How Can I Find Acceleration in Terms of Time Given a Function of Position?

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Homework Help Overview

The discussion revolves around a differential equation problem in physics, specifically focusing on the relationship between acceleration and position, given by the function a = -1.5*s. The participants are tasked with finding acceleration in terms of time.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between acceleration and position, noting that acceleration can be expressed as the second derivative of position with respect to time. There are attempts to manipulate the equation and separate variables, but some participants question the validity of these methods for second-order differential equations. Others suggest using an Ansatz to solve the equation.

Discussion Status

The discussion is active, with participants sharing their attempts and insights. Some have provided guidance on potential methods, such as using an Ansatz or Laplace transforms, while others express uncertainty about their approaches. There is a recognition of the challenges faced due to a lack of formal training in differential equations.

Contextual Notes

Participants mention that the problem is situated within a physics class context, and there is a noted absence of formal instruction on solving differential equations. Some participants are relying on online resources for assistance.

yoamocuy
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differential equation problem :(

Homework Statement


I'm given a function acceleration a=-1.5*s, where s is a position. I need to find a in terms of t.


Homework Equations





The Attempt at a Solution


I know that a is also equal to d^2*s/dt^2. Therefore d^2*s/dt^2 is equal to -1.5*s. By dividing by s and multiplying by dt^2 I get d^2*s/s=-1.5*dt^2. At this point I'm not sure what to do. If I can figure out how to get rid of the dt^2 and the d^2*s, then I can probably solve the rest of the problem. I imagine I need to integrate but am not sure how that would work out.
 
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yoamocuy said:

Homework Statement


I'm given a function acceleration a=-1.5*s, where s is a position. I need to find a in terms of t.


Homework Equations





The Attempt at a Solution


I know that a is also equal to d^2*s/dt^2. Therefore d^2*s/dt^2 is equal to -1.5*s. By dividing by s and multiplying by dt^2 I get d^2*s/s=-1.5*dt^2. At this point I'm not sure what to do. If I can figure out how to get rid of the dt^2 and the d^2*s, then I can probably solve the rest of the problem. I imagine I need to integrate but am not sure how that would work out.
You cannot simply 'divide' and 'multiply' differentials like that. I know many non-mathematical sciences courses will tell you that it is fine to do so, but it is simply wrong.

I see that you are attempting to separate the variables, however, this method cannot generally be used for second order differential equations. What methods have you learned for solving second order, homogeneous ODE's?
 


Well this is actually just a physics class so we haven't learned any methods for solving differential equations. I've taken calculus classes but won't be taking differential equations until next semester so I've just been looking through notes online trying to figure out how to solve this question.
 


yoamocuy said:
Well this is actually just a physics class so we haven't learned any methods for solving differential equations. I've taken calculus classes but won't be taking differential equations until next semester so I've just been looking through notes online trying to figure out how to solve this question.
The standard method of solving such problems is to take an Ansatz of the form

[tex]s = Ae^{\lambda t}[/tex]

and substitute that into the ODE.

I'm sorry that I don't know of any good online references for DE's, but I'm sure someone here will be able to suggest a suitable reference.
 


Ok, thanks for the help. I'll keep searching and working with it.
 


yoamocuy said:
Ok, thanks for the help. I'll keep searching and working with it.
If you would like to post your working, I'd be more than happy to help you with it. All you need to do is substitute that Astatz I gave you into the ODE and you should find that you have a rather simple equation to solve.
 


Ok, I got an equation that looks like s^2*e^((lambda)*t)=-1.5, but when I try to solve for the roots it isn't possible. Am I still completely off on my attempt?
 


yoamocuy said:
Ok, I got an equation that looks like s^2*e^((lambda)*t)=-1.5, but when I try to solve for the roots it isn't possible. Am I still completely off on my attempt?
I'm not sure how you've managed to get that. Can you work out what

[tex]\frac{d^2 s}{dt^2} = \frac{d^2}{dt^2} Ae^{\lambda t}[/tex]

is?
 


Ug I don't even know what I'm supposed to be solving for there. On the sites I've been looking at, it looks like they are just taking a Laplace transform and multiplying each differential by e^-((lambda)*t)
 
  • #10


yoamocuy said:
Ug I don't even know what I'm supposed to be solving for there. On the sites I've been looking at, it looks like they are just taking a Laplace transform and multiplying each differential by e^-((lambda)*t)
I was simply asking what is the second derivative of [itex]Ae^{\lambda t}[/itex] with respect to t?

Of course, you can use Laplace transforms if you like, but it is much more straightforward (if a little inelegant) to us an Anstatz.
 
  • #12


I took the second derivative of that and got A*(lambda)^2*e^((lambda)*t). Would lambda be -2 for that?

I also tried using Laplace transforms and ended up with s^2+1.5=0, which has imaginary roots.
 
  • #13


yoamocuy said:
I took the second derivative of that and got A*(lambda)^2*e^((lambda)*t). Would lambda be -2 for that?
You're on the right lines. So let's take a look at the over all equation,

[tex]A\lambda^2 e^{\lambda t} = -1.5Ae^{\lambda t}[/tex]

Hence,

[tex]\lambda^2 = -1.5[/tex]

Do you agree?
 
  • #14


Yea, I agree with that, but if I solve for lambda I get j1.22. What can I do with that?
 
  • #15


yoamocuy said:
Yea, I agree with that, but if I solve for lambda I get j1.22. What can I do with that?
Simple, you have determined lambda, so you now have the [general] solution,

[tex]s\left(t\right) = A\exp\left(i\sqrt{1.5}t\right)[/tex]

That is it. You have now found the general solution to the ODE. You can use Euler's relation to write it in a 'nicer' form, but you have found a valid solution.
 

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