Timetable

If a train travelling between two cities increases its scheduled speed by 5 km/h, it arrives 20 minutes early. If it decreases its speed by 5 km/h, it arrives 25 minutes late.

How long is the route between the cities?

Final Answer:

x = 150 km

Step-by-step explanation:

t_{1} = 20 min \to h = 20 : 60 h = 0.33333 h t_{2} = 25 min \to h = 25 : 60 h = 0.41667 h Δ v = 5 km/h x = v t x = (v + Δ v) \cdot (t - t_{1}) x = (v - Δ v) \cdot (t + t_{2}) v t = (v + Δ v) \cdot (t - t_{1}) = v t - v t_{1} + Δ v t - Δ v t_{1} Δ v t = v t_{1} + Δ v t_{1} v t = (v - Δ v) \cdot (t + t_{2}) = v t + v t_{2} - Δ v t - Δ v t_{2} Δ v t = v t_{2} - Δ v t_{2} v \cdot t_{1} + Δ v \cdot t_{1} = v \cdot t_{2} - Δ v \cdot t_{2} v \cdot t_{1} - v \cdot t_{2} = - Δ v \cdot t_{1} - Δ v \cdot t_{2} v \cdot (t_{2} - t_{1}) = Δ v \cdot (t_{1} + t_{2}) v = Δ v \cdot \frac{t _{1} + t _{2}}{t _{2} - t _{1}} = 5 \cdot \frac{0 . 3 3 3 3 + 0 . 4 1 6 7}{0 . 4 1 6 7 - 0 . 3 3 3 3} = 45 km/h t = \frac{v \cdot t _{1} + Δ v \cdot t _{1}}{Δ v} = \frac{4 5 \cdot 0 . 3 3 3 3 + 5 \cdot 0 . 3 3 3 3}{5} = \frac{1 0}{3} ≐ 3.3333 h x = v \cdot t = 45 \cdot 3.3333 = 150 km

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Timetable

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