Question

СРОЧНО!!!!! Допоможіть розв‘язати задачу!!!! Із двох міст, відстань між якими 529,6км, одночасно назустріч один одному виїхали два потяги й зустрілися через 4год. Швидкість одного з потягів 65,2км/год. Яка швидкість другого поїзда?

Answer 1

Ответ:

ось

Пошаговое объяснение:

Надіюсь, все гарненько видно. Удачі :)

Answer 2

I can help you solve this problem. The user's message is in Ukrainian and it means: "Answer in detail. URGENT!!!!! Help me solve the problem!!!! From two cities, the distance between which is 529.6 km, two trains left at the same time towards each other and met after 4 hours. The speed of one of the trains is 65.2 km/h. What is the speed of the other train?"

To solve this problem, we need to use the formula for the relative speed of two objects moving towards each other:

$$v_r = v_1 + v_2$$

where $v_r$ is the relative speed, $v_1$ is the speed of the first object, and $v_2$ is the speed of the second object.

We also need to use the formula for the distance traveled by an object moving with a constant speed:

$$s = vt$$

where $s$ is the distance, $v$ is the speed, and $t$ is the time.

We know that the distance between the two cities is 529.6 km, and the two trains met after 4 hours. Therefore, we can write the equation:

$$529.6 = (v_1 + v_2) \times 4$$

We also know that the speed of the first train is 65.2 km/h. Therefore, we can substitute $v_1$ with 65.2 and solve for $v_2$:

$$529.6 = (65.2 + v_2) \times 4$$ $$\frac{529.6}{4} = 65.2 + v_2$$ $$132.4 - 65.2 = v_2$$ $$v_2 = 67.2$$

The speed of the second train is 67.2 km/h.

You can check your answer by plugging in the values into the original equation:

$$529.6 = (65.2 + 67.2) \times 4$$ $$529.6 = 132.4 \times 4$$ $$529.6 = 529.6$$

The equation is true, so the answer is correct.

I hope this helps you understand how to solve this problem. If you need more help with math problems, you can use [this online calculator](https://urok.in.ua/content/ss-13-tekstovi-zadachi-na-ruh-519-548) or [this online textbook](https://ua.onlinemschool.com/math/assistance/).