VCE Specialist 3/4 Question Thread!

[M_BONG]

When two particles coincide, does it mean they collide or they cross paths?

As far as I understand, collide means same position at the same time whereas cross paths means same position, not necessarily same time.

So is coincide collide or cross path?
Thanks!

[lzxnl]

When two particles coincide, does it mean they collide or they cross paths?

As far as I understand, collide means same position at the same time whereas cross paths means same position, not necessarily same time.

So is coincide collide or cross path?
Thanks!

I would say collide

[Cort]

Right. I'm really struggling to finish this projectile motion chapter right now. IN the Mathsquest Textbook.

I can understand how they got the height, but why on earth is it equal to (V^2*sin^2)/2*g? It's not equal to the range equation.

I can understand the basics..but this..just what.

[Phenomenol]

Right. I'm really struggling to finish this projectile motion chapter right now. IN the Mathsquest Textbook.

I can understand how they got the height, but why on earth is it equal to (V^2*sin^2)/2*g? It's not equal to the range equation.

I can understand the basics..but this..just what.

Okay, so I'm not entirely sure about what you mean by "how they got the height". But what you've given is a rearrangement of the equation you need to use to answer a:
v^2 = u^2 + 2as

Subbing in u = Vsin(35) (initial vertical component of velocity), v = 0, a = - g, you get s = the expression you've provided. As you know that s = (8 - 1.5) = 6.5 (displacement) in this case, you can find the initial speed V.

If you're talking about the range equation that I think you're talking about, you can't use it in this case as your initial and final positions are not equal in height.

[tiff_tiff]

can someone please help???


Two compounds A and B are put together in a solution where they react to form a compound X . To form
compound X , twice as much compound A is needed as compound B . The rate at which X is formed is
proportional to the unused amounts of A and B present at time t. If initially 5 grams of A and 8 grams of B are
put together, set up a differential equation for the amount of X present at time t, assuming that no X is present
initially.

[lzxnl]

Gotta love chemical kinetics

I'm going to let the remaining amount of A at any time be a
And that of B to be b
I think there's an implicit assumption that the chemical equation is 2A + B => X

So, let the amount of X be x.
dx/dt = k(ab)
But at the same time, for every two amounts of A used up, one amount of X is formed => a = 5 - x/2. Think about that. Initial amount is 5 and the amount of X formed is x
Similarly, b = 8 - x
So dx/dt = k(5 - x/2)(8 - x)

[Brunette15]

Can someone please help me understand how to tackle this question? 

[tlbdrummerkid]

Differentiate r to find v and equate i and j components then diff v and sub in the equations you got.

[Conic]

We have and .  So from this, we know that .
Using the definition of acceleration, it follows that

[allstar]

Two cars A and B, each moving with constant acceleration, are travelling in the same
direction along the parallel lanes of a divided road. When A passes B, the speeds are 64
and 48 km/h respectively. Three minutes later, B passes A, travelling at 96 km/h. Find the instant at which both are moving with the same speed, and the distance between them at this time

please help

[dcc]

Let A and B be the accelerations of A and B respectively, then after converting everything to metres and seconds, you have

a_b(t) = B
v_b(t) = (40/3) + Bt
x_b(t) = (40/3)t + (B/2)t^2

[40/3 m/s = 48 km/h]

a_a(t) = A
v_a(t) = (160/9) + At
x_a(t) = (160t/9) + (A/2)t^2

[160/9 m/s = 64 km/h]

Without loss of generality, I've taken the position of A and B
at t=0 to be 0.

-----

Since you know the speed of B at 3 minutes, you get the acceleration of B

v_b(180) = 96km/h = (80/3)m/s
= (40/3) + B*180
--> B = 2/27

-----

Since you know that A and B are at the same place at 3 minutes, you can also get
the acceleration of A

x_a(180) = x_b(180), so
3600 = 3200 + 16200A
--> A = 2/81 = B/3

-----

If v_a(t) = v_b(t), then
t = (480-640)/(3*(A-B))
-> t = 90 seconds

At which point the distance between them is

d = |x_b(90)-x_a(90)|
  = 200 metres

[allstar]

What does the a_b (t) mean?

[keltingmeith]

What does the a_b (t) mean?

.
If written normally (i.e. without LaTeX), people write it as you see above. Doesn't mean anything, really, just a method of labelling different variables. For example, could refer to the acceleration of particle b.

[IndefatigableLover]

Would anyone know what is meant by 'Functions of a Complex Variable and Euler's Formula' just out of curiosity?

[kinslayer]

Would anyone know what is meant by 'Functions of a Complex Variable and Euler's Formula' just out of curiosity?

Functions of a complex variable are functions that take complex numbers as arguments. It is the main area of study of Complex Analysis:

http://en.wikipedia.org/wiki/Complex_analysis

Euler's formula is