Please answer a trip Olympiad question, please help solve a trip Olympiad question

Let the hydrostatic velocity be x m seconds, the first encounter is m m from a ground, and the second encounter is n m from a ground n.

Downstream m m (x+2), countercurrent (100000-m) (100000-m) (x-2).

m/(x+2)=(100000-m)/(x-2)m=100000*(x+2)/2x

Because it takes the same time to reach the end after the second encounter.

Therefore, there is n (x-2)=(100000-n) (x+2)n=100000*(x-2) 2x

m-n=4*100000 2x=20*1000x=10 m seconds.

i.e. the speed of two boats in still water is 10 meters per second.

A and B have a speed difference of 2+2=4 meters and seconds.

Ship A arrives at place B earlier than ship B arrives at place A: 100,000 4 = 25,000 seconds.

Ship A arrived at place B and returned against the current for 25,000 seconds before ship B arrived at place A.

Because the first ship returned to the current for 25,000 seconds in advance, which just eliminated the distance difference caused by the speed difference, the two ships met again at the midpoint of A and B.

Therefore, the first time the two ships meet, the distance from place A = 50 + 20 = 70 km.

If the hydrostatic velocity of the two ships is v m seconds and the time taken for the first encounter is equal, then 70000 (v+2) = 30000 (v-2).

Solution: v=5 msec.

After the second encounter, the journey continues and returns at the same time, indicating that the distance up the river is 20 kilometers longer than the distance against the water.

The whole journey is 2 times the distance of the still water, and 20 is the distance of 2 times the water flow, then the still water speed is 100 20 = 5 times the water flow speed, so the still water speed is 2 5 = 10 meters per second.

According to the title, if the distance from the park to the museum is exactly 2 kilometers, then it will take the same amount of time to go home to pick up the car and go directly to the museum.

Take 1 time to walk from the park to the museum

Then the time it takes to ride from the park to the museum is 1 to 4

Walking home from the park and then cycling to the park takes 1-1 4 = 3 4 The round trip is the same, and the speed ratio is 1:4

The time taken ratio is 4:1

The time it takes to walk home from the park is: 4 (1+4) 3 4=3 5The walk home from the park is 3 5 The distance from the park to the museum is: 2 3 5=km.

Solution: If the distance from the entrance of the park to their house is a kilometer, the walking speed is x km h, and the cycling speed is 4xkm h, then: 2*2 x>a x+(2*2+a) (4x)4>a+1+a 4

3>5/4*a

a< km

That is, the distance from the entrance of the park to their home is less than a kilometer.

As you can see from the title, when the distance from the park to the museum is 2 km, the time required for both methods is the same, so if the distance to their home is xkm, and the walking speed is 1 part, then the cycling speed is 4 parts. Then there is:

x+(2+x)/4=2

x = 6 5 km.

If you don't want to list equations, you can analyze it like this. For the same distance, the time required to ride a bicycle is 1 4 of the walk, so it actually took only 1 4 time to get to the museum, and the rest is the time to walk to the house and cycle back, so there is:

Distance to home = (1-1 4) (1+1 4)*2=6 5 km.

First of all, the best way is to try it yourself (I don't think it's possible, hehe), in fact, it's also very simple, first set the problem x, the distance between the park and the museum y, the speed to 4, 1, and then list 3 inequality equations, and then linear planning, solve. Simple!!!

The distance from Mr. Liu's home to his unit is x kilometers.

Then there is: [(x 3) 8+(2x 3) 16-1 2]*(3 8*8+5 8*16)=x

x/12-1/2=x/13

x=12*13/2

x=78km

Solution: The speed ratio of cycling to riding is 1:2

Going to the unit is less than going home and riding the whole journey (1-5 8)-1 3=1 24 car, so the ratio of cycling time to riding on the bus during this journey is 2:1, [note that the distance remains the same, and the time is inversely proportional to the speed].

During this time, I took a car when I went and rode a bicycle when I came back, so the time ratio before and after this journey was 1:2, so it took 2 minutes to take a car when I went to this distance, [obtained by the proportional relationship (2 (2-1))*1].

So this journey takes 2 (1 24) = 48 minutes, so the total distance is 16 * 48 60 = km.

3/8-1/3=1/24 ……The distance to work is less than that to go home (i.e., 1 24 This section is a ride when you go to work, but you ride a bike when you go home).

8:16=1:2 ……The speed ratio of riding to riding.

2:1 ……The ratio of the time spent riding to the car when traveling the same distance.

2 (2-1) = 2 minutes ......The time of each serving.

2*2=4 minutes ......The time it takes to ride a bicycle for this 1 24 distance.

8*4 60 (1 24) = km ......The distance from Mr. Liu's home to the unit.

or 16 * 2 60 (1 24) = km.

Let Xiao Ming's speed be 4x, Xiaoguang's speed be x, and the car's speed be y, and the time interval is tyt+30*4x=30y

YT 2 cars distance car + 30 minutes Xiao Ming's distance = 30 minutes car distance YT + 12x = 12Y

Type 2 * 10-1 type.

9yt=90y

t=10 (minutes).

The special feature of this problem is that the length between the two cars is unknown, and the other distance is unknown.

Distance difference speed difference = chasing time, you can set the distance between the two cars as a value: 1, that is, the distance difference between the chasing time is 1

That is, at each minute, the speed of the vehicle - small light speed = 1 30, at each minute, the speed of the vehicle - small speed of light = 1 12;

Then use 1 12-1 30 = 1 20, (that is, Xiao Ming is faster than Xiao Guang per minute), 1 20 (4 1) 1 60 ......the speed of a small light;

1/60×4＝1/15, …Xiao Ming's speed;

1/12+1/60=1/10……the speed of the car; (or use 1 15 + 1 30 = 1 10).

1 1 10 10 (min) ......There is an interval of 10 minutes between the two adjacent buses.

In addition, the actual value (distance between two cars) can be set for easy understanding.

The distance between the two cars below is 1200 meters. The answer process is omitted).

First walk 4 parts, 40 parts of the car line, the difference is 40-4 = 36 parts, then walk another 36 (50 + 4) 4 = 8 3 parts of walking, a total of 4 + 8 3 = 20 3 parts.

Full length 40 + 20 3 = 140 3 parts.

The ratio of walking distance to the whole journey is 20 3:140 3 = 1:7

Let the velocities of the three be v A, v B, V C, the distance between C and O is s, the time for B and C to meet is t, and the time for A and C to meet is t2

From "C sets out from town C and meets B at the same time as A and B at a distance of 15 kilometers from town O", we get v B*t=30+15=45 v C*t=s-15 and v B v C =45 (S-15) Equation 1.

From "when C arrives at town O and then goes to town A, and meets A at a distance of 6 kilometers from town O", we get V A*T2=30+20+20-6=64 V C*T2=S+6 to get V A V C=64 (S+6) Formula 2.

It is known that the velocity ratio of A and B is 8 9, and the equations 1 and 2 obtain v A v B = 64*(s-15) {45*(s+6)} =8 9 to obtain s=50 km.

So the distance between the towns of O and C is 50 km.

Solution: When B and C met, B traveled 45 meters and A traveled 40 meters. When A and C meet, A travels 10+20-6=24 meters, and C travels 15+6=21 meters.

The velocity ratio of A,C is 8:7The velocity ratio of A, B, C is 8:9:7, so when row B is 45 meters, row C is 35 meters, so the distance between the two towns is 35+15=50 meters.