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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?