There are two kinds of fast and slow cars at a certain station, and the departure station is 7 2km 3

The travel time of the slow train is xmin, the distance traveled by the fast train is Y1km, and the distance traveled by the slow train is Y2km.

y1=y2=

Seek to meet, that is, the distance is equal.

x = two trains meet in 8 minutes, and they meet at the time of departure station.

The speed of the slow train (in km min).

The functional relation of the distance traveled by the slow train on the travel time of the slow train is: y= x and the definition domain is [0,16].

The speed of the express train is.

The domain of the functional relation y= -3) x on the travel time of the fast train is [3,13].

Solve systems of equations. y= y= -3）

There is y=x=8

The two trains met 8 minutes after the departure of the slow train, and they met a kilometer away from the departure station.

Meet after 5 minutes of travel by express train, from the starting station.

Slow speed: Fast speed:

Set up the express train to meet after t min.

If you want to find the time of encounter, the column formula is as follows.

Time = Distance Speed.

10 hours. Station A and Station B are 1500k meters apart, and the slow train departs from Station A at a speed of 60km h. The express train departs from Station B at a speed of 90km, and it takes 10 hours for the two Bizhu trains to meet.

Solution: Set two teasing mills x hours to meet, 60 + 90) x 1500150 x 1500

x 10 two cars for 10 hours.

Answer: When the two cars meet, the mountain bucket express car travels 300km longer than the slow car.

The two cars met when they drove out at the same time after 10 small cars.

Dress up as above. Journey topics.

10 (hours) of dullness.

Answer: The two cars drove out of the two stations in the same spring, and after 10 hours of picking up and shooting.

Set the distance between the two cars after x hours by 48km.

60×(32/60+x)+48+100x=448

x = hours.

1.Let's say x hours meet.

60x+100x=448 Solution: x= two cars meet in opposite directions 2How much time to chase Then the two cars should go in the opposite direction Suppose it takes y hours to chase to 1) car A chases car B because car A is faster than car B speed, car A will not catch up with car B, 2) car B chases car A.

100y-60y=448

Solution y= both cars set off at the same time and drove for hours in pursuit.

Where did the LS 100 come from? I don't understand! I think so.

1) Suppose x hours meet then:

80x+60x=448 can be solved x = hours) Answer: Two cars are traveling in opposite directions and meet after hours.

2) Suppose car B catches up with car A after x hours then:

60x+448=80x can be solved x = hour) Answer: Two cars start at the same time Car B catches up with car A after hours.

(1) 448 (60+80)= So the hours meet in the opposite direction.

2) In the same direction, if the fast car is chasing the slow car, 448 (80-60) = so it will catch up after an hour, but if the slow car is chasing the fast car, it will never catch up.

Solution 1: Let the distance between the two cities of A and B be xkm, and the equation is x 100 + x 80 + 30 60 = 5

x÷100+x÷80=

180x=450×80

x=200km

Answer: 2 solutions:

1) Let the equation of the meeting train x hours after departure.

60x+100x=448

160x=448

x = hours. 2) Let the express train meet the equation x hours after it leaves.

60x+100x=448-60×（32÷60）160x=448-32

160x=412

x = hours.

2 5 time, (400 100*2 5) (100+140) time.

(140-100) 10 o'clock.

Meet the time of distance and speed and.

Catch up with the time distance and speed difference.

I didn't see the equation.

1 set for x hours.

140x+100(x+2/5)=400

x = set x hours.

140x－100x＝400

x=10

When they met, the fast train traveled 32,000 meters more than the slow car.

Coupled with the 27 kilometers of the slow train, after the departure of the fast train, it is more than the slow train to take a total of 32 + 27 = 59 kilometers.

After the express train departed, the two cars traveled together:

59 (5-4) (5+4) = 531 km.

AB distance is: 531 + 27 = 558 km.

Analysis: Because it is an encounter problem, the time for the fast and slow cars to travel at the same time is the same, so from the start of the fast train, the sliding distance ratio of the fast and slow cars is equal to the speed ratio is equal to 5:4, so.

If the travel time of the express train is X, the distance between the express train and station B is 5x kilometers, and the distance between the slow train and station A is (4x+27) kilometers.

5x-（4x+27）=32

x=59 distance: 59x5+59x4+27=558 (1,000 letters) Answer: AB is 558 kilometers apart. Limbs.

Assuming that the distance of the meeting point B is x meters, then the distance of the infiltration meeting point is X - 32 meters. If the speed of the fast train is 5a, the speed of the slow train is 4a. According to the time used at the time of the encounter, the equation is (x-32) 5a=(x-37) 4a x=263 meters, and the distance between the two bases and the ground is 2x-32=484 meters, which can be understood by drawing a diagram from the analysis.

48÷（12-8）×（12+8）

240 km.

A: The distance between the two stations is 240 kilometers.

Solution: Solve this problem according to the speed ratio = distance ratio.

Velocity ratio = 12:8 = 3:2

Let the fast car go x then the slow car go x-48

x:(x-48)=3:2

2x=3x-144

x=144 so the distance between the two places = 144 * 2-48 = 240 kilometers Answer: The distance between the two places is 240 kilometers.

Let the distance be xkm

Express speed x 8

Slow train x 12

Encounter time x (x 8 + x 12) = 24 5 per hour the fast train travels more than the slow train x 8-x 12 = x 24, so 48 = (x 24) * (24 5).

Solution x=240

So, when they meet, the distance traveled by the express train is (x-27)*(5 9); The distance traveled by the slow train is (x-27)*(4 9)+27

According to the title, it is: [(x-27)*(5 9)]-x-27)*(4 9)+27]=32

=> (x-27)*(1/9)=59

=> x-27=59*9=531

=> x=558

That is, AB is 558 km apart.

Method 1: When encountering, the fast train travels 32 kilometers longer than the slow train.

Coupled with the 27 kilometers of the slow train first, after the express train departs, a total of 32 + 27 = 59 kilometers more than the slow train after the departure of the express train The two cars travel together:

59 (5-4) (5+4) = 531 km.

AB distance is: 531 + 27 = 558 km.

Method 2: If two cars start at the same time, then only the fast ones are the slow cars: 4 5 = 4 5

The title says that the fast train travels 32 kilometers more than the slow train, which is under the premise that the slow train goes 27 kilometers first, so after the slow train goes 27 kilometers first, the distance of the two cars at the same time, the fast train travels more than the slow train: 32 + 27 = 59 kilometers.

So when we met, the train was fast: 59 (1-4 5) = 295 km, and the distance between the two stations was: 295 + (295-32) = 558 km.