Two trucks

Two trucks left cities A and B against each other and met after an hour. The first car came to B 27 minutes later than the second car to A. Calculate the car speed if the distance between cities A and B is 90 km.

Correct answer:

v₁ = 40 km/h
v₂ = 50 km/h

Step-by-step explanation:

s = 90 km t_{1} = 1 h t_{2} = 27 min \to h = 27 : 60 h = 0.45 h s_{1} + s_{2} = s v_{1} \cdot t_{1} + v_{2} \cdot t_{1} = s v_{1} \cdot (t + t_{2}) = s v_{2} \cdot t = s 1 \cdot v_{1} + 1 \cdot v_{2} = 90 v_{1} \cdot (t + 27 / 60) = 90 v_{2} \cdot t = 90 v_{1} + v_{2} = 90 v_{1} \cdot t + v_{1} \cdot 27 / 60 = 90 t = 90 / v_{2} v_{1} + v_{2} = 90 v_{1} \cdot 90 / v_{2} + v_{1} \cdot 27 / 60 = 90 v_{1} \cdot 90 + v_{2} \cdot v_{1} \cdot 27 / 60 = 90 \cdot v_{2} (90 - v_{2}) \cdot 90 + v_{2} \cdot (90 - v_{2}) \cdot 27 / 60 = 90 \cdot v_{2} (90 - x) \cdot 90 \cdot 60 + x \cdot (90 - x) \cdot 27 = 60 \cdot 90 \cdot x (90 - x) \cdot 90 \cdot 60 + x \cdot (90 - x) \cdot 27 = 60 \cdot 90 \cdot x - 27 x^{2} - 8370 x + 486000 = 0 27 x^{2} + 8370 x - 486000 = 0 27 = 3^{3} 8370 = 2 \cdot 3^{3} \cdot 5 \cdot 31 486000 = 2^{4} \cdot 3^{5} \cdot 5^{3} GCD (27, 8370, 486000) = 3^{3} = 27 x^{2} + 310 x - 18000 = 0 a = 1; b = 310; c = - 18000 D = b^{2} - 4 a c = 31 0^{2} - 4 \cdot 1 \cdot (- 18000) = 168100 D > 0 x_{1, 2} = \frac{- b \pm D}{2 a} = \frac{- 3 1 0 \pm 1 6 8 1 0 0}{2} x_{1, 2} = \frac{- 3 1 0 \pm 4 1 0}{2} x_{1, 2} = - 155 \pm 205 x_{1} = 50 x_{2} = - 360 v_{2} = x_{1} = 50 v_{1} = 90 - v_{2} = 90 - 50 = 40 km/h

Our quadratic equation calculator calculates it.

v_{2} = 50 = 50 km/h

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