Distance, time and speed are the three basic quantities of travel problem, and their relationships are as follows:

Distance = time × speed,

Time = distance/speed,

Speed = distance/time.

This lecture is to deepen the understanding of these three basic quantitative relations through examples.

Example: 1 A motorcade slowly passed a 200-meter-long bridge at a speed of 4 meters per second, and it took * * * 1 15 seconds. It is known that each car is 5 meters long and the distance between two cars is 10 meter. Q: How many cars are there in this convoy?

Analysis and solution: If you want to know how many cars there are in the fleet, you need to know the length of the fleet first. The length of the fleet is equal to the distance of the fleet of 1 15 seconds minus the length of the bridge. From "distance = time× speed", it can be found that the motorcade distance of 1 15 seconds is 4× 1 15=460 (meters).

So the length of the motorcade is 460-200=260 (meters). From the problem of planting trees, it can be concluded that the motorcade has (260-5) ÷ (5+10)+1=18 (vehicles).

Example 2 rides a bike from A to B at the speed of 10 km/h and arrives at 1 in the afternoon; Drive at the speed of 15km/h and arrive at 1 1 in the morning. What speed should I drive if I want to arrive at noon 12?

Analysis and solution: this problem has no departure time, and there is no distance between A and B, that is to say, there is neither time nor distance, and it seems impossible to find the speed. This requires finding out the time and distance through known conditions.

Suppose A and B leave from A to B at the same time, A travels 10 km per hour and arrives1in the afternoon; B every hour 15km, and arrive at 1 1 in the morning. When B arrives at B, A is still 10×2=20 (km) away from B, which is the distance that B travels from A to B more than A. Because B has more 15- 10=5 (km) lines per hour than A, the time from A to B is therefore.

20( 15- 10)= 4 (hours).

Therefore, A and B start at 7 am, and the distance between A and B is

15×4=60 (km).

If you want to arrive at noon 12, that is, to drive 60 kilometers at (12-7=)5 o'clock, the speed should be

60( 12-7)= 12 (km/h).

Example 3 Two racing schemes were discussed before the rowing competition. The first scheme is to row halfway at the speed of 2.5 m/s and 3.5 m/s respectively; The second scheme is to arrange half the competition time at the speed of 2.5 m/s and 3.5 m/s respectively. Which of the two schemes is better?

Analysis and solution: When the distance is fixed, the faster the speed, the shorter the time required. The speed of these two schemes is not fixed, so it is not easy to compare them directly. In the second scheme, because the paddling time of the two speeds is the same, the paddling distance at the speed of 3.5m/s is longer than that at the speed of 2.5m/s .. The paddling distance at the speed of 2.5m/s is represented by a single line, and the paddling distance at the speed of 3.5m/s is represented by a double line, so a comparison diagram of the two schemes is drawn as shown in the following figure. Where section A+section B = section C. ..

In paragraphs A and C, the two schemes take the same time; In section B, because the distance is the same, the second scheme is faster than the first scheme, so the second scheme takes less time than the first scheme.

To sum up, of the two schemes, the second scheme takes less time than the first scheme, that is, the second scheme is better.

Example 4 Xiaoming climbs the mountain, walking 2.5 kilometers per hour when going up the mountain, walking 4 kilometers per hour when going down the mountain, and it takes 3.9 hours to go back and forth. Q: How many kilometers did Xiao Ming walk back and forth?

Analysis and solution: Because the distance up the mountain and down the mountain is the same, if we can find out the time required to go up the mountain 1 km and down the mountain 1 km, we can find out the total distance up the mountain and down the mountain.

Because it takes 1 km to go up and down the mountain.

So the total distance up and down the mountain is

In the travel problem, there is also a concept of average speed: average speed = total distance ÷ total time.

For example, the average speed of uphill and downhill in Example 4 is

An ant crawls along three sides of an equilateral triangle. If three sides crawl 50 cm, 20 cm and 40 cm per minute, how many centimeters do ants crawl per minute on average?

Solution: Let the side length of an equilateral triangle be l cm, then the time required for an ant to crawl for one week is

Ants crawl once a minute on average for a week.

There is a kind of "drifting boat" problem in the travel problem. When using the relationship between distance, time and speed to solve such problems, we should pay attention to the significance and relationship of various speeds:

Downstream velocity = still water velocity+water velocity,

Countercurrent velocity = still water velocity-water flow velocity,

Static water velocity = (downstream velocity+countercurrent velocity) ÷2,

Water velocity = (downstream velocity-countercurrent velocity) ÷2.

Here, the still water speed, downstream speed and countercurrent speed refer to the speed of the ship in still water, downstream and countercurrent respectively.

The distance between two wharves is 4 18km. Motorboat needs 65,438+065,438+0 to travel downstream and 65,438+09 to travel upstream. Find the velocity of the river.

Solution: water velocity = (downstream velocity-countercurrent velocity) ÷2

=(4 18÷ 1 1-4 18÷ 19)÷2

=(38-22)÷2

=8 km/h

The current flow rate of this river is 8 kilometers per hour.

Exercise 24

1. Xiaoyan rides a bike at school and walks when she goes home. The journey takes 50 minutes. If you walk back and forth, it takes 70 minutes. How long does it take to go back and forth by bike?

2. Someone wants to go to a farm 60 kilometers away. At first, he walked at a speed of 5 kilometers per hour. Later, a tractor with a speed of 18 km took him to the farm, which took 5.5 hours. Q: How far did he go?

3. It is known that the length of the railway bridge is 1 000m, and a train passes through the bridge. It is measured that it takes 1.20 seconds for the train to get off the bridge completely from the beginning, and the time for the whole train to stay on the bridge completely is 80 seconds. Find the speed and length of the train.

4. Xiaohong takes a break every 30 minutes when going up the mountain 10 minute, and takes a break every 30 minutes when going down the mountain for 5 minutes. It is known that Xiaohong's downhill speed is 0.5 times of 65438+ uphill speed. If it takes 3: 50, how long will it take to go downhill?

The car traveled from A to B at a speed of 72km/h, and immediately returned to A at a speed of 48km/h upon arrival. Find the average speed of the car.

6. The distance between the two places is 480 kilometers, and there is a ship sailing in it. Downstream 16 hours, countercurrent for 20 hours, and find the flow rate.

7. When a ship sails between two wharves of a river, it is 6 hours downstream and 8 hours upstream, and the current speed is 2.5 km/h. Find the speed of the ship in still water.

Exercise 24

1.30 points.

Tip: Cycling saves 70-50=20 minutes compared with one-way walking.

2. 15km。

Solution: If he walks X kilometers, then x÷5+(60-x)÷ 18=5.5.

The solution is x= 15 (km).

3. 10 m/s; 200 meters.

Solution: Let the train be x meters long. According to the train speed, (1000+x) ÷120 = (1000-x) ÷ 80.

X=200 (m), the train speed is (1000+200) ÷120 =10 (m/s).

4.2: 15.

Solution: It took 60×3+50=230 (minutes) to climb the mountain, from 230 ÷ (30+ 10) = 5...30, and I got five breaks and walked 230-10× 5 =/kloc-0. According to 120÷30=40, we broke three times on the way down, so we use 120+5×3= 135 (minutes) =2: 00 15.

5.57.6 km/h.

6.3 km/h.

Solution: (480÷ 16-480÷20)÷2=3 (km/h).

7. 17.5 km/h.

Solution: Let the distance between two docks be X kilometers. From the speed of water flow

The solution is x= 120 (km). So the speed of the ship in still water is120 ÷ 6-2.5 =17.5 (km/h).

Lecture 25 Travel Problems (2)

This lecture focuses on the problems of encounter and pursuit. In these two problems, the relationship between distance, time and speed is as follows:

In practical problems, we always know two of distance, time and speed, and find the other one.

Example: 1 car a has a speed of 40 kilometers per hour and car b has a speed of 60 kilometers per hour. Two cars leave from a and b at the same time, in opposite directions. At 3 o'clock after the encounter, a car arrived at B. Find the distance between A and B.

Analysis and solution: first draw a schematic diagram as follows:

Point c in the figure is the meeting place. Because it is 3: 00 a.m. from point C to point B, the distance from C to B is 40×3= 120 (km).

This 120km the second car drove 120÷60=2 (hours), which means that when they met, the two cars had been driving for 2 hours each, so the distance between A and B is (40+60)×2=200 (kilometers).

Xiaoming leaves home for school on time every morning, and Uncle Li goes out for a walk regularly every morning. They walked in opposite directions. Xiaoming walks 60 meters every minute, and Uncle Li walks 40 meters every minute. They meet at the same time every day. One day Xiaoming went out early, so he met Uncle Li 9 minutes earlier than usual. How many minutes does Xiaoming leave home earlier than usual?

Analysis and solution: Because we met 9 minutes in advance, it shows that when Uncle Li went out, Xiaoming walked 9 minutes longer than usual, that is, (60+40)×9=900 (meters).

So Xiao Ming left 900÷60= 15 (minutes) earlier than usual.

Xiao Gang is walking along the road beside the railway. His walking speed was 2m/s, when a train was approaching. It took him 18 seconds to get from the front to the rear. The total length of the known train is 342 meters. Find the speed of the train.

Analysis and solutions:

In the above picture, A is where Xiao Gang met the train and B is where Xiao Gang got off the train. According to the meaning of the question, it took 18 seconds for Xiao Gang to walk from A to B, and for the motorcycle to walk from A to C, because C to B was just the length of the train, and it took 18 seconds for Xiao Gang to follow the train. It is inferred that the speed sum of Xiao Gang and the train is 342 ÷18 =19.

Therefore, the train speed is 19-2= 17 (m/s).

There is a railway road next to the railway line, and a tractor on the road is running at a speed of 20 kilometers per hour. At this time, a train came from behind at a speed of 56 km/h, and it took 37 seconds for the train to pass the tractor from front to back. Find out the full length of the train.

Analysis and solutions

It is similar to Example 3, except that the problem of meeting in the opposite direction becomes the problem of catching up in the same direction. As can be seen from the above figure, within 37 seconds, the locomotive goes from B to C and the tractor goes from B to A, and the distance traveled by the train is longer than that of the tractor. Using meters as the unit of length and seconds as the unit of time, the length of the train can be obtained by the following formula

Speed difference × catch-up time

= [(56000-20000)÷3600]×37

= 370 (meters).

Example 5, as shown on the right, forms a square with a side length of 300 meters along the path outside the wall of a unit, and both parties start from two diagonal counterclockwise directions at the same time. It is known that A walks 90 meters per minute and B walks 70 meters per minute. Q: How long will it take for A to meet B at least?

Analysis and solution: A can only see B when A and B are on the same side (including the end point). Catch up with Otsuichi once, that is, catch up with 300 meters.

300(90-70)= 15 (minutes). At this time, the distance between Party A and Party B is a side length. Party A walks 90× 15÷300=4.5 (side), which is located at the midpoint of one side and Party B is located at the midpoint of the other side, so Party A can see Party B after walking 0.5 side, that is, Party A can see Party B after walking 5 sides, which requires * * *.

The hounds chased the hare for 30 meters. The hound has a big step. It goes in four steps. The rabbit runs seven steps, but the rabbit runs very fast. A hound can run four steps out of three. How far can hounds run to catch up with rabbits?

Analysis and solution: the conditions of this problem are relatively hidden, and the time and speed are not obvious. In order to find out the speed relationship between rabbits and hounds, we change the condition to the situation that hounds run 12 steps (think about why this changes):

(1) The distance that the hound runs 12 step is equal to the distance that the rabbit runs 2 1 step;

(2) The time for hounds to run 12 step is equal to the time for rabbits to run 16 step.

Therefore, in the time when the hound runs 12 step, the hound can run 12 step, which is equivalent to the rabbit running.

That is to say, every time the hound runs 2 1 m, the rabbit runs 16 m, and the hound needs to run 21× [30 ÷ (21-kloc-0/6)] =126 (.

Exercise 25

1.Village A and Village B are 2800 meters apart. Xiaoming walked from village A for 5 minutes, Xiaojun rode from village B. 10 minutes later, the two met again. It is known that Xiaojun walks more per minute by bike than Xiaoming130m. How many meters does Xiaoming walk every minute?

2. Two cars, A and B, leave from A and B at the same time. When they meet, they are 8 kilometers away from the centers of A and B. It is known that the speed of car A is 0.2 times that of 65438+ car B. Find the distance between A and B. ..

Xiaohong and Xiao Qiang set out from home at the same time and walked in the opposite direction. Xiaohong walks 52 meters per minute, and Xiao Qiang walks 70 meters per minute. They met at a station on the road. If Xiaohong leaves 4 minutes early, but the speed remains the same, and Xiao Qiang walks 90 meters every minute, then the two will still meet at point A.. How far is Xiaohong and Xiao Qiang's home?

An express train and a local train are in the opposite direction. The length of the express train is 280 meters and the length of the local train is 385 meters. The time for people sitting on the express train to see the local train pass is 1 1 sec, and how many seconds for people sitting on the local train to see the express train pass?

5.A and B go from A to B at the same time. A cycling is 250 meters per minute, and you must rest for 20 minutes after every 10 minute; B walking intermittently, walking every minute 100 meters. As a result, when A was about to rest, two people arrived at B at the same time. Q: How far is A and B?

6. Party A and Party B walk counterclockwise at the same time from two opposite vertices of a square pool with a circumference of 1600 meters, walking 50 meters and 46 meters per minute respectively. How long did it take them to walk on the same side for the first time after they set off?

7. A hunting dog is chasing a rabbit 20 meters ahead. It is known that dogs jump 3 meters, rabbits jump 2. 1 meter, dogs jump 3 times and rabbits jump 4 times. How far will the rabbit run out and be chased by the hounds?

Exercise 25

1.60m.

Solution: (2800-130×10) ÷ (10× 2+5) = 60 (m).

2. 176 km.

3.2196m.

Solution: Because Xiaohong's speed and meeting place are the same, Xiaohong's time of walking twice is the same. It is inferred that Xiao Qiang walked 4 points less than the first time. Multiply (70×4)÷(90-70)= 14 (minutes),

It is inferred that Xiao Qiang's second walk 14 points and his first walk 18 points. The distance between their homes is (52+70)× 18=2 196 (meters).

4.8 seconds.

Tip: people on the express train watch the slow train at the same speed as people on the slow train watch the express train.

(seconds).

5.10000m.

Solution: After departure 10 minutes, the distance between them is (250-100) ×10 =1500 (m).

Mi, need

B travels * * * 100 minutes, so the distance between A and B is100×100 =10000 (meters).

6. 104.

Solution: A needs 400÷(50-46)= 100 (minutes) to catch up with Otsuichi (400 meters).

At this time, A walked 50× 100=5000 (meters) and was located at the midpoint of an edge. It takes four points to walk 200 meters to reach the front vertex, so after starting, 100+4= 104 (minutes), they walk on the same side for the first time.

7.280 meters.

Solution: When the dog runs 3×3=9 (meters), the rabbit runs 2. 1×4=8.4 (meters), and when the dog catches up with the rabbit, the rabbit runs 8.4×[20÷(9-8.4)]=280 (meters).

Lecture 26 Travel Issues (3)

In the travel problem, we often encounter comprehensive problems such as meeting, catching up, the relationship between time, distance and speed. This kind of problem is difficult, and it often needs drawing to help clarify the relationship between quantity and quantity, decompose the comprehensive problem into several individual problems, and then solve them one by one.

The two roads cross. A go south from the intersection 1800 meters, go straight north, and b go straight east from the intersection. 12 After Party A and Party B leave at the same time, the distance from them to the intersection is equal; After 75 minutes of departure, the distance between them and the intersection is equal again. How many meters are they from the intersection at this time?

Analysis and solution: As shown in the left figure below, after 12 minutes, Party A will arrive at point B from point A, and Party B will arrive at point C from point O, with OB=OC. If B changes to the south, then this condition is equivalent to the meeting problem of "two people meet at a distance of 1800 meters, 12 minutes", and then two people make a line of1800150 (meters) every minute.

As shown at the top right, after 75 minutes of departure, Party A arrives at point E from point A, and Party B arrives at point F from point O, OE=OF. If B changes to go north, then this condition is equivalent to the catching-up problem of "the distance between two people is 1800 meters, and A catches up with B after 75 minutes", so the distance difference between two people walking per minute is 1800÷75=24 (meters).

From the sum-difference problem, we can find that the line that B walks every minute is (150-24)÷2=63 (meters).

75 minutes after departure, 63×75=4725 (meters) away from the intersection.

Example 2 The speeds of cars, vans and buses are 60km/h, 48km/h and 42km/h respectively. The car and the bus leave from A and B at the same time, and the van meets the bus 30 minutes later. Q: How far is the distance between A and B?

Analysis and solution: As shown in the figure below, the van and the car meet at point A. When the bus arrives at point B, it takes 30 minutes for the bus and the van to drive to BA.

From the problem of bus and van meeting, we know that BA = (48+42) × (30 ÷ 60) = 45 (km);

The time required for a car to travel 45 kilometers more than a bus is 45÷(60-42)=2.5 (hours) from the point of catching up;

During these 2.5 hours, a car and a van travel one-way to A and B, and the distance between A and B can be found from the encounter problem (60+48)×2.5=270 (km).

From the example 1 and example 2, it can be seen that the more complex comprehensive problem can be decomposed into several single problems, which can achieve the purpose of turning the difficult into the easy.

Example 3 Xiaoming walked home at a constant speed along a bus route after school, and the bus on this road kept running at a constant speed. Every nine minutes, a bus overtook him from behind, and every seven minutes, he met an oncoming bus. Q: How many minutes does this bus leave?

Analysis and solution: this is a very difficult problem, which has a very hidden quantitative relationship. There are many solutions, but most of them are complicated and difficult to understand. In order to find out the relationship between quantities, we modify the subject conditions appropriately.

Suppose Xiaoming walked on the road for 63 minutes, then immediately turned around and walked for 63 minutes, and returned to his original place. 63 is chosen here because [7,9] = 63. At this point, in the first 63 minutes, he met 63÷7=9 (car) head-on, and in the second 63 minutes, 63÷9=7 (car) caught up with him, so in two 63 minutes, he * * * met a 16 car coming in the same direction, and the departure time interval was

Example 4 A and B swim straight back and forth in a 30-meter-long pool. The speed of A is 1m/s, and the speed of B is 0.6 m/s. They started from both ends of the pool and swam back and forth 1 1 min. If you don't count the turning time, during this time, they * *.

Analysis and solution: A takes 30÷ 1=30 (seconds) for one way, and B takes 30÷0.6=50 (seconds) for one way. A has five one-way trips and B has three one-way trips, all of which arrive at different destinations before leaving. The time of this process is 150 seconds, that is, 2.5 minutes, during which they meet five times (see the figure below), and the intersection of the real broken line and the imaginary broken line indicates the meeting point.

Taking 2.5 minutes as a cycle, 1 1 minute includes four cycles and 1 minute, but within 1 minute of a cycle, it can be seen from the figure that the two meet twice, so a * * * meets 5×4+2=22 (times

Example 4 Using the method of drawing, we can intuitively see the number of encounters in a cycle, which shows the importance of drawing.

Example 5 A and B started to climb the mountain from the foot of the mountain at the same time, and went down immediately after reaching the top of the mountain. Both of them went down the mountain at twice their own speed. When A reached the top of the mountain, B was 400 meters away. When A returned to the foot of the mountain, B just came down halfway up the mountain. Find the distance from the foot of the mountain to the top of the mountain.

Analysis and solution: the difficulty of this problem lies in the different speed of going up and down the mountain. If the speed of going up and down the mountain can be the same and the meaning of the problem is not changed, then the problem may become easier.

If two people go down the mountain at the same speed as themselves, then when A returns to the foot of the mountain.

The distance from the top of the mountain is