Answer: \(V=55\), acceleration \(0\); \(v=3\sin2t+10\), \(s=-\frac32\cos2t+10t+\frac32\).

(a)(i) The distance travelled is the area under the velocity-time graph.

From \(0\) to \(20\) seconds, the triangular area is

\(\frac12(20)(30)=300.\)

From \(20\) to \(60\) seconds, the rectangular area is

\(40(30)=1200.\)

From \(60\) to \(90\) seconds, the trapezium area is

\(\frac12(30)(30+V)=15(30+V).\)

The total distance is \(2775\), so

\(300+1200+15(30+V)=2775.\)

Thus

\(1950+15V=2775,\)

and hence

\(V=55.\)

(a)(ii) At \(t=40\), the graph is horizontal, so the acceleration is

\(0.\)

(b) Since acceleration is

\(a=6\cos2t,\)

integrate to find velocity:

\(v=\int 6\cos2t\,dt=3\sin2t+C.\)

When \(t=0\), \(v=10\), so \(C=10\). Therefore

\(v=3\sin2t+10.\)

Now integrate velocity to find displacement:

\(s=\int(3\sin2t+10)\,dt =-\frac32\cos2t+10t+C.\)

When \(t=0\), \(s=0\), so

\(0=-\frac32+C.\)

Hence \(C=\frac32\), and

\(s=-\frac32\cos2t+10t+\frac32.\)