Solution:

1. Time Ratios: The time ratios are given as $1:8:1$. So, if the total time is $t$, the train spends $\frac{t}{10}$ accelerating, $\frac{8t}{10}$ at constant velocity, and $\frac{t}{10}$ decelerating.
2. Velocity-Time Graph:
   • The graph forms a trapezium.
   • The train accelerates linearly from 0 to 60 km/h, holds constant at 60 km/h, and then decelerates back to 0.

   Key points:

   • The area under this graph represents the total distance traveled.
   • The height (maximum velocity) = 60 km/h.
   • The time intervals are in the ratio $1:8:1$.
3. Average Speed:
   The area of the trapezium is given by:

   $$\text{Area} = \frac{1}{2} \times (t_{\text{total}}) \times (\text{initial velocity} + \text{final velocity})$$For the constant velocity portion:

   $$\text{Average speed} = \frac{60 \times (1 + 8 + 1)}{10} = 54 \, \text{km/h}$$

Thus, the average speed is 54 km/h.