The train was delayed by 60 minutes and the delay on the stretch of 36 km was eliminated

| March 20, 2021 | education

The train was delayed by 60 minutes and the delay on the stretch of 36 km was eliminated, increasing the speed by 4 km / h. What is the initial speed of the train?

These tasks: Δt is the delay time of the presented train (Δt = 60 min = 1 h); L – haul length, track (L = 36 km); ΔV – increase in train speed (ΔV = 4 km / h).

To find out the initial speed of the presented train, consider the expression: L / Vк – L / Vн = Δt and L / (Vн + ΔV) – L / Vн = Δt.

Let’s calculate: 36 / (Vn + 4) – 36 / Vn = 1 | * Vн.

36Vn / (Vn + 4) – 36 = Vn | * (Vн + 4).

36Vn – 36 * (Vn + 4) = Vn * (Vn + 4).

36Vn – 36Vn + 144 = Vn2 + 4Vn.

Vn ^ 2 + 4Vn – 144 = 0.

D = 4 ^ 2 – 4 * 1 * (-144) = 592.

t1 = (-4 – √592) / (2 * 1) ≈ -14.17 km / h (incorrect).

t2 = (-4 + √592) / (2 * 1) ≈ 10.17 km / h (correct).

Answer: The initial speed of the presented train, according to the calculation, was 10.17 km / h.



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