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!

Displacement under varying acceleration

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data
    A 7.00 kg object starting from rest falls through a viscous medium and experiences a resistive force = -b , where is the velocity of the object. The object reaches one half its terminal speed in 6.14 s.

    terminal speed = 86.898m/s

    How far has the object traveled in the first 6.14 s of motion?
    3. The attempt at a solution

    i don't understand why i couldn't simply use an integration of V to find the displacement since the acceleration is not constant.

    the method i've tried.
    v=vt(1-e(-t/T))
    T=m/b
    =7/b
    =vt/9.81
    v=86.898(1-e(-t/8.858))

    [tex]\int^{t=6.14}_{t=0}[/tex] vt(1-e(-t/T))dt
    [tex]\int^{t=6.14}_{t=0}[/tex] (86.898-86.898e-t/8.858)dt
    533.55-(43.448-86.898)
    577m

    my answer is wrong, but i don't know why it is wrong. hope somebody could explain to me
     
  2. jcsd
  3. Sep 19, 2009 #2
    Check your definite integration - the numbers look funny.
     
  4. Sep 19, 2009 #3
    where it goes wrong?
     
  5. Sep 19, 2009 #4
    [tex]\int 86.898 - 86.898 e^{-t/8.858} dt = 86.898t + (8.858) 86.898 e^{-t/8.858} + C[/tex]
    I can't tell exactly where you went wrong since your integration process and final integrated form is not shown, but from your values, it seems that you must have messed up the integration of the exponential part.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Displacement under varying acceleration
  1. Varying acceleration? (Replies: 2)

Loading...