(ii) 
Answer:
(i)
Distance-time graph
Average speed 
In this problem, total distance travelled = 50 km
Total time taken 10:00 AM to 11:30 AM
= 1 hour 30 minutes 
Now average speed 
(iii) We know, speed = slope of distance-time graph.
The greater the slope, the greater is the speed.
From the graph it is clear that slope of distance-time
graph is maximum between 10:00 AM to 10:30 AM, so the train was travelling at the
highest speed during this interval of time.
(iv) The part CD of the graph has minimum slope, so the
train had minimum speed between 11:00 AM and 11:15 AM. Thus, the train had
slowed down between 40 km and 42 km.
(v) Speed between 10:40 AM to 11:00 AM
(1x5)
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