Higher Derivatives - Acceleration help

[Slimsta]

Homework Statement

A particle moves along the x-axis, its position at time t is given by
$\displaystyle \Large x(t)=\frac{7 t}{6+5 t^2}, \ t\geq 0,$
where t is measured in seconds and x is in meters. The acceleration of the particle equals 0 at time t = ____ and ____ seconds.

Homework Equations

$\displaystyle \Large x(t)=\frac{7 t}{6+5 t^2}, \ t\geq 0,$
v(t) = x'(t)
a(t)=v'(t)=x''(t)

The Attempt at a Solution

once i find a(t), which is the 2nd derivative of the original function, i get:
$\displaystyle \Large a(t)=\frac{7t-7}{2t(6+5t^2)}

so when acceleration=0, $\displaystyle \Large a(t)=\frac{7t-7}{2t(6+5t^2)}=0
t=1
but what the other value?

 

[zcd]

You solved the derivatives incorrectly

 

[Slimsta]

You solved the derivatives incorrectly

oh my bad. it should be:
-440t / (6+5t^2)

now what?

 

[Slimsta]

someone?

 

[David Gould]

You still do not have the correct second derivative. (I think).

 

[Slimsta]

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:
$\displaystyle \Large x''(t)=\frac{ -70t(6+5t^2)-20t(42-35t^2) }{(6+5t^2)^2} \ $$\displaystyle \Large =\frac{ -1260t-1050t^3 }{(6+5t^2)^2} \ $

maybe I am stuck on the factoring of (42-35t^2)... but i just don't know aaaah

 

[David Gould]

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.

 

[zcd]

2nd derivative is still wrong
\frac{dx}{dt}=-7\frac{5t^{2}-6}{(5t^{2}+6)^{2}}
\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}}

 

[Slimsta]

2nd derivative is still wrong
\frac{dx}{dt}=-7\frac{5t^{2}-6}{(5t^{2}+6)^{2}}
\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}}

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