Reconstruction of the corridor

Calculate how many minutes will be reduced to travel a 167 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 0.9 m/s2 and from station accelerates 0.35 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 =  20.875 min
t2 =  21.3037 min
w1 =  95.4916 km/h
w2 =  119.8177 km/h

Step-by-step explanation:

v1=120 km/h v2=160 km/h s=167 km n=6+1=7  t1=60 (s/v1s/v2)=60 (167/120167/160)=8167 min=20.875 min
a1=0.9 36002/1000=11664 km/h2 a2=0.35 36002/1000=4536 km/h2  t11=v1/a1=120/11664=48650.0103 h t21=v2/a1=160/11664=729100.0137 h t12=v1/a2=120/4536=18950.0265 h t22=v2/a2=160/4536=567200.0353 h  s11=21 a1 t112=21 11664 0.01032=81500.6173 km s12=21 a1 t212=21 11664 0.01372=7298001.0974 km  s21=21 a2 t122=21 4536 0.02652=631001.5873 km s22=21 a2 t222=21 4536 0.03532=56716002.8219 km  T1=(n (t11+t12)+(sn (s11+s12))/v1)=(7 (0.0103+0.0265)+(1677 (0.6173+1.0974))/120)1.5488 h T2=(n (t21+t22)+(sn (s21+s22))/v2)=(7 (0.0137+0.0353)+(1677 (1.5873+2.8219))/160)1.1938 h t2=60 (T1T2)=60 (1.54881.1938)=21.3037 min
t8=2 min h=2:60  h=0.03333 h  w1=s/(T1+(n1) t8)=167/(1.5488+(71) 0.0333)=95.4916 km/h
w2=s/(T2+(n1) t8)=167/(1.1938+(71) 0.0333)=119.8177 km/h

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