Reconstruction of the corridor

Calculate how many minutes will be reduced to travel 187 km long railway corridor, where the maximum speed increases from 120 km/h to 160 km/h.

Calculate how many minutes will shorten travel time if we consider that the train must stop at 6 stations, each station with the equally slow 0.8 m/s2 and from station accelerates 0.5 m/s2?

What is the average speed before and after reconstruction if the train stays for 2 minutes in every station?

Consider that the intermediate stops are so long that the train will always reach the speed limit.

Correct answer:

t1 =  23.375 min
t2 =  20.9444 min
v1 =  0 km/h
v2 =  0 km/h

Step-by-step explanation:

t1=60 (187/120187/160)=23.375 min
a1=0.8 3600 3600/1000=10368 a2=0.5 3600 3600/1000=6480 t11=120/a1=120/10368=43250.0116 t21=160/a1=160/10368=32450.0154 t12=120/a2=120/6480=5410.0185 t22=160/a2=160/6480=8120.0247  s11=a1 t112/2=10368 0.01162/2=36250.6944 s21=a1 t212/2=10368 0.01542/2=9101.1111  s21=a2 t122/2=6480 0.01852/2=9101.1111 s22=a2 t222/2=6480 0.02472/2=811601.9753  T1=(7 (t11+t12)+(1877 (s11+s21))/120)=(7 (0.0116+0.0185)+(1877 (0.6944+1.1111))/120)1.6637 T2=(7 (t21+t22)+(1877 (s21+s22))/160)=(7 (0.0154+0.0247)+(1877 (1.1111+1.9753))/160)=4806311.3146 t2=60 (T1T2)=60 (1.66371.3146)=20.9444 min
v1=187/(T1+6/30)=187/(1.6637+6/30)100.3403=0 km/h
v2=187/(T2+6/30)=187/(1.3146+6/30)123.4663=0 km/h

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