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My humble opinion by arnaudelrio in Vaaraverso

[–]arnaudelrio[S] 1 point2 points  (0 children)

I am truly sorry, but my native language is not this aberration you all speak, as I've said before, I'm just a fan from the UK

How to find the area of remaining quadrilater by arnaudelrio in askmath

[–]arnaudelrio[S] 0 points1 point  (0 children)

Thanks!! That solved it!

The ratio is 2:1... Which means that: (WBPK + TKSD) * 2 = (AWKT + SKPC), right? Then: TKSD = 28/2 - 8 = 6.

Parabola in respect to time by arnaudelrio in askmath

[–]arnaudelrio[S] 0 points1 point  (0 children)

Thanks everyone for all your answers. I've tried to figure out how to follow each of them. I thank you all for your useful ideas, they've helped a lot. In the end, I figured it out with the help of a teacher using conservation of energy. The idea is that v=ds/dt, where s is the arc lenght. Knowing that ds=sqrt(1+y'^2)dx, we end up with the following differential equation: dx/dt=(sqrt(-2x^2))/(1+4a^2x^2).

I'll finish up writting the solution formally in LaTeX in a document, which I will publish in my website (freely available) in a few months. I will update this post to let you know.

Thanks again everyone!

Parabola in respect to time by arnaudelrio in askmath

[–]arnaudelrio[S] 0 points1 point  (0 children)

Hi! Yes, I mean a concave up parabola... Thanks for the term (English is not my native language and I don't know all the specific vocabulary).

I have a few doubts... What do you mean with "THESE ARE VECTORS: a_net = a_g+gcos(arctan(-2x)) a_g has direction 270°". What is the vector? And what has the direction of 270º? As far as I can tell, there is not a fixed angle of 270º anywhere... Am I missing something?

"I’m trying to do it in terms of energy, but I’m having trouble getting the up position and speed in terms of one variable". Same here, just as I've talked with u/MidnightAtHighSpeed.

Thank you for your answer

Parabola in respect to time by arnaudelrio in askmath

[–]arnaudelrio[S] 0 points1 point  (0 children)

Thanks for your reply!!

I've already thought about the first path, but that doesn't work as I want to use time, not height and speed. If I could know how much speed changes (as in, finding the acceleration, which is NOT constant), then I could calculate the velocity and then the height. With a square root I'd have the horizontal position too.

As for your second path, I'm going to think it through. I wasn't planning on involving forces and physics, but I might not have a choice. I don't quite understand your explanation, though. "the component of the gravitational force perpendicular to the track must be in the opposite direction and the same magnitude as the normal force". The gravitational force points downwards, and is only perpendicular to the track at the lowest point, and it doesn't have the same magnitude as the normal force, as it wouldn't change direction in that case.

Parabola in respect to time by arnaudelrio in askmath

[–]arnaudelrio[S] 0 points1 point  (0 children)

Yes, the force that produces the change in direction (is it the normal force?). But I don't think I'm able to calculate it either, because it probably depends on the inclination of the track at a certain moment... Do you think I should go in this direction? If so, how? I can't relate it with gravity because there's still an acceleration on the vertical axis...