(i) The distance travelled by the elevator is the area under the velocity-time graph. The graph consists of a trapezium with a base of 24 s and a height of 0.4 m/s. The area is calculated as:

\(s = \frac{1}{2} \times 5 \times 0.4 + 19 \times 0.4 + \frac{1}{2} \times 4 \times 0.4 = 9.4 \text{ m}\)

(ii) The acceleration during the first stage is the slope of the line from 0 to 5 s:

\(a = \frac{0.4}{5} = 0.08 \text{ m/s}^2\)

The deceleration during the third stage is the slope of the line from 24 to 28 s:

\(a = \frac{0.4}{4} = 0.1 \text{ m/s}^2\)

(iii) Using Newton's second law for the elevator and box system:

\(T - (800 + 100)g = (800 + 100)a\)

\(T - 900g = 900a\)

For the first stage, \(a = 0.08 \text{ m/s}^2\):

\(T = 900 \times 9.8 + 900 \times 0.08 = 9072 \text{ N}\)

For the second stage, \(a = 0\):

\(T = 900 \times 9.8 = 9000 \text{ N}\)

For the third stage, \(a = -0.1 \text{ m/s}^2\):

\(T = 900 \times 9.8 - 900 \times 0.1 = 8910 \text{ N}\)

(iv) For the force on the box, using \(R - 100g = 100a\):

Greatest force (first stage):

\(R = 100 \times 9.8 + 100 \times 0.08 = 1008 \text{ N}\)

Least force (third stage):

\(R = 100 \times 9.8 - 100 \times 0.1 = 990 \text{ N}\)