Question

Путь из А в В велосипедист проехал с постоянной скоростью. На обратном пути он вдвое увеличил скорость, но, проехав половину пути, из-за дождя вынужден был снизить скорость в 4 раза. Какой путь — из А в Б или обратно — он проехал быстрее?

Answer 1

Путь из А в В
S
V-скорость
$t= \frac{S}{V}$

Путь из В в А
первая половина пути
2V-увеличенная вдвое скорость
$\frac{S}{2}$ -половина пути
$t1= \frac{ \frac{S}{2} }{2V} = \frac{S}{4V} = \frac{1}{4} * \frac{S}{V} =0.25 \frac{S}{V} =0.25 t$
вторая половина пути
$\frac{2V}{4} = \frac{V}{2}$-скорость на второй половине пути,сниженная в 4 раза
$\frac{S}{2}$ -половина пути
$t2= \frac{ \frac{S}{2} }{ \frac{V}{2} } = \frac{S}{V}t$

Время,затраченное на поездку из В в А
$t1+t2 \\ \\ t1+t2=0.25t+t=1.25t$

Сравнивая,видим
1.25 t>t

Делаем выводы:
на обратный путь он затратил времени больше.

Ответ:путь из А в В он проехал быстрее.

Answer 2

Problem Analysis

We are given that a cyclist traveled from point A to point B at a constant speed. On the return journey, the cyclist doubled their speed but had to reduce their speed by a factor of 4 when they reached the halfway point due to rain. We need to determine whether the cyclist traveled faster from A to B or from B to A.

Solution

To solve this problem, we can compare the time taken for each leg of the journey. The cyclist will travel faster on the leg where they took less time.

Let's assume the distance from A to B is D.

1. Time taken from A to B: Since the cyclist traveled at a constant speed, the time taken from A to B can be calculated using the formula: time = distance / speed. Let's denote the speed on this leg as S1.

2. Time taken from B to A: On the return journey, the cyclist doubled their speed. So the speed on this leg is 2S1. However, when the cyclist reached the halfway point, they had to reduce their speed by a factor of 4. So the effective speed for the second half of the journey is (2S1) / 4 = S1/2. The time taken from B to A can be calculated as the sum of the time taken for the first half of the journey at a speed of 2S1 and the time taken for the second half of the journey at a speed of S1/2.

To determine which leg was faster, we need to compare the time taken from A to B with the time taken from B to A.

Calculation

Let's calculate the time taken for each leg of the journey.

1. Time taken from A to B: - Distance: D - Speed: S1 - Time taken: D / S1

2. Time taken from B to A: - Distance for the first half of the journey: D/2 - Speed for the first half of the journey: 2S1 - Time taken for the first half of the journey: (D/2) / (2S1) = D / (4S1) - Distance for the second half of the journey: D/2 - Speed for the second half of the journey: S1/2 - Time taken for the second half of the journey: (D/2) / (S1/2) = D / S1

Comparing the time taken from A to B (D / S1) with the time taken from B to A (D / S1), we can see that both legs took the same amount of time. Therefore, the cyclist traveled at the same speed in both directions.

Conclusion

The cyclist traveled at the same speed from point A to point B and from point B to point A.