Answer: \(v\gt 0\); acceleration \(=-\dfrac1{108}\text{ m s}^{-2}\); \(s=2-\dfrac{2}{(t+1)^2}\); distance \(\dfrac9{200}\text{ m}\).

For \(t\geq0\), \((t+1)^3\gt 0\), so

\(v=\frac4{(t+1)^3}\gt 0.\)

Therefore the particle never changes direction.

The acceleration is

\(a=\frac{dv}{dt}.\)

Since \(v=4(t+1)^{-3}\),

\(a=-12(t+1)^{-4}.\)

When \(t=5\),

\(a=-12(6)^{-4}=-\frac{12}{1296}=-\frac1{108}.\)

So

\(\boxed{a=-\frac1{108}\text{ m s}^{-2}}.\)

For displacement, integrate the velocity:

\(s=\int4(t+1)^{-3}\,dt.\)

This gives

\(s=-2(t+1)^{-2}+C.\)

At \(t=0\), the particle is at \(O\), so \(s=0\). Hence

\(0=-2+C,\)

so \(C=2\). Therefore

\(\boxed{s=2-\frac{2}{(t+1)^2}}.\)

The fourth second is from \(t=3\) to \(t=4\). The distance travelled is

\(s(4)-s(3).\)

Thus

\(\left(2-\frac2{25}\right)-\left(2-\frac2{16}\right) =\frac2{16}-\frac2{25}.\)

So

\(\frac18-\frac2{25}=\frac{25-16}{200}=\frac9{200}.\)

Hence the distance travelled in the fourth second is

\(\boxed{\frac9{200}\text{ m}}.\)