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 at 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:

v_{1} = 120 km/h v_{2} = 160 km/h s = 167 km n = 6 + 1 = 7 t_{1} = 60 \cdot (s / v_{1} - s / v_{2}) = 60 \cdot (167 / 120 - 167 / 160) = \frac{1 6 7}{8} min = 20.875 min

a_{1} = 0.9 \cdot 360 0^{2} / 1000 = 11664 km/h^{2} a_{2} = 0.35 \cdot 360 0^{2} / 1000 = 4536 km/h^{2} t_{11} = v_{1} / a_{1} = 120 / 11664 = \frac{5}{4 8 6} ≐ 0.0103 h t_{21} = v_{2} / a_{1} = 160 / 11664 = \frac{1 0}{7 2 9} ≐ 0.0137 h t_{12} = v_{1} / a_{2} = 120 / 4536 = \frac{5}{1 8 9} ≐ 0.0265 h t_{22} = v_{2} / a_{2} = 160 / 4536 = \frac{2 0}{5 6 7} ≐ 0.0353 h s_{11} = \frac{1}{2} \cdot a_{1} \cdot t_{11}^{2} = \frac{1}{2} \cdot 11664 \cdot 0.010 3^{2} = \frac{5 0}{8 1} ≐ 0.6173 km s_{12} = \frac{1}{2} \cdot a_{1} \cdot t_{21}^{2} = \frac{1}{2} \cdot 11664 \cdot 0.013 7^{2} = \frac{8 0 0}{7 2 9} ≐ 1.0974 km s_{21} = \frac{1}{2} \cdot a_{2} \cdot t_{12}^{2} = \frac{1}{2} \cdot 4536 \cdot 0.026 5^{2} = \frac{1 0 0}{6 3} ≐ 1.5873 km s_{22} = \frac{1}{2} \cdot a_{2} \cdot t_{22}^{2} = \frac{1}{2} \cdot 4536 \cdot 0.035 3^{2} = \frac{1 6 0 0}{5 6 7} ≐ 2.8219 km T_{1} = (n \cdot (t_{11} + t_{12}) + (s - n \cdot (s_{11} + s_{12})) / v_{1}) = (7 \cdot (0.0103 + 0.0265) + (167 - 7 \cdot (0.6173 + 1.0974)) / 120) ≐ 1.5488 h T_{2} = (n \cdot (t_{21} + t_{22}) + (s - n \cdot (s_{21} + s_{22})) / v_{2}) = (7 \cdot (0.0137 + 0.0353) + (167 - 7 \cdot (1.5873 + 2.8219)) / 160) ≐ 1.1938 h t_{2} = 60 \cdot (T_{1} - T_{2}) = 60 \cdot (1.5488 - 1.1938) = 21.3037 min

t_{8} = 2 min \to h = 2 : 60 h = 0.03333 h w_{1} = s / (T_{1} + (n - 1) \cdot t_{8}) = 167 / (1.5488 + (7 - 1) \cdot 0.0333) = 95.4916 km/h

w_{2} = s / (T_{2} + (n - 1) \cdot t_{8}) = 167 / (1.1938 + (7 - 1) \cdot 0.0333) = 119.8177 km/h

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