Can I get some help with questions? (2 Viewers)

[jssicachoi]

1. From a point P, a person observes that the angle of elevation of the top of a cliff A is 40°. After walking 100 m towards A along a straight road inclined upwards at an angle of 15° to the horizontal, the angle of elevation of A is observed to be 50°. Find the vertical height of A above P

Could I please get a diagram along with the answer
I will add extra question on this thread

 

[jimmysmith560]

Not sure about the diagram though, sorry

 

[jssicachoi]

View attachment 30644

Not sure about the diagram though, sorry

all good, thanks

 

[jssicachoi]

Can I please get help with question a. I know how to solve it but i feel the working out will not make sense.

 

[jimmysmith560]

Would this help?

 

[jssicachoi]

could i know the diagram for this question?

 

[jimmysmith560]

I believe this is what it would look like

 

[CM_Tutor]

@jssicachoi , has the diagram drawn by @jimmysmith560 helped you to solve the problem?

 

[jssicachoi]

really easy trigonometry problem but i am dumb:

0° < x < 360°
sin x = sin y (find all possible values for x)

 

[jimmysmith560]

Probability density functions weren’t in the curriculum when I was in high school

I don't remember studying something like that in 2U at all either, maybe it's from 2020 and beyond.

 

[jimmysmith560]

I'm surprised this thread is still going. Hopefully, you've relearnt/bettered your understanding Hahahaha.

Well this definitely seems like a misconception that could be quite damaging if taken into OP’s trial and HSC exam (still, one cannot blame OP for being taught an incorrect formula). I’m glad to see the concept being clarified and the problem rectified though. This thread came about just in time and hopefully other students who were using OP’s original approach will become aware of the problems associated with it.

 

[jimmysmith560]

View attachment 31165

Would this help?

 

[jimmysmith560]

What is that symbol you are adding and subtracting onto 3, in the last 3 lines

That symbol is called epsilon and represents a variable, like x or y. I'm not entirely sure of the reason it's used here though.

 

[quickoats]

That symbol is called epsilon and represents a variable, like x or y. I'm not entirely sure of the reason it's used here though.

Epsilon is not a variable in the sense like x or y that we change - here you can think of it as a placeholder for a small increment. means a small nudge to the right (and we can observe that the acceleration is negative). In the HSC, you are probably more comfortable using notation like (it means the same thing for this purpose).

 

[jimmysmith560]

Correct me if I'm wrong but isn't maximum velocity reached when the particle stops accelerating? The graph shows the acceleration of the particle and the particle stops accelerating at t = 3, which is consistent with the answer.

 

[Lith_30]

The particle accelerates until t = 3 so it can't be t = 1 or 2, and once it passes t = 3 it decelerates so it can't be t = 5. So I think the answer is C

 

[quickoats]

The particle accelerates until t = 3 so it can't be t = 1 or 2, and once it passes t = 3 it decelerates so it can't be t = 5. So I think the answer is C

On the right track with the physical reasoning. Perfectly fine for multiple choice.

A mathematical approach to maximise v would be to solve dv/dt = 0. i.e. where a = 0. We have 2 possible values where dv/dt = 0, namely 3 and 5. To check if a point is a max or a min, you check the value of dv/dt either side of where it's 0.

 

[jimmysmith560]

View attachment 31167

Consider the velocity function. It will equal 0 at t = 11 and t = 13. You are trying to show that the velocity function must have a stationary point in between t = 11 and t = 13. Assume that it doesn't have a stationary point. Therefore, the velocity function must either be always increasing, or always decreasing from t = 11 to t = 13 as it can never switch. However, if this were the case, the velocity function can never come back to 0, meaning this is impossible and it must have a stationary point.

This tells us that the answer cannot be A. The answer is not B either because we are talking about a stationary point in terms of the velocity function and not the displacement function. The answer cannot be D either because similarly to B, we are talking about a stationary point in terms of the velocity function and not the acceleration function, leaving C as the correct answer. I believe that such reasoning is viable.

I hope this helps!

 

[CM_Tutor]

@Hivaclibtibcharkwa, another couple of ways to approach this problem...

Firstly, forget about the particle and pretend we are talking about you. So, at some point in time (t = 11, as it happens) you are stopped. 2 minutes later, you are stopped again. Knowing only those facts, which MUST be true?

A: At some point in those 2 minutes, you changed direction in your movement.

B: If you plot your displacement from some starting point, at some point in those two minutes it reaches a maximum or minimum (or horizontal POI).

C: At some point in those 2 minutes, you have an acceleration of zero.

D: If you plot your acceleration, at some point in those two minutes it reaches a maximum or minimum (or horizontal POI).

All of these COULD be true... but only one has to be true.

Secondly, there is nothing that says there is any movement in these 2 minutes. You could have sat down at your desk to study at the time that happened to be t = 11. 2 minutes later, you could (and hopefully would) still be there. For the whole 2 minutes, you could have been stationary. Thought of that way, three of these obviously won't have happened.

Maths problems can be presented in an abstract way but you can often reconsider them from a more concrete perspective. I hope this helps!

 

[jimmysmith560]

View attachment 31215View attachment 31216

The triangle is being subtracted from the total area of the curve and is used to prove that the area of the region R is equal to the given expression. It's using the formula for the area of a triangle , where the base is , which means subtracting the x-coordinate where tangent meets the x-axis from 1, and the height being 1 because the y-coordinate of (1,1) is 1.

I hope this helps!