Question

Моторная лодка прошла против течения реки 140км и вернулась в пункт отправления, затратив на обратный путь на 4 часа меньше. Наядите скорость лодки в неподвижной воде, если скорость течения равна 2 км в час

Answer 1

Обозначим  скорость лодки в неподвижной воде за x, тогда скорость лодки по течения = x+2, скокрость против течения = x-2.

Время, за которое лодка прошла против течения реки 140км: t1=140/(x-2).

Время на обратный путь: t2=140/(x+2)

Так как обратно она потратила на 4 часа меньше времени, то t1-t2=4. Получаем уравнение:

140/(x-2)-140/(x+2)=4. Приведем к общему знаменателю:

$\frac{140x+280-140x+280}{(x-2)(x+2)}$=4 приведем подобные слагаемые в числителе: $\frac{560}{(x-2)(x+2)}$=4

Умножим обе части на знаменатель дроби:

$560=4x^{2}-16$ 

$4x^{2}=576$ 

$x^{2}=144$ 

x1=12, x2=-12. Однако скорость не может быть отрицательным числом, поэтому остается x=12.

Ответ: скорость лодки в неподвижной воде 12 км/ч

Answer 2

Problem Analysis

We are given that a motorboat traveled 140 km against the current of a river and then returned to the starting point, spending 4 hours less on the return journey. We need to find the speed of the boat in still water, given that the speed of the current is 2 km/h.

Solution

Let's assume the speed of the boat in still water is v km/h. Since the boat is traveling against the current, its effective speed will be reduced by the speed of the current. Therefore, the boat's speed against the current is v - 2 km/h.

We are given that the boat traveled 140 km against the current and spent 4 hours less on the return journey. Let's calculate the time taken for the return journey.

The time taken for the forward journey is given by: time_forward = distance / speed_forward where distance = 140 km and speed_forward = v - 2 km/h.

The time taken for the return journey is given by: time_return = distance / speed_return where distance = 140 km and speed_return = v + 2 km/h.

We are given that the time taken for the return journey is 4 hours less than the time taken for the forward journey. Therefore, we can write the equation: time_return = time_forward - 4

Substituting the values, we get: 140 / (v + 2) = 140 / (v - 2) - 4

Simplifying the equation, we get: 140 / (v + 2) - 140 / (v - 2) = 4

To solve this equation, we can cross-multiply and simplify: 140(v - 2) - 140(v + 2) = 4(v + 2)(v - 2)

Simplifying further, we get: 140v - 280 - 140v - 280 = 4(v^2 - 4)

Simplifying again, we get: -560 = 4v^2 - 16

Rearranging the equation, we get a quadratic equation: 4v^2 = 544

Dividing both sides by 4, we get: v^2 = 136

Taking the square root of both sides, we get: v = √136

Therefore, the speed of the boat in still water is approximately 11.66 km/h.

Answer

The speed of the motorboat in still water is approximately 11.66 km/h.