Distance travelled while accelerating = 4.3 km
Vmax = sqrt(2aX) = 74.01 m/s
Average velocity = 37.0 m/s
Total travel time between stops
= 8600 m/37.0 m/s = 232 s
Average velocity including one stop:
(8600 m)/(232 + 50) = 30.5 m/s
A subway train is designed with a maximum acceleration of ±0.637 m/s2 , which allows for both passenger safety and comfort.
(a) If subway stations are 8.6 km apart, what is the maximum velocity that can be obtained between stations?
(b) How long does it take to travel between two stations?
(c) The train stops for a total of 50 s at each station. What is the overall average velocity of the train from station to station?
I got part a right with an answer of 74.01 m/s.
When I did parts b & c, my answers were incorrect.
Any help would be greatly apreciated!
1 answer
1 answer