Question

Два велосипедиста одновременно отправились в 220 километровый пробег. Первый ехал со скоростью на 1км/ч больше й, чем скорость второго,и прибыл к финишу на 2 часа раньше второго. Найдите скорость велосипедиста, пришедшего к финишу вторым.

Answer 1

Скорость 2-ого = х (км/ч)
Скорость 1-ого = (х + 1) км/ч
Время 2-ого = (220 /х) час
Время 1-ого = 220/ (х + 1) час
Уравнение:
220 /х - 220/(х + 1) = 2
220*(х + 1) - 220х = 2х (х + 1) 
220х + 220 - 220х = 2х^2 + 2x-
2x^2 + 2x - 220 = 0
x^2 + x - 110 = 0
D = 1 - 4*(-110) = 1 + 440 = 441;   √D = 21
x1 = (- 1 + 21)/ 2 = 20/2 = 10
x2 = (-1 - 21)/2 = - 11 (не подходит по условию)
Ответ: 10 км/час - скорость 2-ого велосипедиста, пришедшего к финишу
                               вторым.

Answer 2

Problem Analysis

We are given that two cyclists simultaneously started a 220-kilometer race. The first cyclist traveled at a speed that was 1 km/h faster than the second cyclist. The first cyclist arrived at the finish line 2 hours earlier than the second cyclist. We need to find the speed of the cyclist who arrived second.

Solution

Let's assume the speed of the second cyclist is x km/h. Therefore, the speed of the first cyclist is x + 1 km/h.

We can use the formula distance = speed × time to calculate the time taken by each cyclist to complete the race.

For the first cyclist: - Distance = 220 km - Speed = x + 1 km/h - Time = 220 / (x + 1) hours

For the second cyclist: - Distance = 220 km - Speed = x km/h - Time = 220 / x hours

According to the problem, the first cyclist arrived at the finish line 2 hours earlier than the second cyclist. Therefore, we can set up the following equation:

220 / (x + 1) = 220 / x + 2

To solve this equation, we can cross-multiply and simplify:

220x = 220(x + 1) + 2x(x + 1)

Simplifying further:

220x = 220x + 220 + 2x^2 + 2x

Combining like terms:

0 = 220 + 2x^2 + 2x

Rearranging the equation:

2x^2 + 2x + 220 = 0

Now we can solve this quadratic equation to find the value of x.

Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a), where a = 2, b = 2, and c = 220.

Calculating the discriminant: b^2 - 4ac = 2^2 - 4(2)(220) = 4 - 1760 = -1756

Since the discriminant is negative, the quadratic equation has no real solutions. This means there is no valid speed for the second cyclist that satisfies the given conditions.

Therefore, there is no solution to this problem.

Conclusion

The problem does not have a valid solution. The given conditions of the race cannot be satisfied, and there is no speed for the second cyclist that would result in the first cyclist arriving 2 hours earlier at the finish line.