A Ball accelarating down a hill

How fast would a ball travel down a hill that follows a function (like f(x)=x^3-9x^2+27x-20) where x and y are both length? x being distance from the end point, y being the height, and y=0 being the ground. (not part of the 'hill') Assume the ball is let go at a point (x,y).
If it were a straight ramp, the velocity would be sine of the angle of elevation, but few hills in real life follow a straight line.

Would you some how combine the derivative of the hill's function with sine to get the angle at any given point, and work off of that? :f6:

Take the derivative of the inclination, dy/dx, then take the arctg(dy/dx) for the angle of inclination, then multiply g = 9,81m/s² by the -sine(arctg(dy/dx)) and you have your acceleration.

That's without considering friction and considering the ball to be a point and not a rotating object.

Thought if you are not considering friction and a rotating system, you might as well use the conservation of mechanical energy.

mV²/2 + mgh = constant

Take the derivative of the inclination, dy/dx, then take the arctg(dy/dx) for the angle of inclination, then multiply g = 9,81m/s² by the sine(arctg(dy/dx)) and you have your acceleration.

by 'arctg', do you mean arctan, tan^-1, etc?

g*sin[arctan(dy/dx)] is the accelaration at any given point... so you would integrate from the initial point to the current point to find the velocity at any given distance x? and again to find the distance it traveled in a certain time...?
I think I just lost myself....

g*sin[arctan(dy/dx)] is the accelaration at any given point... so you would integrate from the initial point to the current point to find the velocity at any given distance x? and again to find the distance it traveled in a certain time...?
I think I just lost myself....

You can only do straight integration/differentiation to convert between displacement, velocity and acceleration when integrating/differentiating with respect to t(ime).

*Edit* Are you trying to find the velocity? Do what was said above.

You can only do straight integration/differentiation to convert between displacement, velocity and acceleration when integrating/differentiating with respect to t(ime).

*Edit* Are you trying to find the velocity? Do what was said above.

And that's the velocity at any given value of x... or something else?
I think I worded my first question poorly/didn't have a solid idea of what I was asking, so I'll try re-wording it.

A hill is modeled by a continuous function f(x), where y is the height above level ground. If you let an object freely fall down, starting at the point (x,y) on the function, how fast will it accelerate, how fast will it move(velocity), and how much distance/displacement will it travel*, relative to t. Ignoring friction, drag, and a rotation vs sliding.

*Would this just be the length of the line making the function, or something else. Like if the function gets too steep for the object to be held onto it via gravity... Though that would depend on previous velocities... this is getting complicated...

Relative to t and t only? I'm not sure whether you can change a path equation back into two components.

[IMG]http://codecogs.com/eq.latex?a=g\sin (\arctan f'(x))[/IMG]

[IMG]http://codecogs.com/eq.latex?v=\int a~dt=at+c=gt\sin (\arctan f'(x))[/IMG]

[IMG]http://codecogs.com/eq.latex?s=\int v~dt=\frac{1}{2}at^2+c=\frac{1}{2}gt^2\sin (\arctan f'(x))[/IMG]

That's as best as I can think of, in terms of t. Hope it helps.

A hill is modeled by a continuous function f(x), where y is the height above level ground. If you let an object freely fall down, starting at the point (x,y) on the function, how fast will it accelerate, how fast will it move(velocity), and how much distance/displacement will it travel*, relative to t. Ignoring friction, drag, and a rotation vs sliding.

It's a basic Energy Conservation exercise with the use of a function that represents position of movement instead of directly giving you the angle of movement. So work that out and you can carry on normally.

- f(x) = [insert function here] represents the position of the ball and its changes over time (run and rise coordinates).
- f'(x) = [insert derivative here] = tan[angle-xOd] represents the direction of the movement at a specific time t. (a.k.a the 'rate' at which the position is changing, up to you how to understand/visualize it).
=> arctan[angle-xOd] = angle-xOd (degrees) = arctan[f'(x)]

From here on it should be basic physics:
- a = g*sin(angle-of-movement)
- speed = ⌠a dt = at + c
- distance = ⌠[speed] dt = ⌠at dt = (at²)/2 + c

Tempus formulas should be a perfect summary so I'm just trying to un-wiki things.

The problem that you two above have is that x is also a function of t, so you can't just take it out from integrals. The way it was set up:

for your function f(x) = x³-9x²+27x-20 -> df(x)/dx = 3x² - 18x + 27 = (3x²-18x+27)
a(x) = -g*sin(arctan(df(x)/dx)) = -g*df/dx/(1+(df/dx)²)^0.5

Disconsider, dv/dx is not a(x). I don't know how to generate the differential equation for this problem... =P

If you're differentiating w.r.t t, wouldn't you get [IMG]http://codecogs.com/eq.latex?x'(t)f'(x(t))[/IMG] (from chain rule)?

I must admit that I've not really done much on differential equations...

f(x) = [insert function here] represents the position of the ball and its changes over time (run and rise coordinates).

But if f(x) is the height of the hill, and x is the distance from the end point, where does t come from?
You could say y=f(t), and x=t, or you could say y=f(2t), and x=2t, or any number of combinations, so how would you choose which one to use...?

y=f(x) and x=f(t).

I'll see if I can find something, my previous thought were wrong...