Bad weather

An aircraft was slowed down during a 600 km flight due to bad weather. Its average speed was reduced by 200 km/hr from its usual speed, and the flight time increased by 30 minutes. Find the scheduled duration of the flight.

Correct answer:

t = 1 h

Step-by-step explanation:

s = 600 km Δ = 200 km/h T = 30 min \to h = 30 : 60 h = 0.5 h v_{2} = v_{1} - Δ t_{2} = t + T t = s / v_{1} t_{2} = s / v_{2} = t + T = s / v_{1} + T s / v_{2} = s / v_{1} + T s / (v_{1} - Δ) = s / v_{1} + T s \cdot v_{1} = s \cdot (v_{1} - Δ) + T \cdot (v_{1} - Δ) \cdot v_{1} s \cdot x = s \cdot (x - Δ) + T \cdot (x - Δ) \cdot x 600 \cdot x = 600 \cdot (x - 200) + 0.5 \cdot (x - 200) \cdot x - 0.5 x^{2} + 100 x + 120000 = 0 0.5 x^{2} - 100 x - 120000 = 0 a = 0.5; b = - 100; c = - 120000 D = b^{2} - 4 a c = 10 0^{2} - 4 \cdot 0.5 \cdot (- 120000) = 250000 D > 0 x_{1, 2} = \frac{- b \pm D}{2 a} = \frac{1 0 0 \pm 2 5 0 0 0 0}{1} = 100 \pm 5001 x_{1, 2} = 100 \pm 500 x_{1} = 600 x_{2} = - 400 v > 0 v_{1} = x_{1} = 600 km/h v_{2} = v_{1} - Δ = 600 - 200 = 400 km/h t = s / v_{1} = 600 / 600 = 1 h Verifying Solution: t_{2} = s / v_{2} = 600 / 400 = \frac{3}{2} = 1.5 h

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