
Из села вышел пешеход со скоростью 4 км/ч . Через 3 часа вслед за ним выехал велосипедист со
скоростью 10 км/ч . Через сколько часов велосипедист догонит пешехода?

Ответы на вопрос

10-4=6(км/ч) скорость сближения
12/6=2(ч)
Ответ:через 2 часа



Problem Analysis
We are given that a pedestrian is walking at a speed of 4 km/h, and after 3 hours, a cyclist starts riding at a speed of 10 km/h. We need to determine how many hours it will take for the cyclist to catch up to the pedestrian.Solution
To solve this problem, we can set up an equation based on the distance traveled by both the pedestrian and the cyclist.Let's assume that the time it takes for the cyclist to catch up to the pedestrian is t hours.
The distance traveled by the pedestrian in t hours is given by: Distance_pedestrian = Speed_pedestrian * t The distance traveled by the cyclist in t hours is given by: Distance_cyclist = Speed_cyclist * (t - 3) Since the cyclist catches up to the pedestrian, the distances traveled by both of them are equal. Therefore, we can set up the equation:
Speed_pedestrian * t = Speed_cyclist * (t - 3)
Now, we can substitute the given values into the equation and solve for t.
4 * t = 10 * (t - 3)
Simplifying the equation:
4t = 10t - 30
6t = 30
t = 5
Therefore, it will take the cyclist 5 hours to catch up to the pedestrian.
Answer
The cyclist will catch up to the pedestrian after 5 hours.


Calculating When the Cyclist Overtakes the Pedestrian
To calculate when the cyclist will overtake the pedestrian, we can use the formula: \[ \text{time} = \frac{\text{distance}}{\text{relative speed}} \]
The relative speed is the difference between the speeds of the cyclist and the pedestrian.
Given: - Pedestrian's speed = 4 km/h - Cyclist's speed = 10 km/h
We can calculate the relative speed: \[ \text{Relative speed} = 10 - 4 = 6 \text{ km/h} \]
Now, we can calculate the time it takes for the cyclist to overtake the pedestrian: \[ \text{time} = \frac{12}{6} = 2 \text{ hours} \]
Therefore, the cyclist will overtake the pedestrian in 2 hours.


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