Hutchoo's spesh thread.

[Hutchoo]

Hey guys, I don't get some of the following questions.

For the bullet q, I know how to do a), I just don't get the 'thickness' part :/

I don't know how to do the golfball q (it's a simple one, but I don't get it).

For the helicopter q, I keep getting it wrong. This is probably the most annoying one because I know it's super easy.

And I don't understand (c) from the last attachment.

Thanks in advance!

[BubbleWrapMan]

The thickness of the soil is the area under the graph during that time interval (because the distance travelled by the bullet equals the thickness of the soil).

For the golf question, it gives you a = -2 and u = 8 so you need to use v^2 = u^2 + 2as, keeping in mind v = 0 at the end.

For the helicopter, the average velocity should be (h(2) - h(0))/(2 - 0), is that how you did it?

For the last question, the magnitude of displacement is equal to the displacement because it is never negative. So it's asking you for the maximum value of x for 0 ≤ t < 3

[Hutchoo]

Ohhhhhhhhh, I see!
I interpreted the first question wrong. I didn't realise that the thickness = distance travelled x_x (idiotic, I know).
Turns out for the helicopter one, I had made a calculation error (bad mental arithmetic, gotta work on speed + mental arithmetic)

And for the golf ball one, I didn't realise that velocity decreasing = -a. Is that true for all cases?

Btw, I've seen your work on other peoples threads you're a cool guy/girl =)
Thanks for the help

[BubbleWrapMan]

For straight line motion, if the magnitude of velocity (the speed) is decreasing then the acceleration has an opposite sign to velocity. So for example if an object's velocity changed from -8 to 0 then it would have had a positive acceleration (assuming it was constant), because that's in the opposite direction to what the velocity was. For the golf question you should get the same answer if you let u = -8 and a = 2 (but your distance will have the opposite sign).

If you draw a diagram with arrows (since acceleration and velocity are vectors) it should make more sense.

[golden]

Ohhhhhhhhh, I see!
I interpreted the first question wrong. I didn't realise that the thickness = distance travelled x_x (idiotic, I know).
Turns out for the helicopter one, I had made a calculation error (bad mental arithmetic, gotta work on speed + mental arithmetic)

And for the golf ball one, I didn't realise that velocity decreasing = -a. Is that true for all cases?

Btw, I've seen your work on other peoples threads you're a cool guy/girl =)
Thanks for the help

And dominated in all three mathematics!

[Hutchoo]

I found these questions to be confusing.

[BubbleWrapMan]

1) Use v^2 = u^2 + 2as, the unknown being a (you have s = 0.1 m, u = 200 m/s, v = 0 m/s). Then use the equation again with the value of a you found to find v after a distance of 0.05 m.

2) 400 kg wt = 400g

T - 320g = 320a

The maximum tension T is 400g so we sub that into the above equation for maximum acceleration

400g - 320g = 320a

3) Let's say after 20 s, the train has travelled 0.5 km = 500 m, then after 20 + 30 = 50 s, it has travelled 1000 m. Consider the initial time and distance to both be 0. Use the equation s = ut + 1/2at^2. Your unknowns are u and a, which you can find since you have two sets of values for s and t (i.e. two equations). Once you have u and a, use v^2 = u^2 + 2as to find s, then subtract 1000 m.

[Hutchoo]

You're awesome, ClimbTooHigh.
It has been a while since i've done these sort of questions:

Solve for z:
Thanks

[b^3]

Haven't done this in a while (or used cis in a while) so I might have dragged this out and overcomplicated it a bit (maybe), but anyway.

[Jenny_2108]

You're awesome, ClimbTooHigh.
It has been a while since i've done these sort of questions:

Solve for z:
Thanks

I know b^3 already solved it. I just give you another method











For

For



You use quadratic formula, then you obtain

[Hutchoo]

2006 spesh exam 1 - VCAA.
Question 4b.
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2006specmath1-w.pdf

When I did this question, I thought of it in terms of vector components, like making a triangle and adding them together with the use of pythag.

I don't think this method is right, but it does work for the question:
I constructed a triangle (look at the attachment) and as you can see, it yields the wrong answer.
Now, my question is this:

1) Is this method actually a valid one?
2) If so, how do you know which ways to put the vector triangle together?

As you can see from the attachment, the first method I did was wrong. My 2nd method (after revising the exam) is right though. How can I make sure that I'm putting the correct values for the triangle?


I hope this question makes sense :/

[Bhootnike]

2006 spesh exam 1 - VCAA.
Question 4b.
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2006specmath1-w.pdf

When I did this question, I thought of it in terms of vector components, like making a triangle and adding them together with the use of pythag.

I don't think this method is right, but it does work for the question:
I constructed a triangle (look at the attachment) and as you can see, it yields the wrong answer.
Now, my question is this:

1) Is this method actually a valid one?
2) If so, how do you know which ways to put the vector triangle together?

As you can see from the attachment, the first method I did was wrong. My 2nd method (after revising the exam) is right though. How can I make sure that I'm putting the correct values for the triangle?


I hope this question makes sense :/

i did this exact question with my tutor yesterday....!
i dont know if your diagrams are right, mine was as attached.
using pythag , you can solve for T

[Bhootnike]

wait a second. your first diagram is right?





 

and thats the ans.

[Hutchoo]

I see. Yeah, I understand it now.
Thanks

Another question:


Question 8 - VCAA 2006 (same link http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2006specmath1-w.pdf)
I did a fob method, but I don't really understand why my fob method doesn't work.

I separated the fraction to be like this:


and then continued to integrate, which gives the wrong answer.

Why doesn't this method work?
Thanks

[Bhootnike]

I see. Yeah, I understand it now.
Thanks

Another question:


Question 8 - VCAA 2006 (same link http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2006specmath1-w.pdf)
I did a fob method, but I don't really understand why my fob method doesn't work.

I separated the fraction to be like this:


and then continued to integrate, which gives the wrong answer.

Why doesn't this method work?
Thanks

nearly.
you should have written it like this:


then you have a arcsin, and a easy substitution

i think you made a mistake in simplifying your integral, because i cant see how you can just get 2 by itself?