VCE Specialist 3/4 Question Thread!

[Dominatorrr]

If P(z) is a polynomial of degree 4 with all of its coefficients real with ai, bi (a,b = R) as two of its zeros, then the term that doesn't contain z is:

A. ab
B. a-b
C. a+b
D. a^3b^3
E. a^2b^2

[abd123]

E)

EDIT: sorry man latex doesn't work, don't know whats really going on with the latex, and solution doesn't look to clear without latex.

[abd123]

Kinematics - Rectilinear Motion.

An object is dropped from rest so that the acceleration due to air resistance is 0.2v where is the speed of the object. The acceleration due to gravity is g m/s^2

a) show that

b) Find in terms of

c) Find the , that is, the maximum veloctiy

d) Find the distance fallen after 5 seconds (to the nearest metre)

[kamil9876]

Lol @ "acceleration due to air resistance" whose phrase is this? I'm sure the better way of saying it is the air resistance is upwards where is the mass of the object. In that case the total force is Gravity - Air Resistance = mg-0.2vm Hence the acceleration is by Newton's ith law . (note downwards is taken as positive direction)

b) just use the "fact" that and if I remember my calculus correctly (I havn't done that in a while, too busy with maths) you get something with log.

[dc302]

^ Of course calculus isn't maths

[abd123]

Lol @ "acceleration due to air resistance" whose phrase is this? I'm sure the better way of saying it is the air resistance is upwards where is the mass of the object. In that case the total force is Gravity - Air Resistance = mg-0.2vm Hence the acceleration is by Newton's ith law . (note downwards is taken as positive direction)

b) just use the "fact" that and if I remember my calculus correctly (I havn't done that in a while, too busy with maths) you get something with log.

Sorry its meant to be acceleration due to gravity, I'm sorry for not checking it twice.

Anyways Thank you very much Kamil .

[samad]

Kinematics - Rectilinear Motion.

An object is dropped from rest so that the acceleration due to air resistance is 0.2v where is the speed of the object. The acceleration due to gravity is g m/s^2

a) show that

b) Find in terms of

c) Find the , that is, the maximum veloctiy

d) Find the distance fallen after 5 seconds (to the nearest metre)

I think the shortcut to finding terminal velocity is calculating v when acceleration is zero (this can be conceptualised by looking at the velocity- time graph for a falling object):
when terminal velocity is reached, the gradient, and hence the acceleration, is zero.

so: a=g-0.2v
let a=0
g-0.2v=0
therefore terminal velocity=v=g/0.2

check if this is right using calculus 

[paulsterio]

This outflow/inflow question is troublesome for me :/.

Q9. A tank containing 600 litres of water has a sugar solution of concentration 0.1 kg/L pumped in at a rate of 10 L/min. if the mixture is kept uniform and is pumped out at a rate of 8 L/min, set up a differential equation for the amount of sugar, x kg, at any time t. (Do not attempt to solve).

b^3 has provided an explanation already, but I'll add mine because I have a slightly different way of doing it

dx/dt = rate of change of amount with respect to time (kg/min)

Volume at time t = V(t) = 600 + 10t - 8t = 600 + 2t

Rate of change of x (in)= Concentration (kg/L) x Flowing In Rate (L/min) = 0.1 x 10 = 1 kg/min

Rate of change of x (out) = Concentration (kg/L) x Flowing Out Rate (L/min) = x/v(t) * 8 = (8x) / (600 + 2t) = (4x) / (300 + t)

Now you find dx/dt = Rate of change of x (in) - Rate of change of x (out) =

Generally I use this method to do it because it doesn't need formulae (I hate formulae =.= it's more to remember to be honest) - you kind of use the units to guide you

[b^3]

This outflow/inflow question is troublesome for me :/.

Q9. A tank containing 600 litres of water has a sugar solution of concentration 0.1 kg/L pumped in at a rate of 10 L/min. if the mixture is kept uniform and is pumped out at a rate of 8 L/min, set up a differential equation for the amount of sugar, x kg, at any time t. (Do not attempt to solve).

b^3 has provided an explanation already, but I'll add mine because I have a slightly different way of doing it

dx/dt = rate of change of amount with respect to time (kg/min)

Volume at time t = V(t) = 600 + 10t - 8t = 600 + 2t

Rate of change of x (in)= Concentration (kg/L) x Flowing In Rate (L/min) = 0.1 x 10 = 1 kg/min

Rate of change of x (out) = Concentration (kg/L) x Flowing Out Rate (L/min) = x/v(t) * 8 = (8x) / (600 + 2t) = (4x) / (300 + t)

Now you find dx/dt = Rate of change of x (in) - Rate of change of x (out) =

Generally I use this method to do it because it doesn't need formulae (I hate formulae =.= it's more to remember to be honest) - you kind of use the units to guide you

It's still basically the same way, you've worked out the inflow of x per second and the outflow of x per second

EDIT: Explaining the equation
The first part of the equation is giving how much of "x" is following in per second.
The second part of the equation is giving how much of "x" is following out per second.

Its the same thing, just that you do and , this is the way that I did it all the time when there wasn't a changing volume, as it is easier to just remember flow_in-flow_out.

[abd123]

Thanks Paulsterio , I'd stick to more of b^3 workings. I finished differential equations and im almost finishing off kinematics and im heading to vector function/vector calculus.

Thanks again Paulsterio, would look back again soon .

[dc302]

Yeah what b^3 and paulsterio did were the same thing if you ask me

[Dominatorrr]

Q(z) = z^3-3iz^2-3z+4+i

Determine the values of a=C, b=R if Q(z) is of the form Q(z) = (z-a)^3+b

[dc302]

Expand (z-a)^3+b out to get:

z^3 - 3z^2a + 3za^2 - a^3 + b

Equate coefficients.

a = i,b = 4

[abd123]

Question c of Q17 Maths Quest right ?


c)



Equating with

[Dominatorrr]

Thanks to you both and yes Q17c Maths Quest