The particle's velocity is given by the derivative of its position:
[tex]s'(t)=\ln5t+t\dfrac5{5t}=\ln5t+1[/tex]
which will be 0 when
[tex]\ln5t+1=0\implies\ln5t=-1\implies5t=e^{-1}\implies t=\dfrac1{5e}[/tex]
The acceleration is given by the second derivative, so we have
[tex]s''(t)=\dfrac5{5t}=\dfrac1t[/tex]
and at [tex]t=\dfrac1{5e}[/tex], the acceleration will be
[tex]s''\left(\dfrac1{5e}\right)=\dfrac1{\frac1{5e}}=5e[/tex]
