Reconstruction of the corridor

Calculate how many minutes the travel time will be reduced on a 213 km long railway corridor if the maximum speed increases from 120 km/h to 160 km/h.

Calculate how many minutes the travel time will be shortened if the train must stop at 6 stations, decelerating uniformly at 0.8 m/s² into each station and accelerating at 0.45 m/s² out of each station.

What is the average speed before and after the upgrade if the train stops for 2 minutes at every station?

Assume that the distances between stops are long enough for the train to always reach the speed limit.

Final Answer:

t₁ = 26.625 min
t₂ = 24.3745 min
w₁ = 102.0347 km/h
w₂ = 126.6889 km/h

Step-by-step explanation:

v_{1} = 120 km/h v_{2} = 160 km/h s = 213 km n = 6 + 1 = 7 t_{1} = 60 \cdot (s / v_{1} - s / v_{2}) = 60 \cdot (213 / 120 - 213 / 160) = 26.625 min

a_{1} = 0.8 \cdot 360 0^{2} / 1000 = 10368 km/h^{2} a_{2} = 0.45 \cdot 360 0^{2} / 1000 = 5832 km/h^{2} t_{11} = v_{1} / a_{1} = 120 / 10368 = \frac{5}{4 3 2} ≐ 0.0116 h t_{21} = v_{2} / a_{1} = 160 / 10368 = \frac{5}{3 2 4} ≐ 0.0154 h t_{12} = v_{1} / a_{2} = 120 / 5832 = \frac{5}{2 4 3} ≐ 0.0206 h t_{22} = v_{2} / a_{2} = 160 / 5832 = \frac{2 0}{7 2 9} ≐ 0.0274 h s_{11} = \frac{1}{2} \cdot a_{1} \cdot t_{11}^{2} = \frac{1}{2} \cdot 10368 \cdot 0.011 6^{2} = \frac{2 5}{3 6} ≐ 0.6944 km s_{12} = \frac{1}{2} \cdot a_{1} \cdot t_{21}^{2} = \frac{1}{2} \cdot 10368 \cdot 0.015 4^{2} = \frac{1 0 0}{8 1} ≐ 1.2346 km s_{21} = \frac{1}{2} \cdot a_{2} \cdot t_{12}^{2} = \frac{1}{2} \cdot 5832 \cdot 0.020 6^{2} = \frac{1 0 0}{8 1} ≐ 1.2346 km s_{22} = \frac{1}{2} \cdot a_{2} \cdot t_{22}^{2} = \frac{1}{2} \cdot 5832 \cdot 0.027 4^{2} = \frac{1 6 0 0}{7 2 9} ≐ 2.1948 km T_{1} = (n \cdot (t_{11} + t_{12}) + (s - n \cdot (s_{11} + s_{12})) / v_{1}) = (7 \cdot (0.0116 + 0.0206) + (213 - 7 \cdot (0.6944 + 1.2346)) / 120) ≐ 1.8875 h T_{2} = (n \cdot (t_{21} + t_{22}) + (s - n \cdot (s_{21} + s_{22})) / v_{2}) = (7 \cdot (0.0154 + 0.0274) + (213 - 7 \cdot (1.2346 + 2.1948)) / 160) ≐ 1.4813 h t_{2} = 60 \cdot (T_{1} - T_{2}) = 60 \cdot (1.8875 - 1.4813) = 24.3745 min

t_{8} = 2 min \to h = 2 : 60 h = 0.03333 h w_{1} = s / (T_{1} + (n - 1) \cdot t_{8}) = 213 / (1.8875 + (7 - 1) \cdot 0.0333) = 102.0347 km/h

w_{2} = s / (T_{2} + (n - 1) \cdot t_{8}) = 213 / (1.4813 + (7 - 1) \cdot 0.0333) = 126.6889 km/h

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