# Two cars left point A to point B simultaneously. The first drove at a constant speed all the way. The second traveled the first

Two cars left point A to point B simultaneously. The first drove at a constant speed all the way. The second traveled the first half of the journey at a speed lower than the first by 6 km / h, and the second half of the journey at 56 km / h, as a result of which he arrived at point B simultaneously with the first car. Find the speed of the first car if it is known to be greater than 45 km / h.

Since the path traveled by the cars is the same, we will take it as 1. If the speed of the first car v1 is X, then the speed of the second car on the first half of the path v2 is (X -6). Let’s make an equation according to the formula “speed * time = distance”, expressing time: 0.5: (X-6) + 0.5: 56 = 1: X 0.5 * 56 * X + 0.5 X * (X- 6) = 56 * (X-6) 28X + 0.5 X² – ​​3X – 56X + 336 = 0 0.5 X² -31X + 336 = 0 Find the discriminant of the equation: D = b² – 4ac = (-31) ² – 4 * 0.5 * 336 = 961 – 672 = 289 = 17² Discriminant> 0, the equation has 2 roots. X1 = (-b + √D) / 2a X2 = (-b -√D) / 2a X1 = (31 +17) / 2 * 0.5 = 48/1 = 48 X2 = (31 – 17) / 2 * 0.5 = 14/1 = 14 According to the conditions of the problem, the vehicle speed is more than 45 km / h, so X2 is not a solution to the problem. Answer: 48 km / h. – the speed of the first car.

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