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 train stay for 2 minutes in every station?

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

Result

t1 =  23.375 min
t2 =  20.944 min
v1 =  0 km/h
v2 =  0 km/h

Solution:

t1=60 (187/120187/160)=1878=23.375=23.375  min t_{ 1 } = 60 \cdot \ (187/120-187/160) = \dfrac{ 187 }{ 8 } = 23.375 = 23.375 \ \text{ min }
a1=0.8 3600 3600/1000=10368 a2=0.5 3600 3600/1000=6480 t11=120/a1=120/10368=54320.0116 t21=160/a1=160/10368=53240.0154 t12=120/a2=120/6480=1540.0185 t22=160/a2=160/6480=2810.0247  s11=a1 t112/2=10368 0.01162/2=25360.6944 s21=a1 t212/2=10368 0.01542/2=1091.1111  s21=a2 t122/2=6480 0.01852/2=1091.1111 s22=a2 t222/2=6480 0.02472/2=160811.9753  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)=6314801.3146 t2=60 (T1T2)=60 (1.66371.3146)=3771820.9444=20.944  min a_{ 1 } = 0.8 \cdot \ 3600 \cdot \ 3600/1000 = 10368 \ \\ a_{ 2 } = 0.5 \cdot \ 3600 \cdot \ 3600/1000 = 6480 \ \\ t_{ 11 } = 120/a_{ 1 } = 120/10368 = \dfrac{ 5 }{ 432 } \doteq 0.0116 \ \\ t_{ 21 } = 160/a_{ 1 } = 160/10368 = \dfrac{ 5 }{ 324 } \doteq 0.0154 \ \\ t_{ 12 } = 120/a_{ 2 } = 120/6480 = \dfrac{ 1 }{ 54 } \doteq 0.0185 \ \\ t_{ 22 } = 160/a_{ 2 } = 160/6480 = \dfrac{ 2 }{ 81 } \doteq 0.0247 \ \\ \ \\ s_{ 11 } = a_{ 1 } \cdot \ t_{ 11 }^{ 2 }/2 = 10368 \cdot \ 0.0116^{ 2 }/2 = \dfrac{ 25 }{ 36 } \doteq 0.6944 \ \\ s_{ 21 } = a_{ 1 } \cdot \ t_{ 21 }^{ 2 }/2 = 10368 \cdot \ 0.0154^{ 2 }/2 = \dfrac{ 10 }{ 9 } \doteq 1.1111 \ \\ \ \\ s_{ 21 } = a_{ 2 } \cdot \ t_{ 12 }^{ 2 }/2 = 6480 \cdot \ 0.0185^{ 2 }/2 = \dfrac{ 10 }{ 9 } \doteq 1.1111 \ \\ s_{ 22 } = a_{ 2 } \cdot \ t_{ 22 }^{ 2 }/2 = 6480 \cdot \ 0.0247^{ 2 }/2 = \dfrac{ 160 }{ 81 } \doteq 1.9753 \ \\ \ \\ T_{ 1 } = (7 \cdot \ (t_{ 11 }+t_{ 12 })+ (187-7 \cdot \ (s_{ 11 }+s_{ 21 }))/120) = (7 \cdot \ (0.0116+0.0185)+ (187-7 \cdot \ (0.6944+1.1111))/120) \doteq 1.6637 \ \\ T_{ 2 } = (7 \cdot \ (t_{ 21 }+t_{ 22 })+ (187-7 \cdot \ (s_{ 21 }+s_{ 22 }))/160) = (7 \cdot \ (0.0154+0.0247)+ (187-7 \cdot \ (1.1111+1.9753))/160) = \dfrac{ 631 }{ 480 } \doteq 1.3146 \ \\ t_{ 2 } = 60 \cdot \ (T_{ 1 }-T_{ 2 }) = 60 \cdot \ (1.6637-1.3146) = \dfrac{ 377 }{ 18 } \doteq 20.9444 = 20.944 \ \text{ min }
v1=187/(T1+6/30)=187/(1.6637+6/30)100.3403=0  km/h v_{ 1 } = 187/(T_{ 1 }+6/30) = 187/(1.6637+6/30) \doteq 100.3403 = 0 \ \text{ km/h }
v2=187/(T2+6/30)=187/(1.3146+6/30)123.4663=0  km/h v_{ 2 } = 187/(T_{ 2 }+6/30) = 187/(1.3146+6/30) \doteq 123.4663 = 0 \ \text{ km/h }



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