Question

Маша, проиграв соревнование в скорости Мише, предложила соревноваться по-другому:сначала она уедет вперед на 200м,а только потом Миша начнет её догонять. Сможет ли Миша догнать Машу, если весь путь, как и раньше, составит 1км, а скорость передвижения у них будет той же самой: 250 м/мин и 200 м/мин соответственно?

Answer 1

Разница скоростей равна 250 - 200 = 50 м/мин 200 метров Миша проедет за 200/50 = 4 мин . Маша 800 метров проедет за 800 / 200 = 4 мин . Значит они приедут одновременно

Answer 2

1 км = 1000 м
1000 - 200 = 800 м  осталось пробежать Маше
800: 200 = 4 мин порбежит Маша весь путь
1000: 250 = 4 мин пробежит Миша весь путь
Миша догонит Машу на финише 

Answer 3

Problem Analysis

Masha and Misha are competing in a race. Masha suggests a different approach: she will start 200m ahead, and then Misha will try to catch up with her. The total distance of the race is still 1km, and both Masha and Misha maintain the same speeds as before: Masha at 250m/min and Misha at 200m/min. The question is whether Misha will be able to catch up with Masha.

Solution

To determine if Misha can catch up with Masha, we need to compare the time it takes for each of them to cover the respective distances.

Let's calculate the time it takes for Masha to cover the distance of 200m: Time taken by Masha = Distance / Speed = 200m / 250m/min = 0.8 minutes.

Now, let's calculate the time it takes for Misha to cover the remaining distance of 800m (1km - 200m): Time taken by Misha = Distance / Speed = 800m / 200m/min = 4 minutes.

Since Masha starts 0.8 minutes ahead of Misha, we need to compare their times to see if Misha can catch up.

Masha's time: 0.8 minutes Misha's time: 4 minutes

Since Masha's time is less than Misha's time, Misha will not be able to catch up with Masha.

Conclusion

Based on the given information, Misha will not be able to catch up with Masha in the race. Masha starts 200m ahead and maintains a higher speed, allowing her to finish the race before Misha can catch up.

Please note that the sources provided do not contain relevant information for this specific problem.