Usual speed

If the usual speed is reduced by 5 km per hour, a train takes 2 hours more to cover a distance of 300 km. Find the usual speed.

Final Answer:

v = 30 km/h

Δ v = 5 km/h Δ t = 2 h s = 300 km s = v \cdot t = (v - Δ t) \cdot (t + Δ t) (v - Δ v) \cdot (t + Δ t) = s v t - Δ v \cdot t + Δ t \cdot v - Δ t Δ v = s - Δ v \cdot t + Δ t \cdot v - Δ t Δ v = 0 t = \frac{Δ t \cdot v - Δ t Δ v}{Δ v} = \frac{Δ t}{Δ v} \cdot v - Δ t k = \frac{Δ t}{Δ v} = \frac{2}{5} = 0.41 /km t = k \cdot v - Δ t s = v \cdot t = v \cdot (k \cdot v - Δ t) s = v \cdot (k \cdot v - Δ t) 300 = v \cdot (0.4 \cdot v - 2) - 0.39999999999998 v^{2} + 2 v + 300 = 0 0.39999999999998 v^{2} - 2 v - 300 = 0 a = 0.4; b = - 2; c = - 300 D = b^{2} - 4 a c = 2^{2} - 4 \cdot 0.4 \cdot (- 300) = 484 D > 0 v_{1, 2} = \frac{- b \pm D}{2 a} = \frac{2 \pm 4 8 4}{0 . 8} v_{1, 2} = 2.5 \pm 27.5 v_{1} = 30 v_{2} = - 25 v = v_{1} = 30 = 30 km/h Verifying Solution: v_{2} = v - Δ v = 30 - 5 = 25 km/h t_{1} = s / v = 300 / 30 = 10 h t_{2} = s / v_{2} = 300 / 25 = 12 h

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