<s>[INST] A particle moves along the x-axis find t when acceleration of the particle equals 0 A particle moves along the x-axis, its position at time t is given by
$x(t)= \frac{3t}{6+8t^2}$, $t≥0$,
where t is measured in seconds and x is in meters.
Find time at which acceleration equals 0.
I got the answer 0.5 and 0 but apparently i got the answer wrong. I know I need to use derivative twice. Can someone please help me
Calculation
$velocity = \frac {-3(4t^2-3)}{2(4t^2+3)^2}$
$Acceleration = 36x(16x^4+8x^2-3)$
answer 0.5 and 0 was found through wolfram
 [/INST] The acceleration is expressed by the second derivative. 
We have $$x(t)=\frac{3t}{6+8t^2}$$ $$x'(t)=\frac{3(6+8t^2)-3t\cdot 16t}{(6+8t^2)^2}=\frac{18+24t^2-48t^2}{(6+8t^2)^2}=\frac{18-24t^2}{(6+8t^2)^2}$$ $$ x''(t)=\frac{-48t(6+8t^2)^2-2(18-24t^2)(6+8t^2)(16t)}{(6+8t^2)^4}=\frac{(6+8t^2) \left [-48t(6+8t^2)-2(18-24t^2)(16t)\right ]}{(6+8t^2)^4}=\frac{ 16\left [-3t(6+8t^2)-2t(18-24t^2)\right ]}{(2(3+4t^2))^3}=\frac{ 16\left [-6t(3+4t^2)-6t(6-8t^2)\right ]}{8(3+4t^2)^3}=\frac{-12t\left [3+4t^2+6-8t^2\right ] }{(3+4t^2)^3}=\frac{-12t\left [9-4t^2\right ] }{(3+4t^2)^3}$$ 
Therefore, the acceleration of the particle equals $0$ at time $t=0$ seconds and $t=\frac{3}{2}$ seconds. 
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