Question

Помогите решить физику по теме кинематика. Ослик выбежал из домика через 30 минут после Винни пуха и через 10 минут их скорости стали одинаковыми. Через какое время ослик догонит Винни пуха? Движение тел считать равноускоренным из состояния покоя.

Answer 1

Могут быть ошибки, поэтому советую проверить)
Сначала выбегает Винни пух. А через пол часа - ослик.
За 10 минут(или 1/6 часа) он развил скорость, равную скорости Винни пуха.
То есть к 30 минутам прибавляем 10 минут и получаем 40 минут, то есть то время, за которое Винни пух набрал скорость 2/3a₁. По условию, Именно эту скорость развил ослик за 10 мин. Теперь можно найти ускорение a₂ ослика: (2/3a₁)/(1/6 ч) = 4a₁. То есть ускорение ослика в 4 раза больше ускорения Винни пуха.
Ослик догонит Винни пуха тогда, когда пути пройденные ими обоими сравняются, то есть 4a₁t²/2 = a₁t²/2 + 0,5²a₁/2 ( выражение 0,5²a₁/2 - это тот путь, который прошел Винни пух, когда ослик еще не стартовал)
Получаем: 4a₁t² = a₁t²+0,25a₁;
4a₁t² = a₁(t²+0,25)
a₁ никакого вклада не вносит, поэтому сокращаем его:
4t² = t²+0,25
3t² = 0,25
t² = 1/12
t ≈ 0,29 ч
Или, что примерно равно, 17 минут

Answer 2

Problem Analysis

To solve this problem, we need to determine the time it takes for the donkey (ослик) to catch up to Winnie the Pooh (Винни пух). We are given that the donkey starts running 30 minutes after Winnie the Pooh and that their speeds become equal 10 minutes later. We also know that the motion of the objects is uniformly accelerated.

Solution

Let's break down the problem step by step:

1. Convert the given time intervals into seconds for easier calculations: - The donkey starts running 30 minutes after Winnie the Pooh, which is equal to 30 * 60 = 1800 seconds. - The speeds become equal 10 minutes later, which is equal to 10 * 60 = 600 seconds.

2. Let's assume that the initial velocity of the donkey is 0 m/s (since it starts from rest) and the acceleration is constant for both the donkey and Winnie the Pooh.

3. Let's denote the time it takes for the donkey to catch up to Winnie the Pooh as t.

4. Using the equations of uniformly accelerated motion, we can set up the following equations:

- For the donkey: - s_d = u_d * t + (1/2) * a * t^2 (Equation 1)

- For Winnie the Pooh: - s_w = u_w * (t + 1800) + (1/2) * a * (t + 1800)^2 (Equation 2)

- Since the speeds become equal after 10 minutes, we can also set up the equation: - u_d = u_w (Equation 3)

- We need to solve these equations to find the value of t.

5. Substituting Equation 3 into Equations 1 and 2, we get: - s_d = u_w * t + (1/2) * a * t^2 (Equation 4) - s_w = u_w * (t + 1800) + (1/2) * a * (t + 1800)^2 (Equation 5)

6. Rearranging Equation 5, we get: - s_w = u_w * t + 1800 * u_w + (1/2) * a * t^2 + a * t * 1800 + (1/2) * a * 1800^2

7. Simplifying Equation 5, we get: - s_w = u_w * t + 1800 * u_w + (1/2) * a * t^2 + 1800 * a * t + (1/2) * a * 1800^2

8. Subtracting Equation 4 from Equation 5, we get: - s_w - s_d = 1800 * u_w + 1800 * a * t + (1/2) * a * 1800^2

9. Rearranging the equation, we get: - 1800 * u_w + 1800 * a * t + (1/2) * a * 1800^2 = s_w - s_d

10. Since the speeds become equal after 10 minutes, the distances covered by both the donkey and Winnie the Pooh are the same. Therefore, s_w - s_d = 0.

11. Substituting this into the equation, we get: - 1800 * u_w + 1800 * a * t + (1/2) * a * 1800^2 = 0

12. Rearranging the equation, we get: - 1800 * u_w + 1800 * a * t = - (1/2) * a * 1800^2

13. Dividing both sides of the equation by 1800, we get: - u_w + a * t = - (1/2) * a * 1800

14. Since the donkey starts from rest, u_w = 0. Therefore, the equation becomes: - a * t = - (1/2) * a * 1800

15. Dividing both sides of the equation by a, we get: - t = - (1/2) * 1800

16. Simplifying the equation, we get: - t = -900

17. Since time cannot be negative, we discard this solution.

Conclusion

Based on the given information, it is not possible for the donkey to catch up to Winnie the Pooh. The negative time value obtained in the calculations indicates that the donkey will not be able to catch up to Winnie the Pooh.

Please note that the solution assumes constant acceleration and uniform motion, as stated in the problem.