Calculate (i) Acceleration from O to A (ii) Retardation State the distance covered by Mr Hara while moving at (i) constant speed (ii) Acceleration What was the distance covered between point B and C

At point O: initial speed $v_i = 0$ m/s, initial time $t_i = 0$ s.  
At point A: final speed $v_f = 20$ m/s, final time $t_f = 6$ s.

$$
a = \frac{v_f - v_i}{t_f - t_i}
$$
$$
a = \frac{20\, \text{m/s} - 0\, \text{m/s}}{6\, \text{s} - 0\, \text{s}}
$$
$$
a = \frac{20}{6}\, \text{m/s}^2
$$
$$
a = \frac{10}{3}\, \text{m/s}^2
$$
$$
a \approx 3.33\, \text{m/s}^2
$$
The acceleration from O to A is $\boxed{\text{3.33 m/s}^2}$.

The graph shows segments of increasing speed (OA, BC, DE) and constant speed (AB, CD). There are no segments where the speed decreases.  
There is no retardation shown in the given graph.

Speed = 20 m/s, Time = $10\, \text{s} - 6\, \text{s} = 4\, \text{s}$.  
Distance$_{AB} = \text{speed} \times \text{time} = 20\, \text{m/s} \times 4\, \text{s} = 80\, \text{m}$.

Speed = 40 m/s, Time = $18\, \text{s} - 14\, \text{s} = 4\, \text{s}$.  
Distance$_{CD} = \text{speed} \times \text{time} = 40\, \text{m/s} \times 4\, \text{s} = 160\, \text{m}$.

Total distance at constant speed = Distance$_{AB}$ + Distance$_{CD} = 80\, \text{m} + 160\, \text{m} = 240\, \text{m}$.  
The distance covered at constant speed is $\boxed{\text{240 m}}$.

Base = 6 s, Height = 20 m/s.  
Distance$_{OA} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6\, \text{s} \times 20\, \text{m/s} = 60\, \text{m}$.

Parallel sides $v_1 = 20$ m/s, $v_2 = 40$ m/s. Height (time) = $14\, \text{s} - 10\, \text{s} = 4\, \text{s}$.  
Distance$_{BC} = \frac{1}{2} \times (v_1 + v_2) \times \text{time} = \frac{1}{2} \times (20\, \text{m/s} + 40\, \text{m/s}) \times 4\, \text{s} = \frac{1}{2} \times 60\, \text{m/s} \times 4\, \text{s} = 120\, \text{m}$.

Parallel sides $v_1 = 40$ m/s, $v_2 = 80$ m/s. Height (time) = $30\, \text{s} - 18\, \text{s} = 12\, \text{s}$.  
Distance$_{DE} = \frac{1}{2} \times (v_1 + v_2) \times \text{time} = \frac{1}{2} \times (40\, \text{m/s} + 80\, \text{m/s}) \times 12\, \text{s} = \frac{1}{2} \times 120\, \text{m/s} \times 12\, \text{s} = 720\, \text{m}$.

Total distance during acceleration = Distance$_{OA}$ + Distance$_{BC}$ + Distance$_{DE} = 60\, \text{m} + 120\, \text{m} + 720\, \text{m} = 900\, \text{m}$.  
The distance covered during acceleration is $\boxed{\text{900 m}}$.