This problem is essentially a combination of encounter model and chase model, and the relationship between the two models should be sought in the process of solving the problem.

Because they start at the same time interval, the distance between the two workshops is the same for all vehicles traveling in the same direction at any time on this section of expressway between the two stations.

The solution is:

■ Because trains leave at regular intervals, the distance between any two adjacent trains in the same direction on the way is equal. We might as well set the unit as "1".

When this man was overtaken by a car coming from behind for the first time, the distance between him and a bus behind him was "1". 12 minutes later, this person caught up with his car (that is, he was overtaken by the same bus for the second time), that is to say, within 12 minutes, the car drove more than the people 1, so the car drove more than the people every minute 1/65438. (Can be regarded as a follow-up question)

When this man meets an oncoming bus for the first time, the distance between him and the oncoming second car is "1". This person met this car four minutes later (that is, he met the same bus for the second time), that is to say, the car and the person traveled a distance of "1" in four minutes, so the car and the person traveled a unit of "1" every minute. (Can be regarded as a meeting problem)

According to the above two results, using the principle of sum and difference, we can get (112+1/4) ÷ 2 =1/6 per minute, so the unit is "65438+.

■ ① Distance between cars = 12 minute car trip-12 minute bike trip.

② Distance between cars = 4-minute journey of cars+4-minute journey of bicycles,

Namely: 3× departure interval distance = tram 12 minute trip+bicycle 12 minute trip,

Solve by sum and difference problem: 12 minutes' drive = vehicle distance *( 1+3)÷2.

That is, the departure interval = 6 minutes' drive.

■ Because of the time interval of meeting: the time interval of catching up? = 1:3;

So (speed? +speed): (speed-? Human speed) = 3:1;

Then you can set (speed+? Man speed) is 3 copies, (speed? -human speed) is 1 serving.

Therefore, it can be found that the speed of the car is 2 copies and the speed of the person is 1 copy.

Speed: (speed? -human speed? 2: 1;

So the departure interval: the catch-up interval? 1:2。

So the departure time interval is: 12÷2= 6 (minutes).

■ A bus leaves every t minutes.

Suppose this person has a bus passing by him for the first time in place A, then a second bus will drive to the place where the first bus passes by him in t minutes (that is, point A). This person walked 12 minutes, and then arrived at point B. At this time, the second bus caught up with this person, so it took the second bus (12-t) minutes to get from point A to point B. Therefore, the distance traveled by this person in 12 minutes is equal to that of the bus in (/kloc-0).

Then, let's still assume that the first time this person meets a bus coming from the opposite side is at point A, and the second bus coming from the same direction passes through point A after t minutes. Now this person has walked for 4 minutes and met the second bus coming from the opposite side at point C, so it takes (t-4) minutes for the bus to travel from point C to point A, which means that the distance traveled by this person in 4 minutes is equal to the distance traveled by the car in (t-4) minutes, so the speed ratio between this person and the car is (t-4): 4.

According to the above two results, we can get:

( 12-t)∶ 12=(t-4)∶4

The solution is t=6.

That is, there is a bus every 6 minutes.

■ steam speed: a person speed: b car every X.

In any case, the distance between the two cars is constant, which is the following two equations.

(a-b) 12=(a+b)4……..………①

(a+b)4 = ax……②

8a= 16b from ①.

a = 2b……③

Substituting ③ into ② gives 12b=2bx.

x=6

■: Let the distance between two cars in the same direction be s (that is, the distance that the cars travel at the same time).

For the oncoming car, then the person and the car go together s,

Because every four minutes, a car comes in front of you, which means people have been walking for four minutes.

If you let the car walk for 4 minutes, the time it takes is t, then the distance between cars is 4+t.

For the car coming from behind, the distance for the car to catch up with the cyclist is s+ (the distance for people to walk). Then it takes 12 minutes for a person to walk this distance, and it takes 3t if a person walks this distance 12 minutes (because the distance of 4 minutes is equivalent to the t minute of a car). The distance between cars is 65438+.

So 12-3t=t+4.

So t=2,

That is, the interval is 4+t=6 minutes.