Calculating velocity with variable acceleration

[scain6043]

This is not so much coursework as it is a thought experiment that has been bugging me for a while now.

1. Homework Statement

Find the equation for velocity either as a function of distance (x) or time (t).

I have an object whose acceleration is given by the rational function a(x) = (a+x)/(b-x) where x is the distance traveled and a and b are some constant.

The actual equation has many more values and constants in it but this simplified version should be fairly similar.

2. The attempt at a solution

First, I resorted to the motion equations but quickly realized that the acceleration is not constant. My second thought was to use motion equations that deal with variable acceleration that use jerk and so forth but soon found that due to the rational nature of the function a(x) that it is not possible to get a derivative that is a constant.

Yesterday, the idea came to me that if I took the derivative of a(x) my resulting function would have the units 1/(sec)² which could be manipulated by taking the square root of the reciprocal to obtain time with respect to distance (x). However, this still does not solve the problem because I'm not sure if the manipulation is mathematically sound and this still does not solve the problem of variable acceleration. Any thoughts?

 

[Nathanael]

Yesterday, the idea came to me that if I took the derivative of a(x) my resulting function would have the units 1/(sec)² which could be manipulated by taking the square root of the reciprocal to obtain time with respect to distance (x). However, this still does not solve the problem because I'm not sure if the manipulation is mathematically sound and this still does not solve the problem of variable acceleration. Any thoughts?

I would be surprised if that gave the correct t(x)

The goal is to solve the differential equation $\frac{d^2x}{dt^2}=\frac{a+x}{b-x}$
I am the opposite of an expert at differential equations, but I'm thinking you might be able to separate it and integrate both sides twice:
$∫∫\frac{b-x}{a+x}d^2x=∫∫dt^2$
The left side would be easier to integrate if you wrote it like this:
$∫∫\frac{b-x}{a+x}d^2x=∫∫\frac{b+a-(a+x)}{a+x}d^2x=∫∫(\frac{b+a}{a+x}-1)d^2x$

I am not sure if this would give the correct solution. I replied to you so that I can also learn from this problem. I'm sure someone more knowledgeable will help soon.

 

[Chestermiller]

I would be surprised if that gave the correct t(x)

The goal is to solve the differential equation $\frac{d^2x}{dt^2}=\frac{a+x}{b-x}$
I am the opposite of an expert at differential equations, but I'm thinking you might be able to separate it and integrate both sides twice:
$∫∫\frac{b-x}{a+x}d^2x=∫∫dt^2$
The left side would be easier to integrate if you wrote it like this:
$∫∫\frac{b-x}{a+x}d^2x=∫∫\frac{b+a-(a+x)}{a+x}d^2x=∫∫(\frac{b+a}{a+x}-1)d^2x$

I am not sure if this would give the correct solution. I replied to you so that I can also learn from this problem. I'm sure someone more knowledgeable will help soon.

Hi Nathanial. You can't integrate it like that. It's not correct mathematically. But, here is something that will work:

By the chain rule:

$a = \frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=\frac{1}{2}\frac{dv^2}{dx}$

Chet

 

[haruspex]

It's not correct mathematically.

In particular $\int d^2x$ is meaningless. It can't be thought of as $\int\int dx^2$.

By the chain rule:
$a = \frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=\frac{1}{2}\frac{dv^2}{dx}$

Yes, a trick worth noting. It's equivalent to working with energy instead of forces, $F.dx = mv.dv$. Typically, it allows you to get v as a function of position in closed form, but not v as a function of time. Fortunately, the OP asks for either.

 

[scain6043]

Thanks for the replies! I understand that using energy I can find the velocity but isn't that only if the change in energy is already known?

 

[Chestermiller]

Thanks for the replies! I understand that using energy I can find the velocity but isn't that only if the change in energy is already known?

No way. Play around with the equations a little and see what you come up with.

Chet