VCE Specialist 3/4 Question Thread!

[Lavar Big BBB Balls]

Hi,

For Q8e) http://imgur.com/a/1oJAi

I'm not sure if the answer is correct? They seem to be ignoring the first 10 seconds? Is my answer correct?

Thanks

[Shadowxo]

Hi,

For Q8e) http://imgur.com/a/1oJAi

I'm not sure if the answer is correct? They seem to be ignoring the first 10 seconds? Is my answer correct?

Thanks

I got the same answer as the book

Distance travelled by motorist is simply 25*t
v₁ (max speed) = 601/20 from a)
The distance the policeman has travelled after 10 seconds is the integral of v_p between t=4 and t=10, which should be 903/10 (should be the answer for d)i )
So distance travelled by policeman = 903/10 + v1*(t-10) which is the distance travelled in the first 10 secs + distance travelled after first 10 secs
Solving 25t = 903/10 + 601/20 * (t-10) gives the book's answer

Let me know if you'd like further clarification

[geminii]

If there is 10g of salt in a solution with volume 200cm^3, how do I find the concentration?

Thanks in advance!!

[Shadowxo]

If there is 10g of salt in a solution with volume 200cm^3, how do I find the concentration?

Thanks in advance!!

Concentration is amount per volume.
So 10/200 = 1/20 g/cm³
You can convert this value into different units if you need g/L etc but usually things are given in similar units

[geminii]

Concentration is amount per weight.
So 10/200 = 1/20 g/cm³
You can convert this value into different units if you need g/L etc but usually things are given in similar units

Thank you!! So if I wanted to convert to g/L I would multiply it by 1000?

[Shadowxo]

Thank you!! So if I wanted to convert to g/L I would multiply it by 1000?

Yep!

[gnaf]

Hi

In vector calculus, one of the questions is to find the cartesian equation of r(t)=2i+5tj

I get x=2 and y=5t

The final cartesian equation is x=2. Why is y=5t rejected?

[gnaf]

Hey

In the first photo attached, what approach do you use for all these questions? Is there a method I can follow?

In the second photo, I don't understand how they graphed the points highlighted in green on the 3D graph. For the first point, x=1. Doesn't this mean it should be 1st or 4th quadrant (not the 2nd quadrant it's currently in)?

Thanks

[VanillaRice]

Hi

In vector calculus, one of the questions is to find the cartesian equation of r(t)=2i+5tj

I get x=2 and y=5t

The final cartesian equation is x=2. Why is y=5t rejected?

The i component (x=2) implies x is ALWAYS zero I.e. It is a vertical line crossing through, x=2. Note for y=5t (assuming t can be any real number), y will be able to take any real value (think about it as a linear graph for t vs y, if dom is all real numbers, then the range is also all real). Therefore, the y=5t component is not actually being rejected, but is implied i.e. the equation is x=2 with an implied range of y for all real numbers.

Hey

In the first photo attached, what approach do you use for all these questions? Is there a method I can follow?

In the second photo, I don't understand how they graphed the points highlighted in green on the 3D graph. For the first point, x=1. Doesn't this mean it should be 1st or 4th quadrant (not the 2nd quadrant it's currently in)?

Thanks

1) This is probably not going to be a typical question you'd see often (if at all) on a SAC/exam, since there are multiple different answers. I think the best way to go about them is practise with a variety of questions asking you to go in the other direction, just so you have an idea of where to go. So really, no hard and fast rule (for me at least). If you need help with any particular question, let us know

2) Which plane are you referring to (x-y, x-z, etc.)? Which quadrant you're referring to will depend on the plane in question.

Hope this helps

[Shadowxo]

Hi

In vector calculus, one of the questions is to find the cartesian equation of r(t)=2i+5tj

I get x=2 and y=5t

The final cartesian equation is x=2. Why is y=5t rejected?

Great answer by VanillaRice. For your first question, x is always 2 and y depends on t, meaning depending on the value of t, y can change. If you think of this as a graph, x is always 2 while y makes the point go up and down, making the line x=2, as y can be anything (depending on domain of t)
for the second photo, you have x,y,z going in different directions. The x axis isn't shown like it is with 2D graphs, instead it's coming towards you at an angle. So to find x=1 you have to follow that line in the positive direction until you reach 1. I don't think you really need to address quadrants in 3D graphs so don't think about assigning it a quadrant, instead just find where it is.

[gnaf]

Thanks so much Shadowxo and Vanilla Rice

Just clarifying a few things-

Since y=5t means that y can be any real number, we only have to include x=2 in our ans?

Plus the question is asking for a cartesian equation but y=5t isn't one since it's not in terms of y and x?

For the 3rd question, I was confused about the x-values i.e.the point with x=1 is on the left of the point with x=-1. Shouldn't the point with x=1 be on the right of the point with x=-1?

[VanillaRice]

Thanks so much Shadowxo and Vanilla Rice

Just clarifying a few things-

Since y=5t means that y can be any real number, we only have to include x=2 in our ans?

Plus the question is asking for a cartesian equation but y=5t isn't one since it's not in terms of y and x?

For the 3rd question, I was confused about the x-values i.e.the point with x=1 is on the left of the point with x=-1. Shouldn't the point with x=1 be on the right of the point with x=-1?

1) Unless the question asks you specify a range, or fully define the function, or something of that description, I would say x=2 is an acceptable answer. However, there's no harm in writing that y is valid for the real field.

2) I think that by simply saying x=2, this will imply that y values included. A Cartesian equation describes the x and y coordinates of a curve, and I would argue that x=2 adequately describes both. Like above, there's no harm in defining your y values to be all real as well.

3) This depiction is using what we refer to as a 'right hand coordinate system', where the positive direction of the x-axis is pointing towards you. This also means the negative direction (i.e. x<0) will be away from you.

Hope this clarifies things 

[Shadowxo]

Thanks so much Shadowxo and Vanilla Rice

Just clarifying a few things-

Since y=5t means that y can be any real number, we only have to include x=2 in our ans?

Plus the question is asking for a cartesian equation but y=5t isn't one since it's not in terms of y and x?

For the 3rd question, I was confused about the x-values i.e.the point with x=1 is on the left of the point with x=-1. Shouldn't the point with x=1 be on the right of the point with x=-1?

1) Yes you only have to include x=2 in your answer. If t > 0 for example though, you'd have to mention that y>0
2) that's right- Cartesian equations don't have t, parametric equations have t. So to convert parametric equations to a Cartesian equation you have to get rid of the t
3) it's just the way 3D graphs are drawn. I think it should be different but that's the way it's done, even though it's counterintuitive, x=1 is to the "left" of x=-1. In reality though, it's just at a different point along the x axis

Hope this clears some things up for you

Edit: VanillaRice beat me but no harm having two responses!

[gnaf]

Thanks so much Shadowxo and VanillaRice- nothing beats your explanations 

In the 1st pic, why is the acceleration always to the left?

How do you do the question in the 2nd pic?

In the 3rd pic, how do you do question 12b)? And shouldn't the answer to Q13b) be accelerating in the opposite direction? the answer says it's decelerating...

[Shadowxo]

Thanks so much Shadowxo and VanillaRice- nothing beats your explanations 

In the 1st pic, why is the acceleration always to the left?

How do you do the question in the 2nd pic?

In the 3rd pic, how do you do question 12b)? And shouldn't the answer to Q13b) be accelerating in the opposite direction? the answer says it's decelerating...

I think you forgot to link the images