The acceleration of the particle is given by the formula mentioned below:
[tex]a=\frac{d^2s}{dt^2}[/tex]Differentiate the position vector with respect to t.
[tex]\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}[/tex]Differentiate both sides of the obtained equation with respect to t.
[tex]\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}[/tex]Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.
[tex]\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}[/tex]The initial position is obtained at t=0. Substitute t=0 in the given position function.
[tex]\begin{gathered} s(0)=-23\times0+65 \\ =65 \end{gathered}[/tex]