dilating from the x-axis, you're 'pulling' the point towards or away from the x-axis, hence, you're changing the y value.

[zvezda]

sorry i probably should have specified. Rather than having "2t" it only has "t"

[polar]

ohh, no that's right, it asks for a possible pair, and have the same shape (path) but different values of to get each point.

[zvezda]

ah ok. thanks for that