Answer:
[tex]a(\frac{1}{5e})=5e[/tex]
Step-by-step explanation:
we are given equation for position function as
[tex]s(t)=tln(5t)[/tex]
Since, we have to find acceleration
For finding acceleration , we will find second derivative
[tex]s'(t)=\frac{d}{dt}\left(t\ln \left(5t\right)\right)[/tex]
[tex]=\frac{d}{dt}\left(t\right)\ln \left(5t\right)+\frac{d}{dt}\left(\ln \left(5t\right)\right)t[/tex]
[tex]=1\cdot \ln \left(5t\right)+\frac{1}{t}t[/tex]
[tex]s'(t)=\ln \left(5t\right)+1[/tex]
now, we can find derivative again
[tex]s''(t)=\frac{d}{dt}\left(\ln \left(5t\right)+1\right)[/tex]
[tex]=\frac{d}{dt}\left(\ln \left(5t\right)\right)+\frac{d}{dt}\left(1\right)[/tex]
[tex]=\frac{1}{t}+0[/tex]
[tex]a(t)=\frac{1}{t}[/tex]
Firstly, we will set velocity =0
and then we can solve for t
[tex]v(t)=s'(t)=\ln \left(5t\right)+1=0[/tex]
we get
[tex]t=\frac{1}{5e}[/tex]
now, we can plug that into acceleration
and we get
[tex]a(\frac{1}{5e})=\frac{1}{\frac{1}{5e}}[/tex]
[tex]a(\frac{1}{5e})=5e[/tex]
