Speed Adjustment

By increasing the speed of his car by 15 km/h, a person covers a 300 km distance in an hour less than before. What was the original speed?

Final Answer:

v = 60 km/h

Δ v = 15 km/h s = 300 km/h Δ t = 1 h s = v \cdot t s = (v + Δ v) \cdot (t - Δ t) s = v \cdot t - v \cdot Δ t + Δ v \cdot t - Δ v Δ t s = s - v \cdot Δ t + Δ v \cdot t - Δ v Δ t Δ v Δ t = - v \cdot Δ t + Δ v \cdot t Δ v Δ t = - v \cdot Δ t + Δ v \cdot (s / v) Δ v Δ t v = - v^{2} \cdot Δ t + Δ v \cdot s v^{2} \cdot Δ t + Δ v \cdot Δ t \cdot v - Δ v \cdot s = 0 v^{2} \cdot 1 + 15 \cdot 1 \cdot v - 15 \cdot 300 = 0 v^{2} + 15 v - 4500 = 0 a = 1; b = 15; c = - 4500 D = b^{2} - 4 a c = 1 5^{2} - 4 \cdot 1 \cdot (- 4500) = 18225 D > 0 v_{1, 2} = \frac{- b \pm D}{2 a} = \frac{- 1 5 \pm 1 8 2 2 5}{2} v_{1, 2} = \frac{- 1 5 \pm 1 3 5}{2} v_{1, 2} = - 7.5 \pm 67.5 v_{1} = 60 v_{2} = - 75 v > 0 v = v_{1} = 60 = 60 km/h Verifying Solution: t_{1} = s / v = 300 / 60 = 5 h t_{2} = \frac{s}{v + Δ v} = \frac{3 0 0}{6 0 + 1 5} = 4 h

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