Answer: \(v=(2t+1)^{1/2}+7\); \(s=\dfrac13(2t+1)^{3/2}+7t+\dfrac{11}{3}\).

(a) Acceleration is

\(a=(2t+1)^{-\frac12}.\)

Since acceleration is \(\dfrac{dv}{dt}\), integrate with respect to \(t\):

\(v=\int(2t+1)^{-\frac12}\,dt.\)

This gives

\(v=(2t+1)^{\frac12}+C.\)

When \(t=0\), \(v=8\), so

\(8=1+C.\)

Hence \(C=7\), and

\(v=(2t+1)^{\frac12}+7.\)

(b) Displacement \(s\) satisfies

\(\frac{ds}{dt}=v.\)

So

\(s=\int\left((2t+1)^{\frac12}+7\right)\,dt.\)

Thus

\(s=\frac13(2t+1)^{\frac32}+7t+C.\)

When \(t=0\), the displacement is \(4\), so

\(4=\frac13+C.\)

Therefore

\(C=\frac{11}{3}.\)

Hence the displacement from \(O\) at time \(t\) is

\(s=\frac13(2t+1)^{\frac32}+7t+\frac{11}{3}.\)