VCE Specialist 3/4 Question Thread!

[gnaf]

Haha woops typical me 

[Shadowxo]

First photo:
The convention is, negative is to the left. So if a particle is to the left of the origin, then it's displacement is negative. If it's travelling to the left, its velocity is negative. If it's accelerating to the left it's negative acceleration.
Second: first you'd find the distance travelled by s=vt where s is the distance travelled (you can use common sense for this formula as well, just speed * distance with a direction). Then use a physics formula to find the acceleration. You know u = 4m/s, s= -48 from earlier (as it travels left to get back to the origin), t=20 s, and a is unknown. Use s=ut + 1/2 at². You should end up with a negative number as acceleration is in the opposite direction
12. a) easiest way to do this is using the information about car B. First you have to convert them to the same units. Metres and seconds are usually the best. To convert k/h to m/s divide by 3.6
U (carB) = 48/3.6 m/s
V = 96/3.6 m/s
T = 3*60 s
Then you want to find s (aka distance / displacement). There should be a formula for this but if it's not (there's one formula that isn't used much) then find acceleration first to solve it.
b) first you'd want to find the acceleration of each. B is easy as outlined above but A you'd want to use
U(carA) = 64/3.6 m/s
S = answer from a), distance travelled is same for both of them
T = 3*60 s
Solve for a using this
13. It should be accelerating downward at a decreasing rate (that is, the velocity approaches a constant value k, as a approaches 0). The particle should be speeding up. They've said that k<0 so a<0. So the velocity should be "decreasing" or speeding up in the downwards direction. They may be calling it "deceleration" as the acceleration is negative.

Hope this helps

[gnaf]

Thanks so much Shadowxo Always to the rescue 

How do you a) in the first photo? I don't understand what the textbook is saying at all...

How did the textbook reach the conclusion that y=2 is also an answer in the 2nd photo?

[Shadowxo]

1. Since there is 2L going out of the container and 1L going in, it means overall the volume of the container is decreasing by 1L per min
So V = 30 -1t (initial volume - 1L/min  * min)
Concentration of the substance = amount / volume
= x/ (30-t)
There are 2 L going out of the container with the substance. The 1L coming in is clean water so doesn't have any substance coming in
So substance leaving per min= concentration * volume
= x/(30-t) * 2
So dx/dt = -2x / (30-t) as the change in amount per time is just the amount leaving per time as nothing is coming in
2. I'm not sure what they mean by that example but solving dy/dx=0 would give y=2 as a solution, and for solving equations in general you have to make sure you don't divide by a solution (eg x^2 + x=0 you could make into x+1=0 giving x=-1 but x=0 is also a solution.

Hope this helps! (Was written on my phone so sorry for the layout)

[gnaf]

I wish I could give you more upvotes Shadowxo! Honestly indebted to you

Thank you so much

[humblepie]

Would really appreciate it if someone could help me with the attached question!

[VanillaRice]

Would really appreciate it if someone could help me with the attached question!

Let r=xi+yj
Substitute r and c into the equation to get a Cartesian equation in terms of a and b.
Note: you may have to complete the square for both x and y to get an equation which looks more familiar

Hope this helps

[humblepie]

Thanks VanillaRice!

[tinagranger]

Hi! P544 24) second part

1st part was 'What force is necessary to accelerate a train of mass 200 tonnes at 0.2 m/s squared against a resistance of 20,000N?'
The bit I don't get is: What will be the acceleration if the train free-wheels against the same resistance?
I have no idea what it means by 'free-wheeling' and how to start!

[sonnyangel]

Hi! P544 24) second part

1st part was 'What force is necessary to accelerate a train of mass 200 tonnes at 0.2 m/s squared against a resistance of 20,000N?'
The bit I don't get is: What will be the acceleration if the train free-wheels against the same resistance?
I have no idea what it means by 'free-wheeling' and how to start!

free-wheeling is kind of letting the vehicle do its thing and not exerting a force to help it move (e.g. standing on a moving skateboard but you're not using your foot to propel it forward, another example is riding a bike down a hill and not pushing the pedals)

this means that there is no force helping the train move forward
---> the only horizontal force acting on the train is the resistance of 20000
acceleration is force/mass
---> a= -20000N/200000kg
         = -0.1m/s/s

hope this makes sense! 

[tinagranger]

Thanks!

1. Also from what I understand, for dynamics, you should always simplify forces, acceleration, etc. to scalars instead of using tildas when writing F = ma or R - mg = ma. That is what my teacher said to do. But what happens when you get a negative answer?

Eg: If I get a (without a tilda, simplifying acceleration to a scalar) = -g/2, how do I give my answer?


2. Should you always define which direction you are taking as up?

3. P550 13b) The values I have are a = -0.3g, u = (2 root 10gx)/5, s = ? But I can't find t so I don't know how to put this info into a constant accel rule.

THanks!

[Shadowxo]

Thanks! Also from what I understand, for dynamics, you should always simplify forces, acceleration, etc. to scalars instead of using tildas when writing F = ma or R - mg = ma. That is what my teacher said to do. But what happens when you get a negative answer?

Eg: If I get a (without a tilda, simplifying acceleration to a scalar) = -g/2, how do I give my answer?


Also should you always define which direction you are taking as up?

THanks!

Hi
I agree you should always simplify to scalars to make it easier to work out.
A negative just indicates the direction. Usually it means left or down.
If you get -g/2, you can usually just say a=-g/2 but if you're unsure you can ask your teacher or say "g/2 m/s² down" for example. You need to have some indication of direction, whether it be with a negative sign or written direction.
I always referred to which direction was positive. Doesn't take long to write "taking up as positive, ..."

[Sine]

Just to add - if you ever feel like your answer isn't clear enough you are always allowed to write a sentence clearly stating your final answer. This is especially useful when you aren't 100% sure about the mathemtaical notation required also this way the assesor won't ever hesitate about giving you the marks.

Thanks!

1. Also from what I understand, for dynamics, you should always simplify forces, acceleration, etc. to scalars instead of using tildas when writing F = ma or R - mg = ma. That is what my teacher said to do. But what happens when you get a negative answer?

Eg: If I get a (without a tilda, simplifying acceleration to a scalar) = -g/2, how do I give my answer?


2. Should you always define which direction you are taking as up?

3. P550 13b) The values I have are a = -0.3g, u = (2 root 10gx)/5, s = ? But I can't find t so I don't know how to put this info into a constant accel rule.

THanks!

3. could you post a screenshot or photo of the question?

[tinagranger]

Ok so I would just say a = -g/2 without the tilda, and they couldn't take off marks?

Here is my 3rd question! (13b in the screenshot)

Also for 15b) in my 2nd screenshot, I can't seem to get the right answer - I keep getting a negative coefficient for friction.
The equations of motion I ended up with were
1. 2.52 - T = 8.4 (because 4.2 sin theta = 4.2 x 0.6 = 2.52)
2. T - mew 3 g = 6

[Sine]

Ok so I would just say a = -g/2 without the tilda, and they couldn't take off marks?

Here is my 3rd question! (13b in the screenshot)

Also for 15b) in my 2nd screenshot, I can't seem to get the right answer - I keep getting a negative coefficient for friction.
The equations of motion I ended up with were
1. 2.52 - T = 8.4 (because 4.2 sin theta = 4.2 x 0.6 = 2.52)
2. T - mew 3 g = 6

yes, I actually never used tildes for dynamics question - unless it was explicitly state in the question (which was not often)

So for 13(b) you actually have 3 variables a = -0.3g, u = (2 root 10gx)/5 and a hidden v = 0 because the question want the distance travelled. This assumes that the particle actually stop -hence a final velocity of 0m/s.

hmm 15b one of the equation should be
0.6 * 4.2 * g - T = 8.4     (1)
the second is fine
T - μ3g = 6             (2)

From 1) T = 0.6 * 4.2 * g - 8.4       => T = 16.296
now sub into (2)
16.296 - μ3g = 6
16.296-6 = μ3g
μ = 10.296/(3g)     =>  μ = 0.35