1. Not finding help here? Sign up for a free 30min 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!

A Simple(?) Differential Equation

  1. Jul 5, 2014 #1
    This is question 1.4 of Chapter 8 of Mary L. Boas's Mathematical Methods in the Physical Sciences, Edition 2. I'm using it as a substitute for my ordinary differential equations class since my textbook has apparently been lost somewhere in the mail.

    1. The problem statement, all variables and given/known data
    Find the distance which an object moves in time t if it starts from rest and has an acceleration d^2/dt^2 = ge^(-kt). Show that for a small t the result is approximately (x = 1/2(gt^2)) and for very large t, the velocity dx/dt is approximately constant. (This problem corresponds roughly to the motion of a parachutist.)


    2. Relevant equations
    The solution manual gives the solution as: x=k^(-1)gt + k^(-2)g(e^(-kt)-1)


    3. The attempt at a solution
    I integrated both sides of d^2/dt^2 = ge^(-kt) twice with respect to t, ending up with x = k^(-2)ge^(-kt) + v_0(t) + x_0.
    However substituting t=0 into the equation gives x = g/(k^2) + x_0, not (1/2)gt^2, so, obviously on the wrong track.

    I put the process in an attached image below since it might be easier to read that than my terrible equation formatting. Thanks!
     

    Attached Files:

  2. jcsd
  3. Jul 5, 2014 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Please post your equations instead of images of them as per forum rules. Your integrations are not correct. For example, if ##x''(t) = ge^{-kt}##, then ##x'(t) = -\frac g k e^{-kt} + C##. When you put ##t=0## in both sides you do not get ##C = v_0##.
     
  4. Jul 5, 2014 #3

    Zondrina

    User Avatar
    Homework Helper

    Here are the equations in a latex format:

    ##\vec a = ge^{-kt}##
    ##\vec v = - \frac{g}{k} e^{-kt} + c_1##
    ##\vec r = \frac{g}{k^2} e^{-kt} + c_1t + c_2##

    Using the initial condition ##r(0) = 0##, you should get something useful.

    ##- \frac{g}{k^2} = c_2##

    The initial condition that ##v(0) = 0## also gives you some more information.

    ##\frac{g}{k} = c_1##

    Then:

    ##\vec r = \frac{g}{k^2} e^{-kt} + \frac{g}{k} t - \frac{g}{k^2}##
     
  5. Jul 5, 2014 #4
    Oh, I see! So that's how to deal with the c's that wind up after integration. Excellent, thank you very much!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: A Simple(?) Differential Equation
Loading...