Question

2. Моторная лодка прошла по течению реки 25 км, затем против течения реки 3 км затратив на весь путь 2 часа. Какова скорость лодки если известно, что она превосходит 2 кмч, а скорость течения реки равна 3 км/ ч​

Answer 1

Ответ:  12 км/ч .

Собственная скорость лодки равна х км/ч .

Скорость лодки по течению = (х+3) км/ч .

Так как по течению лодка прошла 25 км, то время, в течение которого  лодка двигалась по течению, равно   25/(х+3) ч.

Скорость лодки против течения = (х-3) км/ч .

Так как против течения лодка прошла 3 км, то время, в течение которого  лодка двигалась против течения, равно   3/(х-3) ч.

В сумме лодка двигалась в течение 2 часов .

Составим уравнение.

$\displaystyle \frac{25}{x+3}+\frac{3}{x-3}=2\ \ ,\ \ \frac{25}{x+3}+\frac{3}{x-3}-2=0\ \ ,\\\\\\\frac{25(x-3)+3(x+3)-2(x+3)(x-3)}{(x+3)(x-3)}=0\ \ ,\\\\\\25x-75+3x+9-2(x^2-9)=0\ \ ,\ \ \ 2x^2-28x+48=0\ \ ,\\\\x^2-14x+24=0\ \ ,\ \ x_1=2\ ,\ x_2=12\ \ (teorema\ Vieta)$

Так как по условию собственная скорость лодки превосходит 2 км/ч , то выбираем число 12 км/ч .

Answer 2

Ответ: 12 км/ч

Пошаговое объяснение:

собственная скорость лодки х км/ч, скорость по течению (х+3) км/ч, скорость против течения (х-3) км/ч, время по течению

25/(х+3) ч, против течения 3/(х-3) ч,

отсюда уравнение.

25/(х+3) +3/(х-3)=2;

25*(х-3)+3*(х+3)=2*(х²-9); 25х-75+3х+9=2*(х²-9);

14х-33=(х²-9); х²-9-14х+33=0;  х²-14х+24=0; по Виету х=2; не подходит по условию. т.к. она превосходит 2км/ч, значит, х=12- единственный корень. скорость лодки 12 км/ч

Answer 3

Problem Analysis

We are given the following information: - The motorboat traveled downstream for 25 km. - The motorboat then traveled upstream for 3 km. - The total time taken for the entire journey was 2 hours. - The speed of the boat is known to be greater than 2 km/h. - The speed of the river current is 3 km/h.

We need to find the speed of the boat.

Solution

Let's assume the speed of the boat is B km/h.

When the boat is traveling downstream, its effective speed is the sum of its own speed and the speed of the river current. So, the effective speed downstream is B + 3 km/h.

When the boat is traveling upstream, its effective speed is the difference between its own speed and the speed of the river current. So, the effective speed upstream is B - 3 km/h.

We can use the formula distance = speed × time to calculate the time taken for each leg of the journey.

The time taken to travel downstream for 25 km is given by 25 / (B + 3) hours.

The time taken to travel upstream for 3 km is given by 3 / (B - 3) hours.

The total time taken for the entire journey is 2 hours.

Using the above information, we can set up the following equation:

25 / (B + 3) + 3 / (B - 3) = 2

Now, let's solve this equation to find the value of B.

Calculation

To solve the equation, we can start by multiplying both sides of the equation by the least common multiple (LCM) of the denominators to eliminate the fractions.

The LCM of (B + 3) and (B - 3) is (B + 3)(B - 3).

Multiplying both sides of the equation by (B + 3)(B - 3), we get:

25(B - 3) + 3(B + 3) = 2(B + 3)(B - 3)

Expanding and simplifying the equation:

25B - 75 + 3B + 9 = 2(B^2 - 9)

28B - 66 = 2B^2 - 18

Rearranging the equation:

2B^2 - 28B + 48 = 0

Dividing both sides of the equation by 2:

B^2 - 14B + 24 = 0

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

Using the quadratic formula: B = (-b ± √(b^2 - 4ac)) / (2a), where a = 1, b = -14, and c = 24.

Calculating the discriminant: b^2 - 4ac = (-14)^2 - 4(1)(24) = 196 - 96 = 100

Since the discriminant is positive, we have two real solutions for B.

Using the quadratic formula:

B = (-(-14) ± √(100)) / (2(1))

Simplifying:

B = (14 ± 10) / 2

This gives us two possible values for B:

B1 = (14 + 10) / 2 = 12 km/h

B2 = (14 - 10) / 2 = 2 km/h

However, we know that the speed of the boat is greater than 2 km/h, so the valid solution is B = 12 km/h.

Answer

The speed of the boat is 12 km/h.

Please note that the above solution assumes a constant speed for the boat and the river current. It also assumes that the boat maintains the same speed throughout the journey.