Answer: \(v=5+\cos t\gt 0\); distance \(=\dfrac{5\pi}{6}+1-\dfrac{\sqrt3}{2}\approx2.75\text{ m}\); acceleration \(=-\sin4\approx0.757\text{ m s}^{-2}\); \(t=\pi\).

Differentiate displacement to find velocity:

\(v=\frac{dx}{dt}=5+\cos t.\)

Since \(-1\leq\cos t\leq1\),

\(4\leq5+\cos t\leq6.\)

So \(v\) is always positive, and the particle is never at rest.

Because the particle never changes direction, the distance travelled is the change in displacement:

\(x\left(\frac{\pi}{2}\right)-x\left(\frac{\pi}{3}\right).\)

Now

\(x\left(\frac{\pi}{2}\right)=5\left(\frac{\pi}{2}\right)+\sin\frac{\pi}{2} =\frac{5\pi}{2}+1\)

and

\(x\left(\frac{\pi}{3}\right)=5\left(\frac{\pi}{3}\right)+\sin\frac{\pi}{3} =\frac{5\pi}{3}+\frac{\sqrt3}{2}.\)

Therefore the distance is

\(\frac{5\pi}{2}+1-\left(\frac{5\pi}{3}+\frac{\sqrt3}{2}\right) =\frac{5\pi}{6}+1-\frac{\sqrt3}{2}.\)

So

\(\boxed{\frac{5\pi}{6}+1-\frac{\sqrt3}{2}\approx2.75\text{ m}}.\)

The acceleration is

\(a=\frac{dv}{dt}=-\sin t.\)

At \(t=4\),

\(a=-\sin4\approx0.757.\)

Thus

\(\boxed{a\approx0.757\text{ m s}^{-2}}.\)

The velocity \(v=5+\cos t\) is first at its minimum when \(\cos t=-1\), which first occurs at

\(\boxed{t=\pi}.\)