two boys start running straight towards each other from two points that are 100m apart. one runs with a speed of 5m/s, while other moves at 7m/s.how close are they to the slower one's starting point when they meet?
The distance between them is reduced at a rate 12 m/s, and they meet after 100/12 = 8.333 s. At that time, the slower runner is
5 * 8.333 = 41.67 m from his starting point. As a fraction, that is 125/3 meters.
To find out how close the boys are to the slower one's starting point when they meet, we need to determine the time it takes for them to meet and then calculate the distance covered by the slower boy.
Let's denote the distance covered by the slower boy with x (in meters).
We can use the formula: Distance = Speed x Time
The slower boy runs with a speed of 5 m/s, so the distance he covers is 5x.
The faster boy runs with a speed of 7 m/s, so the distance he covers is 7(100 - x), since he starts 100 meters apart from the slower boy, and the combined distance covered by both boys is equal to the initial distance between them (100 meters).
Setting the distances covered by each boy equal to each other, we can write the equation:
5x = 7(100 - x)
Simplifying the equation, we have:
5x = 700 - 7x
Combining like terms, we get:
12x = 700
Dividing both sides by 12, we find:
x ≈ 58.33
Therefore, the slower boy is approximately 58.33 meters away from his starting point when they meet.