Car a and car B are facing each other from ab. car a starts 15 minutes earlier than car B. car a travels 20 km more per hour than car B. when they meet, car a travels 55 km more than car B,

Car a and car B are facing each other from ab. car a starts 15 minutes earlier than car B. car a travels 20 km more per hour than car B. when they meet, car a travels 55 km more than car B,


Suppose that when Party B walks SKM for the first time, Party A walks 1.5 SKM, and the distance between Party A and Party B is 2.5 SKM. After each meeting, the total distance between the two places increases by 2, that is, 5 SKM. Then, when they meet for the third time, they walk 2 * 5S 2.5s = 12.5 SKM. Because the speed of Party A is 1.5 times that of Party B, Party A walks 12.5s * 3 / 5 = 7.5 SKM, B walked 5skm, and the meeting place was in B. at the fourth meeting, they walked 3 * 5S 2.5s = 17.5skm. Because a's speed was 1.5 times that of B, a walked 17.5s * 3 / 5 = 10.5skm. B walked 7skm. At this time, the meeting place was 0.5s away from a, that is, 2skm away from B, so 2S = 20km, 2.5s = 25km, and AB was 25km away



The speed ratio of a and B is 2:3. When they meet, a walks 6 km less than B. It is known that B walks for 1 hour and 30 minutes. The speed of a and B and the distance between two places are calculated


Suppose that the speed of a and B is 2xkm / h and 3xkm / h respectively. From the meaning of the question, we get 2x · (112 + 14) = 3x · 32-6, and the solution is x = 6, 2 × 6 = 12km / h, 3 × 6 = 18km / h, 12 × (112 + 14) + 18 × 32 = 21 + 27 = 48km. A: the speed of a and B is 12km / h and 18km / h respectively, and the distance between the two places is 48km