Answer: \(a=-\frac{1+\sqrt3}{2}\); displacement \(=\sqrt2-1\); \(a=-s-1\).

Set up the integral using the correct limits or antiderivative first. When definite integration is used, evaluate the upper limit and lower limit separately before subtracting.

Differentiate \(v=\cos t-\sin t\):

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

At \(t=\frac{\pi}{3}\),

\(a=-\frac{\sqrt3}{2}-\frac12=-\frac{1+\sqrt3}{2}.\)

The particle changes direction when \(v=0\), so

\(\cos t-\sin t=0\), giving \(\tan t=1\).

The first such time after \(t=0\) is \(t=\frac{\pi}{4}\).

Since \(v=\frac{\mathrm ds}{\mathrm dt}\),

\(s=\int(\cos t-\sin t)\,\mathrm dt=\sin t+\cos t+C.\)

The particle passes through \(O\) when \(t=0\), so \(s=0\) at \(t=0\). Therefore \(0=0+1+C\), and \(C=-1\).

Thus \(s=\sin t+\cos t-1\).

At \(t=\frac{\pi}{4}\),

\(s=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}-1=\sqrt2-1.\)

Finally, since \(s+1=\sin t+\cos t\),

\(a=-\sin t-\cos t=-(s+1)=-s-1.\)

This gives the required answer: \(a=-\frac{1+\sqrt3}{2}\); displacement \(=\sqrt2-1\); \(a=-s-1\).