Question

Из пунк­та А в пункт В, рас­по­ло­жен­ный ниже по те­че­нию реки, от­пра­вил­ся плот. Од­но­вре­мен­но нав­стре­чу ему из пунк­та В вышел катер. Встре­тив плот, катер сразу по­вер­нул и по­плыл назад. Какую часть пути от А до В прой­дет плот к мо­мен­ту воз­вра­ще­ния ка­те­ра в пункт В, если ско­рость ка­те­ра в сто­я­чей воде вчет­ве­ро боль­ше ско­ро­сти те­че­ния реки?

Answer 1

Если скорость течения и плота х, то скорость катера в стоячей воде 4х.
Соответственно, скорость катера сначала против течения 4х-х=3х, затем по течению 4х+х=5х.
Таким образом, до встречи катер прошёл расстояние в 3 раза большее, чем плот. То есть плот прошёл 1/4 расстояния, а катер 3/4. Назад катеру нужно было пройти тоже 3/4, и он уже двигался со скоростью в 5 раз большей, чем у плота. Тогда за это же время плот прошёл расстояние в 5 раз меньше: 3/4 :5=3/20.
всего плот прошёл 1/4+3/20=5/20+3/20=8/20=4/10=0,4 пути.

Answer 2

Problem Analysis

In this problem, a raft is traveling from point A to point B downstream in a river, while a motorboat is traveling from point B to point A upstream. The motorboat encounters the raft, turns around, and heads back to point B. We need to determine the fraction of the distance from A to B that the raft will have traveled by the time the motorboat returns to point B.

Solution

Let's assume that the distance from A to B is represented by the variable `d`. We are given that the speed of the motorboat in still water is four times greater than the speed of the river current.

Let's denote the speed of the river current as `V` and the speed of the motorboat in still water as `Vb`. Therefore, the speed of the motorboat relative to the river current is `Vb - V`.

Since the motorboat turns around and heads back to point B, its speed relative to the river current will be `Vb + V`.

The time it takes for the motorboat to reach the raft and turn around is the same as the time it takes for the raft to travel from A to the point where it meets the motorboat.

Let's denote the time it takes for the motorboat to reach the raft and turn around as `t`. The time it takes for the raft to travel from A to the meeting point is also `t`.

The distance traveled by the motorboat relative to the river current is `(Vb + V) * t`, and the distance traveled by the raft is `V * t`.

Since the distance traveled by the motorboat relative to the river current is equal to the distance traveled by the raft, we can set up the following equation:

`(Vb + V) * t = V * t`

Simplifying the equation, we get:

`Vb + V = V`

Solving for `Vb`, we find:

`Vb = 2V`

Therefore, the speed of the motorboat in still water is twice the speed of the river current.

Now, let's calculate the fraction of the distance from A to B that the raft will have traveled by the time the motorboat returns to point B.

The time it takes for the motorboat to return to point B is `2t`, and the distance traveled by the raft is `V * t`.

Therefore, the fraction of the distance from A to B that the raft will have traveled is:

`(V * t) / d = (V * t) / (2V * t) = 1/2`

So, the raft will have traveled half of the distance from A to B by the time the motorboat returns to point B.

Answer

The raft will have traveled half of the distance from A to B by the time the motorboat returns to point B.