# Acceleration as a function of distance to function of time

1. ### A(s)

58
1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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

Last edited: Sep 14, 2007
2. ### bob1182006

492
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.

3. ### A(s)

58
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?

Last edited: Sep 14, 2007
4. ### bob1182006

492
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.

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?

5. ### Dick

25,913
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.

6. ### A(s)

58
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

7. ### bob1182006

492
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.

8. ### A(s)

58
well the C in both cases is 0 as the starting position and velocity are 0, but im afraid i don't understand the form of your equation Dick

9. ### Dick

25,913
Yes, it doesn't make sense because the solution is wrong. The correct solutions are exponentials of the time variable.

10. ### A(s)

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so, im 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

11. ### A(s)

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im sorry if im being dense, heh i dont have a solid background in physics

12. ### bob1182006

492
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.

13. ### A(s)

58
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

14. ### A(s)

58
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)

15. ### bob1182006

492
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 alot 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.

16. ### A(s)

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

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

17. ### A(s)

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

18. ### A(s)

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

19. ### bob1182006

492
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.

20. ### A(s)

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