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Xiao Ming was running on the road one day. He found a thing. The bus in front of him comes every ten minutes on average, and the bus behind him catches up with one every fifteen minutes on average. How many minutes does the bus stop send the bus on average
On the road, there is a 15 meter bus running from east to west at a speed of 18 kilometers per hour. On the sidewalk beside the road, there are two young people, a and B, who are practicing long-distance running. A runs from east to west and B runs from west to East. At a certain moment, the car catches up with a and leaves a in six seconds. Half a minute later, the car meets B running head-on. After another two seconds, the bus stops The car left B. ask again______ Seconds later, a and B meet
(1) How many meters per second is the speed converted to? 18 × 1000 △ 3600 = 5 (m). (2) find the speed of A. It's a matter of catching up if the car goes in the same direction as a. the distance of 6 seconds between a and B = the distance of 6 seconds - the length of the car body. Therefore, the speed of a = 5 × 6-15, the speed of a = (5 × 6-15) △ 6 = 2.5 (M / s). (3) find the speed of B. when the car goes in opposite direction to B, it's a matter of opposite travel. The distance of 2 seconds between B and B = the length of the car body - the length of 2 seconds (4) what is the time between the car leaving a and B? 5 × 60 + 2 = 32 seconds. (5) when the car leaves B, what is the distance between a and B? 32 × 5-2.5 × 32 = 80 (m) (6) what is the meeting time between a and B? 80 (2.5 + 2.5) = 16 (seconds) a: after another 16 seconds, a and B will meet. So the answer is: 16
Xiao Ming was running on the road one day. He found a thing. The bus in front of Xiao Ming comes every ten minutes on average. The bus behind Xiao Ming catches up with one every 15 minutes. The bus keeps the same speed. The problem is how many minutes does the bus stop send a car on average? The steps should be more detailed
Let the distance between two cars be s, the speed of people be V1, and the speed of cars be v2
s/(v1+v2)=10
s/(v2-v1)=15
The solution is V2 = s / 12
The departure interval T = s / V2 = 12min
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