# Integrate Acceleration to Find Time?

1. Mar 29, 2012

### Trapezoid

This question is from my calculus assignment but I apologize if it belongs on the a physics board regardless.

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

A particle of mass m is attracted towards a fixed point 0 with a force inversely proportional to its instantaneous distance from 0. If a particle is released from rest at a distance L from 0, how long will it take to reach 0?

2. The attempt at a solution
$$F = ma = \frac{k}{d}$$
where d is the distance from 0 and k is a proportionality constant. Therefore, $$a = \frac{d^2x}{dt^2} = \frac{k}{md}$$
I know that integration is required but I don't see how to use it to find time. Could somebody give me a tip as to how to proceed?

Thanks,
Trapezoid

2. Mar 29, 2012

### Robert1986

Well, imagine that the question asked you for an equation that gave the position of the particle as a function of time. E.g. $s(t) = a_0 + a_1t + a_2t^2$. You know how to do this, right? Now, once you have the equation, just put the distance traveled on the LHS and solve for $t$.

3. Mar 29, 2012

### RoshanBBQ

$$x\frac{d^2x}{dt^2} = \frac{k}{m}$$
d is x. Is that what you were forgetting?

4. Mar 29, 2012

### HallsofIvy

Staff Emeritus
You say "d is the distance from 0 and k is a proportionality constant" but you don't say what "x" is! I think you mean x to be also the distance from 0 at time t so your differential equation is really
$$\frac{d^2x}{dt^2}-= \frac{k}{mx}$$

You will have to solve the differential equation for x, then set it equal to 0 and solve for t.

Now, that is not easy to integrate directly but you can use a technique called "quadrature".
Let v= dx/dt so that $d^2x/dt^2= dv/dt$. By the chain rule, $dv/dt= (dv/dx)(dx/dt)= v dx/dt$. So your equation is
$$v\frac{dv}{dx}= \frac{k}{mx}$$
a separable equation.

$$v dv= \frac{k}{m}\frac{dx}{x}$$
Integrating,
$$(1/2)v^2= \frac{k}{m}ln(|x|)+ C$$
Since the object was released "distance L" (x= L) at rest (v= 0) we have
$$(1/2)0^2= \frac{k}{m}ln(L)+ C$$
so
$$C= -\frac{k}{m}ln(L)$$
and the equation for v becomes
$$\frac{1}{2}v^2= \frac{k}{m}ln(x)- \frac{k}{m}ln(L)= \frac{k}{m}ln\left(\frac{x}{L}\right)$$
(I am assuming that L is positive and as x changes from L to 0, it will be positive so we don't need the absolute value.)
$$v= \frac{dx}{dt}= \sqrt{\frac{2k}{m}ln\left(\frac{x}{L}\right)}$$

Integrate that, solve for x as a function of t, set it equal to 0, and solve for t.

5. Mar 29, 2012

### Trapezoid

Thanks for all the help everybody. I'll take a closer look to see if I can't find the answer from here.

6. Mar 29, 2012

### RoshanBBQ

I thought it would just be
$$x\frac{d^2x}{dt^2} = \frac{k}{m}$$
$$\int \int x dxdx = \frac{k}{m} \int \int dtdt$$
$$\frac{x^3}{6}=\frac{k}{m}\frac{t^2}{2}+c$$
$$x^3=3\frac{k}{m}t^2+c$$
Here is a nice junction to find c (assuming x = 0 is at the point and x = L is the initial condition, L > 0):
$$L^3 = 0 + c \rightarrow c = L^3$$
then
$$x^3=3\frac{k}{m}t^2+L^3$$
$$x(t)=\left ( 3\frac{k}{m}t^2+L^3 \right )^{1/3}$$

You then solve for the time for which x(t) = 0. Note, k must be < 0 for such a solution.

7. Mar 29, 2012

### Trapezoid

Thanks HallsofIvy and RoshanBBQ,

Unless I've misunderstood, the method outlined by HallsofIvy leaves me with the integral $\int \frac{dx}{\sqrt{\ln(\frac{x}{L})}}$ which I cannot even begin to solve. With regards to RoshanBBQ's solution, it seems odd to me that k would be negative since constants in examples done in class are always specified to be positive. That being said, a negative k would imply that the force acts in the negative direction, which seems consistent with the problem. Is my thinking here correct?

Trapezoid

8. Mar 29, 2012

### RoshanBBQ

It just depends on how you set up the problem.I'm not entirely comfortable with the answer I found, though. Consider when t -> infinity. You would expect x -> 0. My solution has x become the 3rd root of an increasingly negative number...

Last edited: Mar 29, 2012
9. Apr 5, 2012

### Trapezoid

Hi all,

Sorry to bump but we got the solutions to the assignment so I thought I'd post some more guidance in case anybody else is having trouble with a similar question.

The method outlined by HallsofIvy is correct. Without giving too much away, to solve the last integral, substitute $u = \frac{L}{x}$ and use the gamma function.