Find the distance between A and B. Can someone help me solve this problem

Used as a graphical solution.

The first time they met, A and B walked a total of 1 AB two places. At this time, A walked a distance of 150 meters.

The second time they met, A and B walked a total of 3 AB two places. So, the distance A travels should be 150 3 = 450 meters.

And A is 150 meters away from place A, so the distance between 2 places A and B = 150 + 450. So the distance between A and B is 300 meters.

No problem, the first time they met, A and B walked a total of 1 AB two places. At this time, A walked a distance of 150 meters. For the second encounter, A and B walked a total of 3 AB distances. So, the distance A travels should be 150 3 = 450 meters.

And A is 150 meters away from place A, so the distance between 2 places A and B = 150 + 450. So the distance between A and B is 300 meters.

When A and B met for the first time, A walked 120 meters, A and B walked a total of one AB distance, and A and B walked a total of 3 AB distances when A and B met for the second time, so A walked 3*120=360 meters, and at this time it was 150 meters away from A, that is to say, A went from A to B and then to A, and after walking 360 meters, there were still 150 meters away from A, so the distance of A and B was (360+150) 2=255 meters.

As shown in the figure: A and B set off from A and B at the same time, going back and forth between A and B, the first meeting is 30 kilometers away from A, and the second meeting is 10 kilometers to the right of the first rotten spike: how far apart are the two points A and B? (

a 90 b 75 c 65 d 50 single-shore type", two encounter problems:

Solution: Let the velocity of A be x, then the velocity of B is (

From the meaning of the title, 4 - 30 60) x + 4 - 40 60) ( = 39 3

x = 18

That is, the speed of A is 18km h, and the speed of B is.

Analysis: In addition to the respective stay time of A and B in A and B, we can find that A and B walked exactly three times the distance between AB and B in four hours, and A and B each walked a complete distance between AB, and finally met and added a piece just another AB

AB two places are 39 kilometers apart, A and B are driving in the same direction at the same time, A starts from place A, arrives at place B, stays for half an hour, B departs from place B and arrives at place A, stays for 40 minutes, the two continue to return, meet between the two places of AB, A is faster than B kilometers per hour, the two share 4 hours, find the speed of A and B?

Sorry, the original question is incomplete and cannot be answered.

Please add the original question, thank you!

Half an hour = 1 2 hours, 40 minutes = 2 3 hours, one and four fifths = 9 5 Let B's speed be x kilometers per hour, then A's speed will be x + 9 5 kilometers When they meet, the two of them walk together for 3 ab full journeys.

So (4-1 2) (x+9 5)+(4-2 3) x=39 3 so that x can be solved

And then you can get the speed of A and B.

Have fun.

Solution: Let B's velocity be xkm h, then A's velocity is (x+9 5) km h, B has gone (4-4 6) x km, and A has gone (

In total, they traveled 3x39 km, and the column equation is:

4-4/6)x+(

x= x+9/5=18(km/h)

A: The speed of A is 18km h, and the speed of B is.

Let the velocity of A be x and the velocity of B be y, then there is.

36 (x+y)=4 Equation 1

36-6x=2*(36-6y) Eq. 2

Simplify. x+y=9 Equation 3

2y-x=6 Equation 4

Solve the above equation.

x=4km/h

y=5km/h

The speed of A and B is 4km h and 5km h respectively

Solution: Let the velocity of A x, B y; 4(x+y)=36, (36-6x) (36-6y)=2, the solution is: x=4km h, y=5km h.

Let the two meet after q hours, and the two meet after Q=m (x+y)m (x+y) hours according to the title.

Under normal circumstances, it takes 9 minutes for A and B to meet; In the case mentioned in the question, as long as the time to walk forward reaches 9 minutes, A and B can meet, and at the 17th minute, A and B travel in the forward direction (i.e., two people walking in the opposite direction) for 9 minutes, and the departure time of the two people is 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81 minutes.

Speed sum = (4 + 5) * 1000 60 = 150 m min.

The time it takes.

1350 150 = 9 minutes.

Advance for 1 minute per turn.

1 minute forward 3 minutes reverse 5 minutes forward ,, 33 minutes forward, a total of 9 forward 34*8+17=289 minutes.

Thinking about it as a whole, both of them walked for 2+ hours, and a total of 2 whole journeys.

Therefore, the distance between the two places is 50 11 2 = 275 km.

Let a be x per hour and b is 50-x

2x+x=then the distance is 275 km.

First encounter.

Full course = speed of A.

Second encounter.

Full Journey = Speed of B.

The two encounters add up.

Two full courses = 210 + 240 + 2 50

ab distance = 550 2 = 275 km. The columns are as follows.