Answer
$\text{displacement = distance = $\frac{204}{25}$}$
Work Step by Step
$\text {It is given that}$
\begin{align}
a(t) = \frac{1}{\sqrt{5t+1}}; v_0 = 2m/s; 0\leq t \leq 3
\end{align}
$\text {First, we have to find v(t):}$
\begin{align}
v(t) = \int a(t) \ dt = \int \frac{1}{\sqrt{5t+1}} \ dt
\end{align}
$\text {Apply u = 5t + 1 $\Rrightarrow$ du =5 dt:}$
\begin{align}
v(t) = \frac{1}{5} \int \frac{1}{\sqrt u} \ du = \frac{2}{5} u^{\frac{1}{2}} = \frac{2}{5} \sqrt {5t+1} + C
\end{align}
$\text {By applying the given initial condition:}$
\begin{align}
v(0) = \frac{2}{5} \sqrt 1 + C = 2 \Rrightarrow C = \frac{8}{5} \Rrightarrow v(t) = \frac{2}{5} \sqrt {5t+1} +\frac{8}{5}
\end{align}
$\text {Now we can calculate the dispalcement and distance:}$
\begin{align}
displacement = \int_0^3 \left(\frac{2}{5} \sqrt {5t+1} +\frac{8}{5} \right) \ dt = \int_0^3 \frac{2}{5} \sqrt {5t+1} \ dt + \frac{24}{5}
\end{align}
$\text {Apply u = 5t + 1 $\Rrightarrow$ du =5 dt:}$
\begin{align}
displacement =\frac{2}{25} &\int_1^{16} \sqrt u \ du + \frac{24}{5} = \frac{2}{25} \left[\frac{2}{3} u^{\frac{3}{2}}\right]_1^{16} + + \frac{24}{5} = \\
& = \frac{2}{25} \times 42 + \frac{24}{5} = \frac{204}{25}
\end{align}
\begin{align}
distance = \int_0^3 |\frac{2}{5} \sqrt {5t+1} +\frac{8}{5} | \ dt = \int_0^3 \left(\frac{2}{5} \sqrt {5t+1} +\frac{8}{5} \right) \ dt = \frac{204}{25}
\end{align}
