Answer: distance \(218.5\text{ m}\); final deceleration \(0.32\text{ m s}^{-2}\).

(a) The velocity-time graph has three straight-line sections:

\((0,3)\to(60,3)\to(75,1.6)\to(80,0).\)

So it is horizontal at \(v=3\) for the first 60 seconds, then slopes down to \(v=1.6\) at \(t=75\), and then slopes down to \(v=0\) at \(t=80\).

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

For the first 60 seconds, the area is

\(3\times60=180.\)

For the next 15 seconds, the trapezium has parallel sides \(3\) and \(1.6\), so its area is

\(\frac12(3+1.6)(15)=34.5.\)

For the final 5 seconds, the triangle has height \(1.6\), so its area is

\(\frac12(5)(1.6)=4.\)

Therefore the distance between \(O\) and \(A\) is

\(180+34.5+4=218.5.\)

So the distance is

\(218.5\text{ m}.\)

(c) In the last 5 seconds, the velocity decreases from \(1.6\text{ m s}^{-1}\) to \(0\text{ m s}^{-1}\).

The deceleration is

\(\frac{1.6}{5}=0.32.\)

So the deceleration is

\(0.32\text{ m s}^{-2}.\)