Answer: \(T=3\) and the acceleration during the last 4 seconds is \(-1.5\text{ m s}^{-2}\).

During the first \(T\) seconds, the acceleration is \(2\text{ m s}^{-2}\). Since the graph starts from rest, the velocity at time \(T\) is

\(2T.\)

The total distance travelled is the area under the velocity-time graph. From the diagram, this area is a trapezium with vertical height \(2T\), lower parallel side \(8\), and upper parallel side \(4-T\). Therefore

\(\dfrac12(8+4-T)(2T)=27.\)

Simplifying,

\(T(12-T)=27.\)

So

\(T^2-12T+27=0.\)

Factorising,

\((T-3)(T-9)=0.\)

From the diagram \(T\lt4\), so the valid value is

\(T=3.\)

In the last \(4\) seconds, the velocity decreases from \(2T\) to \(0\). Since \(T=3\), the velocity at the start of the last part is

\(2T=6\text{ m s}^{-1}.\)

The acceleration is the gradient of the velocity-time graph:

\(\dfrac{0-6}{8-4}=-\dfrac64=-1.5.\)

Therefore the acceleration during the last \(4\) seconds is

\(-1.5\text{ m s}^{-2}.\)