Higher Derivatives - Acceleration help

  • Thread starter Slimsta
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  • #1
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Homework Statement


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.

Homework 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)

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?
 

Answers and Replies

  • #2
zcd
200
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You solved the derivatives incorrectly
 
  • #3
190
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You solved the derivatives incorrectly

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

now what?
 
Last edited:
  • #4
190
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someone?
 
  • #5
You still do not have the correct second derivative. (I think).
 
  • #6
190
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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
 
  • #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.
 
  • #8
zcd
200
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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]
 
  • #9
190
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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]

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
 

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