Elementary School Mathematics Special Topic Training: Encounter Problems

1.Through practical demonstrations, understand "reciprocal motion", "encounter" and "speed and".

2. Master the solution method of finding distance in the directional motion: speed and × time = distance.

3.Cultivate students' good habit of carefully reviewing questions. The practical problem of two- or three-step computation related to this will be solved.

4.Develop students' ability to analyze and answer questions.

1. Enable students to master the problem solving methods of the Mid-Autumn Festival journey of the opposite movement.

Two: Understand "speed and".

Example 1. The two ships of A and B departed from the two ports of A and B at the same time, and the first ship traveled 18 kilometers per hour, and the B ship traveled 15 kilometers per hour, and after 6 hours, the two ships met on the way. How many kilometers is the waterway between the two places?

Analysis: How many kilometers of waterway length between the two places is required, first find out the speed sum of ship A and ship B, and then use speed and time of encounter, the problem can be solved.

Solution: (18+15) ×6,

=33×6，

= 198 (km);

Answer: The waterway between the two places is 198 kilometers long.

Example 2. A car and a motorcycle depart at the same time from A and B, which are 900 kilometers apart, respectively, with cars traveling 40 kilometers per hour and motorcycles traveling 50 kilometers per hour. How many kilometers are the two cars apart after 8 hours?

Analysis: There are four situations in this question: (1) Two cars are moving in opposite directions, and the distance between the two cars after 8 hours is equal to the distance between A and B minus the distance of two cars;

(2) Traveling in the opposite direction, the distance between the two vehicles after 8 hours is equal to the distance between A and B plus the distance of the two vehicles;

(3) Motorcycle chase car, the distance between the two places minus the distance of 8 hours of motorcycle chase car, that is, the distance between two cars;

(4) Car chase motorcycle, the distance between the two places plus 8 hours of car chase motorcycle distance, that is, the distance between two cars.

Solution: (1) Go in opposite directions.

900﹣（40+50）×8，

=900﹣720，

= 180 (km);

(2) Walk with your back to your back.

900+（50+40）×8，

=900+720，

= 1620 (km);

A: After 8 hours, the distance between the two vehicles is 1620 kilometers.

(3) Motorcycle chase car.

900﹣（50﹣40）×8，

=900﹣80，

= 820 (km);

A: After 8 hours, the distance between the two vehicles is 820 kilometers.

(4) Ride a motorcycle to chase a motorcycle.

900+（50﹣40）×8，

=900+80，

=980 (km);

A: After 8 hours, the distance between the two vehicles is 980 kilometers.

Example 3. Car A and B depart from the two cities A and B, which are 480 kilometers apart, and travel in opposite directions, it is known that it takes 6 hours for car A to go from city A to city B, and car B takes 12 hours from city B to city A. How many hours after the two cars depart?

Analysis: According to the meaning of the topic, using the speed of the distance ÷ = time, the speed of the two cars of A and B is found, and then according to the encounter time = the total distance÷ the speed and sum, which can be solved.

Solution: velocity of A: 480÷6 = 80 (km / h), B's velocity: 480÷12=40 (km/h), Time of encounter: 480÷ (80 + 40) = 4 (hours);

Answer: The two cars meet 4 hours after departure.

Example 4. Wang Xin and Lu Liang walked in opposite directions from 2,000 meters apart at the same time, Wang Xin walked 110 meters per minute, and Lu Liang walked 90 meters per minute. If a dog and Wang Xin walk in the same direction at the same time, 500 meters per minute, after encountering Lu Liang, immediately turn around and run to Wang Xin; after encountering Wang Xin, turn around and run to Lu Liang. This was constantly going back and forth until Wang Xin and Lu Liang met, how many meters did the dog walk together?

Analysis: According to the meaning of the topic, it can be known that the dog and the owner are walking at the same time, no matter how many times the dog runs between the two people, before the two people meet, the dog has been running, and the time when the dog keeps going back and forth is the time when Wang Xin and Lu Liang set off at the same time to meet the two people, according to the title, you can find out how long it took For Wang Xin and Lu Liang to meet, and then use the speed of the dog to × the time of the encounter to find out how many meters the dog traveled together.

Solution: According to the meaning of the question, it can be found that Wang Xin and Lu Liang set off at the same time to meet the time when the two met:

2000÷（110+90），

=2000÷200，

=10 (cents),

Dogs: 500×10 = 5000 (m);

Answer: The dog traveled for 5000 meters.

Example 5. Teams A and B students set off from two places 18 kilometers apart at the same time. A classmate rides a bicycle at a speed of 15 kilometers per hour to make constant contact between the two teams. Team A travels 5 km per hour and Team B travels 4 km per hour. When the two teams met, how many kilometers did the cyclists travel together?

Analysis: According to the distance ÷ speed and = encounter time, the two people met a total of 18 ÷ (4 + 5) = 2 hours, in these two hours, the cyclist student is always in motion, so when the two teams meet, the cyclist students travel together: 15×2 = 30 km.

Solution: 18÷ (4+5) ×15

=18÷9×15，

=30 (km).

Answer: When the two teams met, the students on bicycles traveled a total of 30 kilometers.

Example 6. A and B are 400 kilometers apart, A and B cars drive from the two places at the same time, A car travels 38 kilometers per hour, B car travels 42 kilometers per hour, a swallow at a speed of 50 kilometers per hour and A car at the same time set off to fly to B car, encounter B car and turn back to fly to A car, so that it has been flying, how many kilometers does the swallow fly, the two cars can meet?

Analysis: How many kilometers the swallow is required to fly, it is necessary to know the time and the speed of the swallow to fly, the speed of the swallow is 50 kilometers per hour, the key problem is to find out the time used by the swallow to fly, the time of the swallow flight is the encounter time of the two vehicles, the encounter time of the two cars is 400 ÷ (38 + 42) = 5 (hours), how many kilometers the swallows flew, the column formula is 50×5, the calculation can be. Solution: The time when the swallows fly is the time when the two cars meet, that is:

400÷（38+42），

=400÷80，

=5 (hours);

The distance the swallow flies:

50×5 = 250 (km);

Answer: The swallows flew 250 kilometers to two cars to meet.

A档

1． Teams A and B traveled in opposite directions from 330 km apart at the same time, with Team A traveling 60 km per hour and Team B travelling 50 km per hour. A man on a motorcycle travels back and forth between two convoys at a speed of 80 km per hour and asks how many kilometres the motorcycle traveled when the two convoys met.

parse:

First of all, clarify: the time it takes for the motorcycle to walk repeatedly is equal to the time when the two teams meet. Encounter time: 330÷ (60 + 50) = 3 (hours). Motorcycle ride: 80×3 = 240 (km).

Solution: 330÷ (60+50) ×80 =3×80， =240(km).

Answer: The motorcycle traveled 240 km.

2． A travels 7 kilometers per hour, B travels 5 kilometers per hour, two people in the same place at the same time opposite each other, one to the east, one to the west, after 5 hours how many kilometers apart the two people?

Analysis: By the meaning of the topic, two people go opposite each other, the speed sum is 7 + 5 = 12 (kilometers) per hour, then after 5 hours, the two people are 12×5 = 60 (kilometers), solve the problem.

Solution: (7+5) ×5 =12×5 =60(km). Answer: After 5 hours, the two were separated by 60 kilometers.

3． A circular track is 400 meters long, Xiaoqiang runs 300 meters per minute, Xiaoxing runs 250 meters per minute, the two set off in the same direction at the same time, how long did Xiaoqiang catch up with Xiaoxing for the first time?

Analysis: When Xiaoqiang first caught up with Xiaoxing, Xiaoqiang's driving distance was 400 meters longer than that of Xiaoxing's circular runway, because Xiaoqiang ran 300-250=50 meters more per minute than Xiaoxing, which can be calculated in column.

Solution: 400÷ (300-250),

=400÷50，

=8 (minutes);

Answer: After 8 minutes, Xiao Qiang caught up with the little star for the first time.

4． Guangming Primary School has a 200-meter-long circular track, where Bright and Jingjing start from the starting line at the same time. Liang Liang ran 6 meters per second, Jingjing ran 4 meters per second, Q: How many meters did Liang Liang run when he first caught up with Jingjing?

Analysis: Because it is a circular runway, when Liang Liang catches up with Jingjing for the first time, Liangliang runs just one week more than Jingjing, and the speed difference between the two is 6-4=2 meters per second, then Liangliang's first time catching up with Jingjing takes 200÷2=100 seconds. At this time, Liang Liang ran 100 × 6 = 600 meters, then Jingjing ran 600-200 = 400 meters.

Solution: 200 ÷ (6-4) ×6

=200÷2×6，

=600 (m);

600-200=400 (m)

Answer: When Liang Liang first caught up with Jingjing, Liangliang ran 600 meters, and Jingjing ran 400 meters.

B-stop

1． A travels 17 kilometers per hour, B travels 24 kilometers per hour, two people in the same place at the same time opposite each other, one to the east, one to the west, a few hours later the two people are 164 kilometers apart?

Analysis: After a few hours, the two people are separated by 164 kilometers, that is, after a few hours, A and B travel 164 kilometers, according to the time = distance÷ speed (the speed of A + the speed of B), you can solve.

Solution: 164÷ (17+24),

=164÷41，

=4 (hours);

Answer: After 4 hours, the two were separated by 164 kilometers.

2． A and B raced around the 1540-meter-long circular square, and it is known that A walked 160 meters per minute, and B's speed was 3 times that of A. Now A is 260 meters behind B, how many minutes does it take B to catch up with A?

Analysis: A walks 160 meters per minute, B's speed is 3 times that of A, then B's speed is 160×3 = 480 m / min, so the speed difference between the two is 480-160 = 320 m / min, now A is 260 meters behind B, because it is a race walk in the circular square, the distance difference between B and A is 1540-260 = 1280 meters, so it takes B to catch up with A÷ 320 = 4 minutes.

Solution: (1540-260) ÷ (160×3-160)

=1280÷（480﹣160）

=1280÷320

=4 (minutes);

Answer: It takes 4 minutes for B to catch up with A.

3． A travels 10 kilometers per hour, B travels 12 kilometers per hour, two people in the same place at the same time opposite each other, one to the south, one to the north, a few hours later the two people are 88 kilometers apart?

Analysis: After a few hours, the two people are separated by 88 kilometers, that is, after a few hours, A and B travel 88 kilometers, according to the time = distance÷ speed (the speed of A + the speed of B), you can solve.

Solution: 88÷ (10+12),

=88÷22，

Answer: After 4 hours, the two were separated by 88 kilometers.

4． A and B trains travel in opposite directions from 700 km apart at the same time, train A travels 85 kilometers per hour, train B travels 90 kilometers per hour, how many hours will the two trains meet?

Analysis: The distance between the two places and the speed of the two cars are known, so according to: the distance ÷ speed and = encounter time are solved

Solution: 700 ÷ (85+90)

=700÷175，

=4 (hours).

Answer: Two trains meet 4 hours later.

5． Two trains depart from two stations at the same time, car A travels 48 kilometers per hour, car B travels 78 kilometers per hour, and after 2.5 hours the two cars meet. How long is the railway between two stations?

Analysis: The speed and encounter time of the two vehicles are known, according to the basic relationship of the encounter problem: speed and × encounter time = distance to solve;

Solution: (48+78) ×2.5

=126×2.5，

=315 (km);

Answer: The length of the railway between the two stations is 315 km.

C gear

1． The two masters and apprentices work together to process 520 parts, the master processes 30 parts per hour, the apprentice processes 20 per hour, and there are still 70 parts that have not been processed after a few hours?

Analysis: This question first finds out the parts that the two masters and apprentices want to work together to process, and then according to the relationship formula: the total amount of work ÷ the sum of work efficiency = working hours.

Solution: (520-70) ÷ (30+20),

=450÷50，

=9 (hours);

Answer: 70 parts are still not machined after 9 hours.

2． The two ships A and B sailed from 654 kilometers apart, and the two ships were 22 kilometers apart in 8 hours. It is known that ship B travels 42 kilometers per hour, and how many kilometers does ship A travel per hour?

Analysis: First find out the distance traveled by ship B for 8 hours, so that the distance traveled by ship A for 8 hours can be found, and then the relationship between distance, speed and time can be solved.

Solution: (654-22-42×8) ÷8,

=296÷8，

=37 (km).

Answer: A ship travels 37 kilometers per hour.

3． A car and a bicycle departed at the same time from A and B, 172.5 kilometers apart, and traveled in opposite directions, and after 3 hours the two cars met. It is known that cars travel 31.5 kilometers more per hour than bicycles, but what is the speed of cars and bicycles?

Analysis: From the title, it can be found that the speed of the car and the bicycle is 172.5÷3 = 57.5 km / h, then the speed of the car is (57.5 + 31.5) ÷ 2 = 89÷ 2 = 44.5 km / h, the speed of the bicycle is easy to find.

Solution: (1) 172.5 ÷3 = 57.5 (km / h);

②（57.5+31.5）÷2，

=89÷2，

=44.5 (km/h);

(3) 44.5-31.5 =13 (km/h).

Answer: The speed of the car and bicycle is 44.5 km/ h and 13 km/h respectively.

4． The distance between the two places is 270 kilometers, and two trains, A and B, depart from opposite places at the same time and meet after 4 hours. It is known that the speed of car A is 1.5 times that of car B, how many kilometers per hour do the two trains of A and B travel?

Analysis: According to the distance ÷ encounter time = speed sum, find out the speed sum of A and B cars, and then according to the knowledge of the problem of the sum, you can find the speed of A and B cars respectively.

Solution: velocity sum: 270÷4 = 67.5 km,

B speed: 67.5÷ (1 + 1.5) = 27 (km),

Vehicle speed: 67.5-27 = 40.5 km

Answer: Trains A and B travel 40.5 km and 27 km per hour respectively.

5． The distance between the two cities A and B is 680 kilometers, the ordinary bus from A city to B city travels 60 kilometers per hour, after 2 hours, the express bus drives from city B to city A, 80 kilometers per hour, and the two cars meet after a few hours.

Analysis: As shown in the figure: the total distance minus the 2-hour journey of the ordinary car, divided by the sum of the speed of the two cars, that is, the time required for the two cars to meet.

Solution: (680-60×2) ÷ (60+80),

=（680﹣120）÷140，

=560÷140，

=4 (hours)

Answer: The two cars meet after the express train drives for 4 hours.

practice:

1． A and B are 3300 meters apart, A and B are walking from the two places at the same time, A walks 82 meters per minute, B walks 83 meters per minute, has been walking for 15 minutes, how many more minutes can we meet?

Analysis: This problem knows the distance between the two places and the speed of the two people, so first according to the distance ÷ speed and = encounter time to find out the encounter time, and then subtract the time that has been done, that is, how much time is needed to meet.

Solution: 3300÷ (82+83)-15

=3300÷165﹣15，

=20﹣15，

=5 (minutes)

Answer: It will take another 5 minutes to meet.

2． A and B trains depart from two places at the same time and go in opposite directions. It is known that car A travels 45 kilometers per hour, car B travels 32 kilometers per hour, and when they meet, car A travels 52 kilometers more than car B. How many kilometers are the distance between A and B?

Analysis: Divide the distance between the two cars by dividing the distance between the two cars by the distance of the car that is more than the car B, find out the time when the two cars meet, and then multiply the speed sum to find the distance between the two places.

Solution: 52÷ (45-32) × (45+32),

=52÷13×77，

=308 (km).

Answer: The distance between A and B is 308 km.

3． The sisters simultaneously traveled from their home to the Children's Palace, a total distance of 770 meters. The younger sister walked 60 meters per minute, and the sister rode a bicycle at a speed of 160 meters per minute to reach the Children's Palace and immediately returned, meeting her sister on the way. How many minutes did my sister walk away?

Analysis: The sister returned and met her sister on the way" When they walked a total of 2 full steps, the meeting time was 770×2÷ (60 + 160) = 7 (minutes). Because neither of them stopped, the sister also walked for 7 minutes.

Solution: 770×2÷ (60+160),

=1540÷220，

=7 (min).

Answer: Sister walked for 7 minutes.

Xiaoming and Xiaohua set off from A and B at the same time, heading in opposite directions. Xiaoming walked 60 meters per minute, Xiaohua rode a bicycle 190 meters per minute, and after a few minutes the two met at 650 meters from the midpoint?

Analysis: It can be seen from the encounter between the two people at 650 meters from the midpoint, at this time Xiaohua has traveled 650 × 2 = 1300 (kilometers) than Xiaoming, and then according to the multi-line distance÷ speed difference = time can be solved.

Solution: 650×2÷ (190-60)

=1300÷130，

=10 (minutes)

Answer: After 10 minutes the two met at 650 meters from the midpoint.

2． A and B are 300 kilometers apart, and two cars depart from both places at the same time and travel in opposite directions. Each arrived at their destination and returned immediately, after 8 hours they met for the second time. It is known that car A travels 45 kilometers per hour, and how many kilometers does car B travel per hour?

Analysis: Because they go in opposite directions, each reaches the destination and then returns immediately, they should be in the express car back to the A place and then on the way to the B place and the slow car back to the A place to meet, so when they meet they traveled 3 full courses, that is, 300 × 3 = 900 (kilometers), the known encounter time is 8 hours, the A car travels 45 kilometers per hour, so the speed of the two cars is 900÷8-45.

Solution: 300×3÷8-45,

=112.5﹣45，

=67.5(km).

Answer: Car B travels 67.5 km per hour.