A plane 3

A plane left 30 minutes later than the scheduled time and in order to reach the destination 1500 km away in time, it has to increase the speed by 250 km/hr from the usual speed. Find its usual speed.

Correct answer:

v = 750 km/h

Step-by-step explanation:

Δ t = 30 min \to h = 30 : 60 h = 0.5 h s = 1500 km Δ v = 250 km/h s = v \cdot t = (v + Δ v) (t - Δ t) s = v \cdot t - v Δ t + Δ v t - Δ v Δ t - v Δ t + Δ v t - Δ v Δ t = 0 - v Δ t + Δ v t = Δ v Δ t - (s / t) Δ t + Δ v t = Δ v Δ t Δ v \cdot t^{2} - s \cdot Δ t = Δ v \cdot Δ t \cdot t 250 \cdot t^{2} - 1500 \cdot 0.5 = 250 \cdot 0.5 \cdot t 250 t^{2} - 125 t - 750 = 0 250 = 2 \cdot 5^{3} 125 = 5^{3} 750 = 2 \cdot 3 \cdot 5^{3} GCD (250, 125, 750) = 5^{3} = 125 2 t^{2} - t - 6 = 0 a = 2; b = - 1; c = - 6 D = b^{2} - 4 a c = 1^{2} - 4 \cdot 2 \cdot (- 6) = 49 D > 0 t_{1, 2} = \frac{- b \pm D}{2 a} = \frac{1 \pm 4 9}{4} t_{1, 2} = \frac{1 \pm 7}{4} t_{1, 2} = 0.25 \pm 1.75 t_{1} = 2 t_{2} = - 1.5 t = t_{1} = 2 h v = s / t = 1500 / 2 = 750 = 750 km/h Verifying Solution: s_{1} = v \cdot t = 750 \cdot 2 = 1500 km s_{2} = (v + Δ v) \cdot (t - Δ t) = (750 + 250) \cdot (2 - 0.5) = 1500 km

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