Answer: (a) \(v=6\) for \(0\leqslant t\leqslant5\), then \(v=-3\) for \(5\leqslant t\leqslant15\). (b) A straight line from \((0,-8)\) through \((10,0)\) to \((20,8)\).

Start by taking the key information from the graph or diagram, such as intercepts, turning points, amplitudes, periods, or intersection points. Then use the relevant algebra to justify the result.

In part (a), during the first 5 seconds the displacement from \(A\) is

\(6\times5=30\text{ m}.\)

The particle returns to \(A\) over the next 10 seconds, so its velocity is

\(\frac{-30}{10}=-3\text{ m s}^{-1}.\)

So the velocity-time graph has a horizontal line at \(v=6\) from \(t=0\) to \(t=5\), followed by a horizontal line at \(v=-3\) from \(t=5\) to \(t=15\).

In part (b), the displacement from \(B\) over the 20 seconds is zero, so the signed area under the velocity-time graph is zero.

Since the acceleration is constant, the graph is a straight line. It starts at \((0,-8)\). For the total signed area to be zero over \(0\leqslant t\leqslant20\), the line must cross the \(t\)-axis halfway at \(t=10\) and end at \(v=8\) when \(t=20\).

Therefore the graph is the straight line through \((0,-8)\), \((10,0)\), and \((20,8)\).

The result has been obtained using the exact condition from the question, so no additional solutions are introduced.