(i) To find the distance travelled in the first 16 s, we divide the motion into two parts: acceleration and constant speed.

For the first 6 s, the sprinter accelerates from rest to 12 m/s. The distance \(s_1\) is given by:

\(s_1 = \frac{1}{2} (0 + 12) \times 6 = 36 \text{ m}\)

For the next 10 s, the sprinter travels at a constant speed of 12 m/s. The distance \(s_2\) is:

\(s_2 = 10 \times 12 = 120 \text{ m}\)

Thus, the total distance for the first 16 s is:

\(36 + 120 = 156 \text{ m}\)

The displacement-time graph is a curve concave up for \(0 < t < 6\) starting at (0, 0) and ending at (6, 36), a straight line with positive gradient from (6, 36) to (16, 156), and a curve concave down from (16, 156) to (20, 200).

(ii) To find \(V\), use the equation for the final 4 s of deceleration:

\(44 = \frac{1}{2} (12 + V) \times 4\)

Solving for \(V\):

\(44 = 2(12 + V)\)

\(44 = 24 + 2V\)

\(20 = 2V\)

\(V = 10\)