The function given in the question is
[tex]v(t)=\frac{1+\sqrt[]{t+9}}{\sqrt[]{t+9}}[/tex]Concept: To calculate the particle position of the function above, we will have to find the anti-derivative of the function above
[tex]x(t)=\int v(t)[/tex]By substituting the values, we will have
[tex]\begin{gathered} x(t)=\int v(t) \\ x(t)=\int \frac{1+\sqrt[]{t+9}}{\sqrt[]{t+9}} \end{gathered}[/tex]Step 1: Spilt the function using the law of integration of addition
[tex]\int \frac{1+\sqrt[]{t+9}}{\sqrt[]{t+9}}=\int \frac{1}{\sqrt[]{t+9}}+\int \frac{\sqrt[]{t+9}}{\sqrt[]{t+9}}[/tex]Step 2: The equation above becomes
[tex]\int \frac{1}{\sqrt[]{t+9}}+\int \frac{\sqrt[]{t+9}}{\sqrt[]{t+9}}=\int \frac{1}{\sqrt[]{t+9}}+\int 1[/tex]Step 3: Integrate the expression below using integration by substitution
[tex]\begin{gathered} \int \frac{1}{\sqrt[]{t+9}}dt \\ \text{let } \\ u=t+9 \\ \frac{du}{dt}=1 \\ du=dt \end{gathered}[/tex][tex]\begin{gathered} \int \frac{1}{\sqrt[]{u}}du=\int u^{-\frac{1}{2}}du=\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=u^{\frac{1}{2}}\div\frac{1}{2}=2u^{\frac{1}{2}}=2\sqrt[]{u}=2\sqrt[]{t+9} \\ \int 1=t \end{gathered}[/tex][tex]\int \frac{1}{\sqrt[]{t+9}}+\int 1=2\sqrt[]{t+9}+t+c[/tex][tex]x(t)=2\sqrt[]{t+9}+t+c[/tex]To calculate the value of constant c., since it is starting from the origin, we will have
[tex](0,0)=(t,x)[/tex][tex]x(0)=0[/tex][tex]\begin{gathered} x(t)=2\sqrt[]{t+9}+t+c \\ 0=2\sqrt[]{0+9}+0+c \\ 0=2\sqrt[]{9}+c \\ 0=6+c \\ c=0-6 \\ c=-6 \end{gathered}[/tex]Replace the value of c in the function of x(t)
[tex]\begin{gathered} x(t)=2\sqrt[]{t+9}+t+c \\ x(t)=2\sqrt[]{t+9}+t-6 \end{gathered}[/tex]At t=16, we will have the particle position be
[tex]\begin{gathered} x(t)=2\sqrt[]{t+9}+t-6 \\ x(16)=2\sqrt[]{16+9}+16-6 \\ x(16)=2\sqrt[]{25}+10 \\ x(16)=2\times5+10 \\ x(16)=10+10 \\ x(16)=20 \end{gathered}[/tex]