1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integrate Acceleration to Find Time?

  1. Mar 29, 2012 #1
    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
    [tex]F = ma = \frac{k}{d}[/tex]
    where d is the distance from 0 and k is a proportionality constant. Therefore, [tex]a = \frac{d^2x}{dt^2} = \frac{k}{md}[/tex]
    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?

  2. jcsd
  3. Mar 29, 2012 #2
    Well, imagine that the question asked you for an equation that gave the position of the particle as a function of time. E.g. [itex]s(t) = a_0 + a_1t + a_2t^2[/itex]. You know how to do this, right? Now, once you have the equation, just put the distance traveled on the LHS and solve for [itex]t[/itex].
  4. Mar 29, 2012 #3
    [tex]x\frac{d^2x}{dt^2} = \frac{k}{m}[/tex]
    d is x. Is that what you were forgetting?
  5. Mar 29, 2012 #4


    User Avatar
    Science Advisor

    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
    [tex]\frac{d^2x}{dt^2}-= \frac{k}{mx}[/tex]

    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 [itex]d^2x/dt^2= dv/dt[/itex]. By the chain rule, [itex]dv/dt= (dv/dx)(dx/dt)= v dx/dt[/itex]. So your equation is
    [tex]v\frac{dv}{dx}= \frac{k}{mx}[/tex]
    a separable equation.

    [tex]v dv= \frac{k}{m}\frac{dx}{x}[/tex]
    [tex](1/2)v^2= \frac{k}{m}ln(|x|)+ C[/tex]
    Since the object was released "distance L" (x= L) at rest (v= 0) we have
    [tex](1/2)0^2= \frac{k}{m}ln(L)+ C[/tex]
    [tex]C= -\frac{k}{m}ln(L)[/tex]
    and the equation for v becomes
    [tex]\frac{1}{2}v^2= \frac{k}{m}ln(x)- \frac{k}{m}ln(L)= \frac{k}{m}ln\left(\frac{x}{L}\right)[/tex]
    (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.)
    [tex]v= \frac{dx}{dt}= \sqrt{\frac{2k}{m}ln\left(\frac{x}{L}\right)}[/tex]

    Integrate that, solve for x as a function of t, set it equal to 0, and solve for t.
  6. Mar 29, 2012 #5
    Thanks for all the help everybody. I'll take a closer look to see if I can't find the answer from here.
  7. Mar 29, 2012 #6
    I thought it would just be
    [tex]x\frac{d^2x}{dt^2} = \frac{k}{m}[/tex]
    [tex]\int \int x dxdx = \frac{k}{m} \int \int dtdt[/tex]
    Here is a nice junction to find c (assuming x = 0 is at the point and x = L is the initial condition, L > 0):
    [tex]L^3 = 0 + c \rightarrow c = L^3[/tex]
    [tex]x(t)=\left ( 3\frac{k}{m}t^2+L^3 \right )^{1/3}[/tex]

    You then solve for the time for which x(t) = 0. Note, k must be < 0 for such a solution.
  8. Mar 29, 2012 #7
    Thanks HallsofIvy and RoshanBBQ,

    Unless I've misunderstood, the method outlined by HallsofIvy leaves me with the integral [itex]\int \frac{dx}{\sqrt{\ln(\frac{x}{L})}}[/itex] 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?

  9. Mar 29, 2012 #8
    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
  10. Apr 5, 2012 #9
    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 [itex] u = \frac{L}{x} [/itex] and use the gamma function.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook