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The distance between a and B is 600 kilometers, and two cars leave from the two places at the same time. Car a travels 120 kilometers per hour, and car B's speed is the same as car a Two thirds of the time, two cars meet?
600 ÷ (120 + 120 × 2 / 3) = 3 (hour) this is the total distance of the encounter problem △ speed sum = time
The speed of car a is 54km / h, and that of car B is 36km / h. The two cars go in opposite directions at 9am at the same place. After 40 minutes, car a immediately turns around to catch up with car B
So when can car a catch up with car B
Using equation
40 minutes = 2 / 3 hours, 2 / 3 hours Party A and Party B travel together for 2 / 3 (54 + 36) = 60 km, the time for car a to catch up with car B is: 60 △ 54-36 = 60 △ 18 = 10 / 3 hours = 3 hours 20 minutes 9 hours + 3 hours 20 minutes = 12 hours 20 minutes car a catches up with car B at 12:20 noon
A. B is 15 kilometers away. One car starts from a at a speed of 30 kilometers per hour, and the other car starts from B at a speed of 40 kilometers per hour,
How long does it take for two cars to travel in opposite directions? I want to solve the equation
Set time t
15 - (30 + 40) t = 3 T = 6 / 35 hours
On the other hand, after meeting, the two cars are 3 kilometers apart
Let t be the meeting time
(30 + 40) t = 15, t = 3 / 14 hours, 3 / (30 + 40) = 3 / 70 hours away from 3 km
When the distance between the two vehicles is 3 km, it takes 9 / 35 hours
Both trains run from a to B. It is known that the speed of car a is x km / h and that of car B is y km / h. After 3 hours, car B is 5 km away from B,
A 3(-x+y)-5 B 3(x+y)-5 C3(-x+y)+5 D 3(x+y)+5
Do me a favor
Well, I admit I've been a little dizzy recently
At this time, the distance between car a and ground B () km?
Suppose the distance between a and B is n, then n = 3Y + 5, n = 3x + the remaining distance of A
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