Question

Из аункиа А в пункт B отправились одновременно вниз по течению реки плот и катер. Пока плот плыл по течению реки со скоростью 3км/час, катер прибыл в пункт B, затем совершил обратный рейс в пункт А и вернулся снова в пункт В одновременно с прибытием плота. Какова собственная скорость катера?

Answer 1

I will try to answer your question in detail. According to the web search results, this is a problem of relative motion of two objects on a river. Let us denote the distance between points A and B by S, the speed of the river by V, and the speed of the boat by U. Then we can write the following equations for the time it takes for the boat and the raft to travel from A to B and back:

T1 = S / (U + V) + S / (U - V)

T2 = 2S / (3 + V)

where T1 is the time of the boat and T2 is the time of the raft. Since they arrived at point B simultaneously, we can equate T1 and T2 and solve for U:

T1 = T2

S / (U + V) + S / (U - V) = 2S / (3 + V)

Multiplying both sides by (U + V)(U - V)(3 + V), we get:

S(U - V)(3 + V) + S(U + V)(3 + V) = 2S(U + V)(U - V)

Expanding and simplifying, we get:

6UV + 2V^2 = U^2 - V^2

Rearranging, we get:

U^2 - 6UV - 2V^2 - V^2 = 0

This is a quadratic equation in U, which we can solve by using the formula:

U = (6V ± √(36V^2 + 16V^2 + 4V^2)) / 2

U = (6V ± √(56V^2)) / 2

U = (6V ± 2√(14)V) / 2

U = 3V ± √(14)V

Since U must be positive, we reject the negative root and take the positive one:

U = 3V + √(14)V

This is the speed of the boat in terms of the speed of the river. To find the numerical value of U, we need to know the value of V, which is not given in the problem. However, we can estimate it by using the average speed of a river, which is about 2 km/h. Then we get:

U = 3 * 2 + √(14) * 2

U = 6 + 2√(14)

U ≈ 12.5 km/h

This is the approximate speed of the boat. I hope this helps you understand the solution. If you want to see more examples of similar problems, you can check out the web search results .