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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
A and B are going from a and B to each other. After meeting, a arrives at B in 15 minutes and B arrives at a in 1 hour. A's speed is several times that of B
Let two cars meet in t minutes. 1 hour = 60 minutes;
The formula of countable proportion is: t ∶ 15 = 60 ∶ T, and the solution is: T = 30 (rounding off the negative value),
It takes T + 15 = 45 minutes for car a to complete the whole journey, and 60 + T = 90 minutes for car B to complete the whole journey,
If the distance is equal, the speed is inversely proportional to the time,
That is: the speed of a is twice that of B
Car a and car B are facing each other from a and B. car a starts 15 minutes earlier than car B. the speed ratio of car a and car B is 2:3. When they meet, car a walks 6 minutes less than car B
v1*15+(v1+v2)*t=s;
v1=2/3*v2
v2*t-v1*(15+t)=6;
The condition is still short
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