The least time will be if our train moves with the acceleration 5 km/h/s . Calculate the distance necessary to accelerate the 🚆 from zero to its maximum speed:
d=2avmax2​​=2⋅5⋅1000/3600(90⋅1000/3600)2​=225 m. This will take time of
t=vmax​/a=90/5=42 s. The rest of the distance the train will do with its maximum speed:
tr​=vmax​D−d​=90⋅1000/360050000−225​=1991 s. The total time will be
T=t+tr​=33.9 min.
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