Answer
The maximum average speed from one start-up to the next is $~~11.7~m/s$
Work Step by Step
As calculated in part (b), the travel time between stations is $49.0~s$
Then the train stops for $20~s$ for a total time of $69.0~s$
We can find the maximum average speed from one start-up to the next:
$v_{ave} = \frac{\Delta x}{\Delta t} = \frac{806~m}{69.0~s} = 11.7~m/s$
The maximum average speed from one start-up to the next is $~~11.7~m/s$
