Answer:
[tex]s(t)=(ln|7|+ln|2|)\,ft[/tex]
Step-by-step explanation:
Acceleration is second derivative of distance and are related as:
[tex]a(t)=\frac{d^2s}{dt^2}\\\\\frac{d^2s}{dt^2}=\frac{-1}{(t+2)^2}\\[/tex]
Integrating both sides w.r.to t
[tex]v(t)=\frac{ds}{dt}=\frac{1}{t+2} +C\\[/tex]
Using initial value
[tex]v(0)=\frac{1}{2}\\\\\frac{1}{2}=\frac{1}{0+2} +C\\\\C=0\\\\\frac{ds}{dt}=\frac{1}{t+2}[/tex]
We have to calculate the distance covered in time interval [0,5], so:
[tex]\int\limits^5_0 \frac{ds}{dt}=\int\limits^5_0 {\frac{1}{t+2}} \, dt\\\\s(t)=[ln|t+2|]^5_0\\\\s(t)=ln|5+2|+ln|0+2|\\\\s(t)=(ln|7|+ln|2|)\,ft[/tex]
