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!

Differential Equations with a body initially at rest

  1. Oct 16, 2011 #1
    1. A body of mass 2kg is initially at rest and is acted upon by a force of (v - 4) newtons where v is the velocity in m/s. The body moves in a straight line as a result of the force.



    2. a. Show that the acceleration of the body is given by dv/dt = (v - 4) / 2
    b. Solve the differential equation in part a to find v as a function of t.




    3. a. I used the formula F = ma where F = (v - 4) and m = 2
    (v - 4) = 2a
    a = (v - 4) / 2

    b. I tried to solve it like any other differential equation with the following initial conditions:
    when t = 0, v = 0
    But I found it very difficult and challenging:
    dv/dt = (v - 4) / 2
    2 dv/dt = v - 4
    2 / dt = (v - 4) / dv
    I want to change the division sign to a multiplication sign so that I can take the integral of both sides, but I don't know how to algebraically manipulate it to be in that form.
     
  2. jcsd
  3. Oct 16, 2011 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    You don't really know how to separate variables.

    [tex] \frac{dv}{dt} = \frac{v-4}{2} \Rightarrow \frac{dv}{v-4} = \frac{1}{2}dt [/tex]

    Now integrate and use the initial condition.
     
  4. Oct 16, 2011 #3
    So after that I do this:
    ∫1 / (v - 4) dv = ∫1/2 dt
    ln|v - 4| = 1/2 t + C
    when t = 0, v = 0
    ln|-4| = C
    C = ln(4)
    So this proves that we need the negative solution of the absolute value:
    ln(4 - v) = 1/2 t + ln(4)
    ln(4 - v) - ln(4) = 1/2 t
    ln((4 - v) / 4) = 1/2 t
    Since it has a base of e and a power of (1/2 t):
    e^(1/2 t) = (4 - v) / 4
    4e^(1/2 t) = 4 - v
    v = 4 - 4e^(1/2 t)
    Is that all correct?
     
  5. Oct 17, 2011 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Yes, it's ok. Excellent.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook