甲、乙两人从 A、B 两地同时出发相向而行，甲每分钟走 60 米，乙每分钟走 50 米。两人在距离中点 120 米处相遇。如果甲、乙两人同时从 A 地出发前往 B 地，且乙先出发 2 分钟，那么甲出发后多少分钟能追上乙？​

视频信息

答案文本 复制

视频字幕 复制

Two people start walking from points A and B simultaneously toward each other. Person A walks at 60 meters per minute, while person B walks at 50 meters per minute. They meet at a point that is 120 meters away from the midpoint between A and B. Since person A walks faster, they cover more distance and meet closer to point B. Let's analyze this mathematically. When two people walk toward each other, together they cover the entire distance between A and B. Since person A walks faster at 60 meters per minute, they cover more ground before meeting. The meeting point being 120 meters from the midpoint means person A walked 120 meters more than half the total distance. If we call the total distance d, then person A walked d over 2 plus 120 meters, while person B walked d over 2 minus 120 meters. Since they walked for the same amount of time, we can set up an equation using time equals distance over speed. Now let's solve the equation step by step. Starting with the time equation where both people walked for the same duration. We cross multiply to get 50 times d over 2 plus 120 equals 60 times d over 2 minus 120. Expanding this gives us 25d plus 6000 equals 30d minus 7200. Rearranging terms, we get 6000 plus 7200 equals 30d minus 25d, which simplifies to 13200 equals 5d. Therefore, d equals 2640 meters. Let's verify this solution: Person A walks 1440 meters in 24 minutes, and Person B walks 1200 meters in 24 minutes. Both take exactly 24 minutes, confirming our answer is correct. Now let's move to the second part of the problem. Both people start from point A and head toward point B, but person B gets a 2-minute head start. After person B has walked for 2 minutes, they will be 50 times 2 equals 100 meters ahead of point A. At this moment, person A starts walking. This creates a chase scenario where person A needs to catch up to person B, with both moving in the same direction toward point B. In chase problems, we use the concept of relative speed. When both objects move in the same direction, the relative speed is the difference between their speeds. Person A moves at 60 meters per minute and Person B at 50 meters per minute, so the relative speed is 60 minus 50 equals 10 meters per minute. This means Person A gains 10 meters on Person B every minute. Since Person B has a 100-meter head start, we can calculate the catch-up time as 100 meters divided by 10 meters per minute, which equals 10 minutes. Let's watch this animation to see how the gap closes over time.