Bazza's 3/4 Question Thread

[WhoTookMyUsername]

given that x=(t-2)^2 (x = displacement)

find the distance travelled from t=0 to t= 2

for something (or a bit more complex, with something that's tougher to differentiate, but the question gives no indication of TP) similar to this, where either a) it's obvious there's no TP within the specified time period, or b) where it's "implied" there's no TP (all the TP quesitons in the textbook lead you through them) do you have to verify it and say there is a turning point only at t = 2, therefore distance traveled is equal to abs(displacement)?
Or is this an extra unnecessary step?

2) given that displacement = -(t-2)^2 + 3
determine when displacement is at a maximum

do we need to do the gradient sign test to verify that it is a maximum? (not so much a problem with this question, but more tricky questions where gradient sign tests can take quite a while)

[WhoTookMyUsername]

given that x=(t-2)^2 (x = displacement)

find the distance travelled from t=0 to t= 2

for something (or a bit more complex, with something that's tougher to differentiate, but the question gives no indication of TP) similar to this, where either a) it's obvious there's no TP within the specified time period, or b) where it's "implied" there's no TP (all the TP quesitons in the textbook lead you through them) do you have to verify it and say there is a turning point only at t = 2, therefore distance traveled is equal to abs(displacement)?
Or is this an extra unnecessary step?

2) given that displacement = -(t-2)^2 + 3
determine when displacement is at a maximum

do we need to do the gradient sign test to verify the TP is a maximum (or velocity = 0) (after differentiation) ? (not so much a problem with this question, but more tricky questions where gradient sign tests can take quite a while)

anyone know ?

thanks!

[Jenny_2108]

why does your class study so fast? Mine just finished Euler's rule

[Somye]

why does your class study so fast? Mine just finished Euler's rule

Yeah that's where ours is up to as well (Bazza's as well), I'm assuming he's going ahead himself though...

[HenryP]

given that x=(t-2)^2 (x = displacement)

find the distance travelled from t=0 to t= 2

for something (or a bit more complex, with something that's tougher to differentiate, but the question gives no indication of TP) similar to this, where either a) it's obvious there's no TP within the specified time period, or b) where it's "implied" there's no TP (all the TP quesitons in the textbook lead you through them) do you have to verify it and say there is a turning point only at t = 2, therefore distance traveled is equal to abs(displacement)?
Or is this an extra unnecessary step?

2) given that displacement = -(t-2)^2 + 3
determine when displacement is at a maximum

do we need to do the gradient sign test to verify that it is a maximum? (not so much a problem with this question, but more tricky questions where gradient sign tests can take quite a while)

In my opinion, when finding distance travelled its always good to verify that there is no turning point within the selected time period so that you can justify that distance travelled=abs(displacement). This may be just me being too careful but I guess its better to be safe than sorry.
As for your second question, I would definitely verify whether the turning point was a maximum or minimum.
If you're unsure whether to do these steps, take a look at how many marks its worth. If its only worth one, you wouldn't need to do any of these extra verification steps.

[WhoTookMyUsername]

thanks for the help

I'm hugely struggling with vectors ATM (trying to go a bit ahead of class)

1) A Ball, just before it hits the ground, has the position vector (where i, j, k are the unit vectors).
At what angle does the ball hit the ground?

2) What's the convention for x, y, z (axes) which 2 are the "ground"? If "z" represents upwards, but "y" represents "upwards" in a 2 D plane, how does this work :S? Does that mean 2D and 3D questions must be considered completely separately in terms of axes?

3) If somehow you use that dot product rule to find the angle, wouldn't letting the other vector = z, or 2z change the angle? (when it's the same angle created)

4) do you lose a mark if you forget the tilde?


I had absolutely no idea how to approach this and still don't know what's going on even after looking at WS

thanks

[b^3]

1) For the angle it hits the ground at you need to consider the velocity vector, that is the direction that it is heading in. So that is .
Now consider the triangle this vector makes with the ground. Now to get the adjacent edge of this triangle, you will need to use pythagoras of the triangle the the component and component make, that is

Then to find the angle we would have


2) For 2D planes you normally take y as upwards and x as across. For 3D, the x-y plane becomes the ground and the z axis becomes upwards. So it depends on the question really.

3) If the vector is parallel, then changing the vector from , to , won't change the angle, try an example, when you do the dot product it will be different, but so will the magnitude that you are dividing by, so it ends up being the same. e.g. Try finding the angle between and then try finding the angle between and using the dot product.

4) We were told that you would lose if for forgetting it as if you don't then you are not denoting a vector. Having a look at past assesors reports they point out that it  is a common problem.

Anyways hope that helps, and is right, I'm a little rusty on this.

[WhoTookMyUsername]

thanks a lot b^3

the textbook WS had invtan(3/sqrt(14)) (they found the 37 degree angle first)
is this coming from the dot product rule where they've taken the second vector as z?
Do they prefer if you use this method? (as it is more "vectorish")

[b^3]

thanks a lot b^3

the textbook WS had invtan(3/sqrt(14)) (they found the 37 degree angle first)
is this coming from the dot product rule where they've taken the second vector as z?
Do they prefer if you use this method? (as it is more "vectorish")

Oh right, yeh they used vectors, they used the velocity vector and the vector tht is straight up, that is they used and , which would give you the other angle, so you would have to minus it from 90 degrees. As to which one is preferred, I always did it the first way, but probably should stick to vectors, but others might have more info on which one is preferred.

EDIT: I checked back with a past exam (not VCAA) and they gave two methods, the original one I used and a vector method that used the dot product of the vel vector and the vector that was in the x-y plane that is in line with the vel vector (for this situation that would be ), all the methods work and should be valid.

[WhoTookMyUsername]

If you have two masses either side of a frictionless pulley ( hanging by a rope) of equal mass

Why is the tension in the string = mg
And not 2mg?

Struggling to understand this concept (what truly "is" tension)

Thanks!

[nina_rox]

Sorry, I have a quick question.
I'm getting confused with signed and unsigned graphs. So for a velocity time graph is the distance the whole area under the graph (disregarding the negative areas and calculating them as positive - so unsigned) while the displacement regards the negative area (so signed)?
Thank you very much

[paulsterio]

Sorry, I have a quick question.
I'm getting confused with signed and unsigned graphs. So for a velocity time graph is the distance the whole area under the graph (disregarding the negative areas and calculating them as positive - so unsigned) while the displacement regards the negative area (so signed)?
Thank you very much

That's correct

[WhoTookMyUsername]

If you have two masses either side of a frictionless pulley ( hanging by a rope) of equal mass

Why is the tension in the string = mg
And not 2mg?

Struggling to understand this concept (what truly "is" tension)

Thanks!

[paulsterio]

If you have two masses either side of a frictionless pulley ( hanging by a rope) of equal mass

Why is the tension in the string = mg
And not 2mg?

Struggling to understand this concept (what truly "is" tension)

Thanks!

Because each mass is only mg

Think about it this way. If you hold a rope, in your hand, and hang a block of m on it, the tension in the rope will be mg

Easiest way to think of tension in this regard is that it's sorta like the normal reaction force - it stops the blocks falling to the ground.

[WhoTookMyUsername]

but isn't there a force of mg (tension) acting on both sides of the rope?
Should you add these together of is tension a force / unit area?