Answer
The distance $=\frac{296}{27} m$
The displacement $=\frac{296}{27} m$
Work Step by Step
$$
\text { You have } a(t)=\frac{1}{\sqrt{1+3 t}}, \quad \frac{4}{3}=v_{0}, \quad 1 \leq t \leq 5
$$
We will find the velocity :
$$
\begin{array}{l}
v(t)=\int \frac{1}{\sqrt{1+3 t}} d t=\frac{2}{3} \sqrt{1+3 t}+C \\
v(0)=\frac{4}{3}=\frac{2}{3}+C \Rightarrow C=\frac{2}{3} \\
\therefore \frac{2}{3} \sqrt{1+3 t}+\frac{2}{3}=v(t)
\end{array}
$$
The displacement is :
$$
s(t)=\int_{1}^{5}\left(\frac{2}{3} \sqrt{3 t+1}+\frac{2}{3}\right), d t=\left.\left[\frac{4}{27} \sqrt{(3 t+1)^{3}}+\frac{2}{3} t\right]\right|_{1} ^{5}=\frac{346}{27}-\frac{50}{27}=\left[\frac{296}{27} m\right]
$$
The distance :
$$
\begin{aligned}
\int_{1}^{5}\left|\frac{2}{3} \sqrt{1+3 t}+\frac{2}{3}\right| d t &=\int_{1}^{5}\left(\frac{2}{3} \sqrt{1+3 t}+\frac{2}{3}\right) d t \\
&=\left[\frac{296}{27} m\right]
\end{aligned}
$$
