Acceleration as a function of distance to function of time

[A(s)]

Homework Statement

I have an equation, A(d)= (9.8d/10 +9.8) / (d/10+1)

It is the acceleration of an object at distance traveled d. (obviously linear and not constant). I was wondering how I can convert this to acceleration, velocity, and or distance as a function of time.

Homework Equations

The Attempt at a Solution

I do not know what to do to thisequations to make them a function of time

by the way, this is not a homework problem, i am not even enrolled in any kind of physics class, it is part of a personal experiment of mine, and has been a major hang-up for me i spent much of last summer pondering how to do this

 

[bob1182006]

Did you make up this problem?

Because I think it's almost impossible to have an equation that gives you the acceleration if you know the distance some particle travels.

 

[A(s)]

Here is the context of the problem

a 1 cubic meter bubble has an acceleration of 9.8 m/s^2
a 3 cubic meter bubble has an acceleration of 29.4 m/s^2

when a 1 cubic meter bubble rises 30 meters, it changes from a 1 cubic meter bubble to a 3 cubic meter bubble (due to decrease in pressure). This is the premise.

So a bubble (that remains intact) rising, allowed to expand, will have an increase in acceleration that depends on the distance it has already traveled

does this make sense?

 

[bob1182006]

So the bubble accelerates AND expands?

well you know the acceleration can you determine the velocity? and then how far it takes for the bubble to travel the 30 m?

that way things can be depending on time.

Also can you see something "interesting" about this:
1 cubic meter a=9.8
3 cubic meter a=29.4

so I'm guessing you're being asked to determine the position/velocity at time t?

 

[Dick]

Ignoring the physics of bubbles, write the equation you want to solve as d''(t)=.98d(t)+9.8. It's a linear ode and is easy to solve (but you don't do it by integrating with respect d). If you haven't done differential equations, your solution is of the form A*exp(k*t)+B*exp(-k*t)+C. Substitute that for d(t) and try to find k and C. A and B are determined by boundary conditions.

 

[A(s)]

well the velocity function i showed above, but the problem comes when you integrate that, it creates a function that doesn't make sense. The position at distance traveled d?

P(d) supposedly = .48x^3/3 + 4.9d^2

but this doesn't make sense. The position at d should = d

 

[bob1182006]

If it is a ODE type problem then Dick's method is what you need to do.
But if you need to find the position/velocity/acceleration of the bubble at a time t, then you need to first find the equation of the non-constant acceleration and then integrate.

 

[A(s)]

well the C in both cases is 0 as the starting position and velocity are 0, but I am afraid i don't understand the form of your equation Dick

 

[Dick]

Yes, it doesn't make sense because the solution is wrong. The correct solutions are exponentials of the time variable.

 

[A(s)]

so, I am at a loss as to what to do

my equation is with respect to distance, i need to somehow convert it to an equation with respect to time

 

[A(s)]

im sorry if I am being dense, heh i don't have a solid background in physics

 

[bob1182006]

Well was that equation A(d)=.98d+9.8 GIVEN to you to solve? or did you figure out that was what you needed to find?

It may be easiest to just type out the entire problem or just the main parts of given data/what you need to find. So we don't get confused too.

 

[A(s)]

well as for the first part, this is not any kind of a homework problem


I derived the equation on my own from something like this:
buoyant force = 9.8m/s^2 *kg
i am assuming a starting 1m^3 bubble
so you get something like this
F=ma


x = a bubble of x cubic meters

Force= 9.8(x)*1000kg/(1m^3 bubble)

so that for every cubic meters, an extra 1000kg is displaced

so now x varies depending on the distance upwards of the bubbles path

x increases by one for every 10 meters of travel, with a starting value of 1m^3

x = 1m^3 (d/10+1)

so now plugging x in, we get Force = 9.8 [(1)(d/10+1)*1000

simplifying we get

(9.8d/10+9.8)*1000kg = Force

| acceleration* Mass| = force

therefore acceleration = .98d+9.8 with respect to distance traveled

 

[A(s)]

i just noticed an error in my final equation, i divided by 1000, but i should have divided by 1000x

So,
(9.8d/10 +9.8) = acceleration
(d/10+1)

 

[bob1182006]

Hm..ok now it's more of a physics problem but I know a bit of physics (lil bit ><).

how did you go from 9.8m/s^2 *kg to 9.8(x)*1000kg/1m^3 bubble ?
1000kg is like 1000000 g O.O which seems like a lot of mass to displace.

Also I think you can change your acceleration from 9.8(x) to a non-constant acceleration that you can determine from the information go gave in post #3.

$$1 m^3; a=9.8\frac{m}{s^2}$$
$$3 m^3; a=29.4\frac{m}{s^2}$$

let's assume you start from time? 0? or 1?
and then try to find some equation for acceleration that will be dependent on t.
so t=0 or 1? a=9.8
t=1 or 2? a=29.4
can you get that somehow to some equation like: ax+b? or maybe ax^2+bx+c I think it's going to be a linear equation for acceleration but Hopefully you can go from here.

 

[A(s)]

holy crap... those cancel, so distance cancels out

9.8(d/10+1) / (d/10+1) = 9.8

 

[A(s)]

well 1 meter cubed is a lot of water, 1000L to be exact, and 1 Liter weights 1 kg

 

[A(s)]

Awesome, I don't think that d is a factor any more, it cancels out. does this look right to you?

 

[bob1182006]

no because then force is constant 9.8 *mass but acceleration is not constant.

First try to find an equation for acceleration since it isn't constant. Then you can move on to the F=ma part of the problem.

 

[A(s)]

wouldnt it be the other way around, acceleration is constant, but force is not

 

[Dick]

If you are assuming acceleration is a linear function of distance d, then I think your fit to the two data points should actually be 0.67*d+9.8. If you believe this then the general solution putting the condition d'(0)=0 is A*cosh(sqrt(0.67)*t)-9.8/0.67. Look up the cosh definition, it's really just a combination of exponentials. You should be able to verify that satisfies d''(t)=0.67d+9.8. Hint: if f(t)=cosh(kt) then f''(t)=k^2*cosh(kt). To find A, you want d(0)=0. cosh(0)=1, so that means A=9.8/0.67. As this is not homework, I'm just giving you all of this so you know what you are getting into. So yes, what you want can be done, but you may not find it very easy. But I'm not terribly sure this problem is physically realistic either. If you want more details on how it's done then keep asking questions.

 

[A(s)]

I think i got it now, thanks for the help. it makes sense that something cannot rise in water faster than 9.8 m/s^2, just as something cannot fall with greater acceleration in air than that. Dispite that, the forces will be different at the top and bottom, because the bubble will displace more water near the surface. If you could answer a few of my questions on the physics of bubbles, i would like to contact you via e-mail if possible

 

[Dick]

Hmm. See I've been wasting my time solving the original equation, since it seems to have changed. Well, good luck with your solution anyway.