It means that two moving objects start in different places at the same time (or start in the same place but not at the same time, or start in different places but not at the same time) and move in the same direction. The back is fast, the front is slow, and in a certain period of time, the back catches up with the front. This application problem is called tracing problem.

Catch-up time = Catch-up distance ÷ (fast-slow)

Catch-up distance = (fast-slow) × catch-up time

Simple problem-solving ideas and methods directly use formulas, and complex problems are modified with formulas.

Example: 1 A good horse walks every day 120km, and a bad horse walks 75km every day. Bad horses go first 12 days. How many days can a good horse catch up with a bad horse?

How many kilometers can a bad horse walk 1 2 days? 75× 12 = 900 km

(2) How many days does a good horse catch up with a bad horse? 900 ÷ (120-75) = 20 (days)

The comprehensive formula is 75×12 ÷ (120-75) = 900 ÷ 45 = 20 (days).

A: A good horse can catch up with a bad horse in 20 days.

Example 2 Xiaoming and Xiao Liang are running on the 200m circular track. Xiao Ming ran for 40 seconds. They started from the same place and ran in the same direction at the same time. Xiaoming ran 500 meters when he first caught up with Liang Xiao. What is the speed of Xiao Liang per second?

When Xie Xiaoming caught up with Liang Xiao for the first time, he ran one lap more than Liang Xiao, that is, 200 meters. At this time, Xiao Liang ran (500-200m). To know the speed of Xiao Liang, you should know the time, that is, it takes Xiaoming to run 500 meters. I also know that it takes 40 seconds for Xiao Ming to run 200 meters and [40× (500 ÷ 200)] seconds for running 500 meters, so the speed of Xiao Liang is

(500-200)÷〔40×(500÷200)〕

= 300 ÷ 100 = 3 (m)

The speed of Xiao Liang is 3 meters per second.

Our People's Liberation Army pursued the fleeing enemy. The enemy began to flee from place A at the speed of10km per hour on16km in the afternoon, and the PLA was ordered to pursue from place B at the speed of 30km per hour at 22pm. As we all know, the distance between A and B is 60 kilometers. How many hours can the PLA catch up with the enemy?

The time difference between the enemy's escape time and the PLA's pursuit time is (22- 16) hours. During this period, the enemy's escape distance is [10× (22-6)] km, and the distance between Party A and Party B is 60 km. Infer from this

Catch-up time = [10× (22-6)+60] ÷ 30-10

= 220 ÷ 20 = 1 1 (hours)

A: The PLA can catch up with the enemy after 1 1 hour.

A bus travels from Station A to bilibili at a speed of 48 kilometers per hour. A truck was traveling from bilibili to Station A at the same time, with a speed of 40 kilometers per hour. Two trucks meet at the midpoint of two stations 16 km, and find the distance between the two stations.

Solving this problem can change from meeting a problem to chasing it. As can be seen from the topic, the bus lags behind the truck (16×2) kilometers, and the time when the bus catches up with the truck is the aforementioned meeting time.

This time is 16× 2 ÷ (48-40) = 4 (hours).

So the distance between the two stations is (48+40) × 4 = 352 (km).

Column into a comprehensive formula (48+40) × [16× 2 ÷ (48-40)]

=88×4

= 352 km

A: The distance between Station A and bilibili is 352 kilometers.

Both brother and sister go to school from home at the same time. My brother walks 90 meters per minute and my sister walks 60 meters per minute. When my brother arrived at the school gate, he found that he had forgotten his textbook. He immediately went home to get it along the original road and met his sister at 0/80 meters away from the school/kloc-. How far is their home from school?

The solution requires that the distance and speed are known, so the key is to find the meeting time. It can be seen from the question that in the same time (from departure to meeting), my brother walked (180×2) meters more than my sister, because my brother walked (90-60) meters more than my sister every minute.

So, the time it takes two people from home to meet is

180× 2 ÷ (90-60) = 12 (minutes)

The distance from home to school is 90× 12- 180 = 900 (meters).

A: My home is 900 meters from school.

Sun Liang plans to go to school five minutes before class. He walks from home to school at a speed of 4 kilometers per hour. When he walked 1 km, he found that his watch was 10 minutes slow, so he immediately ran forward and arrived at school on time. Later, I calculated that if Sun Liang had run away from home, he would have walked to school nine minutes earlier than before. Seek Sun Liang's running speed.

Taking off the watch is slow 10 minutes, which means it is late 10 minutes. If you continue to walk at the original speed, you will be (10-5) minutes late, and you will arrive at school on time for the second run, which means that running takes (10-5) minutes less than walking. If you run from home, it's 9 minutes less than walking. So running takes [9-( 10-5)] minutes less than walking.

therefore

Walking needs 1 km [9-( 10-5)].

= 0.25 (hour)

= 15 (minutes)

Running time 1 km15-[9-(10-5)] =11(minutes).

The running speed is1÷11/60 = 5.5 (km) per hour.

A: The running speed in Sun Liang is 5.5 kilometers per hour.