The first problem analysis: this is a catch-up problem. The liaison and Class Two start at the same time. When they meet, the time for the liaison to ride a bike is equal to the time for Class 2 to catch up 1 class. So we just need to find out the time when Class 2 catches up with Class/kloc-0, and the problem will be solved.

Solution: Let Class 2 catch up with 1 class for x hours.

Equality: When meeting, the distance of Class Two = 1.

Equation: 6X=2+4X

Solve the equation: X= 1

So the contact time is also 1 hour, so the contact distance is:12 *1=12km.

Analysis of the second question: this is an encounter problem, and the sum of speeds = total distance/encounter time; 0.5 hours after meeting, we only need to find out the distance and destination of two people. According to the ideal relationship, the remaining distance of A = twice the remaining distance of B.

Solution: Let Xiao Ming ride at the speed of X km/h.

Xiaoli's riding speed is 36/1-x = 36-xkm/h.

36- 1.5x = 2 *(36- 1.5(36-x))

x= 16

Xiaoming's riding speed16 km/h.

Xiaoli's cycling speed is 36-x = 20km/h.

The third problem analysis: this is a circular runway problem.

One sentence in the known conditions of this problem should be changed to "Xiaohong's speed is 5/3 times that of grandpa", and he can only catch up with the slow.

Circular runway problem: distance of B = distance of A+runway perimeter *N (A is slow, B is fast, and n is the number of chases).

Solution: let grandpa's speed be x meters/minute.

Equality: Xiaohong's distance = Grandpa's distance +400.

Equation: 5*5/3X=5X+400.

Solve the equation: x = 120m/min.

So Xiaohong's speed is 120 * 5/2 = 200m/min.

(1) If Xiaohong catches up with Grandpa and runs in the opposite direction, it will become a problem of passing the car.

Time = 400/(120+200) =1.25min.

(2) If they are 80 meters apart, we need to discuss two possibilities.

The first possibility: Xiaohong is in the front and Grandpa is in the back.

Xiaohong's distance +80= Grandpa's distance +400.

Set Xiaohong to catch up with Grandpa in X minutes.

200X+80= 120X+400

X=4

So when Xiaohong was in front and Grandpa was 80 meters apart, they met four minutes later.

The second possibility: Xiaohong is in the back and Grandpa is in the front.

Xiaohong's distance = Grandpa's distance +80.

Set Xiaohong to catch up with Grandpa in X minutes.

200X= 120X+80

X= 1

So when grandpa is in front and Xiaohong is behind, when they are 80 meters apart, 1 minute later meets.