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!

Higher Derivatives - Acceleration help

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle moves along the x-axis, its position at time t is given by
    [tex]$\displaystyle \Large x(t)=\frac{7 t}{6+5 t^2}, \ t\geq 0,$[/tex]
    where t is measured in seconds and x is in meters. The acceleration of the particle equals 0 at time t = ____ and ____ seconds.

    2. Relevant equations
    [tex]$\displaystyle \Large x(t)=\frac{7 t}{6+5 t^2}, \ t\geq 0,$[/tex]
    v(t) = x'(t)
    a(t)=v'(t)=x''(t)

    3. The attempt at a solution
    once i find a(t), which is the 2nd derivative of the original function, i get:
    [tex]$\displaystyle \Large a(t)=\frac{7t-7}{2t(6+5t^2)}[/tex]

    so when acceleration=0, [tex]$\displaystyle \Large a(t)=\frac{7t-7}{2t(6+5t^2)}=0[/tex]
    t=1
    but what the other value?
     
  2. jcsd
  3. Oct 20, 2009 #2

    zcd

    User Avatar

    You solved the derivatives incorrectly
     
  4. Oct 20, 2009 #3
    oh my bad. it should be:
    -440t / (6+5t^2)

    now what?
     
    Last edited: Oct 20, 2009
  5. Oct 21, 2009 #4
    someone?
     
  6. Oct 21, 2009 #5
    You still do not have the correct second derivative. (I think).
     
  7. Oct 21, 2009 #6
    ok.. i give up. my algebra is good but i have no idea where to go.
    this is what i got for the second derivative after doing it 20 times to make sure i didnt make mistakes:
    [tex]$\displaystyle \Large x''(t)=\frac{ -70t(6+5t^2)-20t(42-35t^2) }{(6+5t^2)^2} \ $[/tex][tex]$\displaystyle \Large =\frac{ -1260t-1050t^3 }{(6+5t^2)^2} \ $[/tex]

    maybe im stuck on the factoring of (42-35t^2)... but i just dont know aaaah
     
  8. Oct 21, 2009 #7
    If that is the correct second derivative - and I have not checked - you are now looking for the points at which acceleration equals zero. So you set the second derivative equal to zero and solve for t.
     
  9. Oct 21, 2009 #8

    zcd

    User Avatar

    2nd derivative is still wrong
    [tex]\frac{dx}{dt}=-7\frac{5t^{2}-6}{(5t^{2}+6)^{2}}[/tex]
    [tex]\frac{d^{2}x}{dt^{2}}=-7\frac{10t(5t^{2}+6)^{2}-20t(5t^{2}+6)(5t^{2}-6)}{(5t^{2}+6)^{4}}=-70t\frac{(5t^{2}+6)-2(5t^{2}-6)}{(5t^{2}+6)^{3}}=-70t\frac{5t^{2}-18}{(5t^{2}+6)^{3}}[/tex]
     
  10. Oct 21, 2009 #9
    oh so the only thing i did wrong is including the 7 in my calculations all the time.. hh tnx man!

    David Gould - yeah i know that but the thing is i got 0 all the time and i needed another value so here i got it.. tnx
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Higher Derivatives - Acceleration help
  1. Higher Derivatives (Replies: 3)

  2. Higher derivatives (Replies: 3)

  3. Higher derivatives (Replies: 42)

Loading...