Answer
To move $4 \mathrm{cm}$ from the starting position, the ant takes 1.27 seconds.
Work Step by Step
The velocity of an ant running along the edge of the shelf has been modeled by the function:
\[
v(t)=\left\{\begin{array}{cc}
6 \sqrt{t}-\frac{1}{t} & 1 \leq t \leq 2 \\
5 t & 0 \leq t<1
\end{array}\right.
\]
We will find the time $t_{0}$ at which the ant is $4 \mathrm{cm}$ from its starting position.
If we assume that $t_{0} \leq 1$
\[
\begin{aligned}
\therefore s(t) &=\int_{0}^{1}(5 t) d t=\left.\frac{5 t^{2}}{2}\right|_{0} ^{1}=\frac{5}{2} \\
\therefore & s\left(t_{0}\right)=\frac{5}{2} \approx 2.5
\end{aligned}
\]
So, the time will be $t_{0}>1$
If $t_{0}>1$
\[
\begin{aligned}
\therefore s(t) &=\int_{0}^{t_{0}} v(t) d t \\
&=\int_{1}^{t_{0}}\left(6 \sqrt{t}-\frac{1}{t}\right) d t+\int_{0}^{1} 5 t d t \\
&=\left.\frac{5 t^{2}}{2}\right|_{0} ^{1}+\left.\left(-\ln t+4 t^{\frac{3}{2}}\right)\right|_{1} ^{t_{0}} \\
&=\frac{5}{2}+\left(-\ln t_{0}+4 t_{0}^{\frac{3}{2}}\right)-4
\end{aligned}
\]
\[
\begin{aligned}
&=\int_{1}^{t_{0}}\left(6 \sqrt{t}+\int_{0}^{1} 5 t d t-\frac{1}{t}\right) d t \\
&=\left.\frac{5 t^{2}}{2}\right|_{0} ^{1}+\left.\left(4 t^{\frac{3}{2}}-\ln t\right)\right|_{1} ^{t_{0}} \\
&=-4+\frac{5}{2}+\left(4 t_{0}^{-\ln t_{0}+\frac{3}{2}}\right) \\
&=-\frac{3}{2}-\ln t_{0}+\left(4 t_{0}^{\frac{3}{2}}\right) \\
s\left(t_{0}\right)=4 &=\left(4 t_{0}^{\frac{3}{2}}-\frac{3}{2}-\ln t_{0}\right) \\
\therefore -\ln t_{0}+4 t_{0}^{\frac{3}{2}} &=\frac{11}{2}
\end{aligned}
\]
We find:
\[
t_{0} \approx 1.27
\]
