
Из двух посёлков в одном направлении вышли два пешехода, расстояние между которыми 540 метров,
первый вышел раньше на 2 секунды со скоростью 15 м/с , скорость первого 30 м/с. Через сколько второй пешеход догонит первого и на каком расстояние?

Ответы на вопрос

Ответ:
270+45 ответ поидее должен быть таким однако это не на 100% я уже все позабыл



Problem Analysis
We have two pedestrians walking in the same direction. The distance between them is 540 meters, and the first pedestrian started walking 2 seconds earlier than the second pedestrian. The first pedestrian has a speed of 15 m/s, while the second pedestrian has a speed of 30 m/s. We need to determine how long it will take for the second pedestrian to catch up to the first pedestrian and at what distance.Solution
To solve this problem, we can use the formula:Distance = Speed × Time
Let's assume that the second pedestrian catches up to the first pedestrian after time 't' seconds. At that time, the first pedestrian would have covered a distance of 15t meters, and the second pedestrian would have covered a distance of 30(t-2) meters.
Since the distance between them is 540 meters, we can set up the following equation:
15t + 540 = 30(t-2)
Now, let's solve this equation to find the value of 't'.
Calculation
15t + 540 = 30(t-2)15t + 540 = 30t - 60
60 + 540 = 30t - 15t
600 = 15t
t = 600 / 15
t = 40
Therefore, it will take the second pedestrian 40 seconds to catch up to the first pedestrian.
To find the distance at which the second pedestrian catches up to the first pedestrian, we can substitute the value of 't' into the equation for the distance covered by the second pedestrian:
Distance = 30(t-2)
Distance = 30(40-2)
Distance = 30(38)
Distance = 1140 meters
Therefore, the second pedestrian will catch up to the first pedestrian after 40 seconds, at a distance of 1140 meters.
Answer
The second pedestrian will catch up to the first pedestrian after 40 seconds, at a distance of 1140 meters.Note: The sources provided did not contain relevant information for this specific problem.


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