Problems involving oncoming traffic (finding time and speed). Oncoming traffic problems
Distance/speed/time problem
Task 1. A car is moving at a speed of 80 km/h. How many kilometers will he travel in 3 hours?
Solution
If a car travels 80 kilometers in one hour, then in 3 hours it will travel three times as much. To find the distance, you need to multiply the speed of the car (80 km/h) by the driving time (3 hours)
80 × 3 = 240 km
Answer: in 3 hours the car will travel 240 kilometers.
Task 2. In a car we traveled 180 km in 3 hours at the same speed. What is the speed of the car?
Solution
Speed is the distance traveled by a body per unit time. A unit means 1 hour, 1 minute or 1 second.
If in 3 hours a car traveled 180 kilometers at the same speed, then dividing 180 km by 3 hours we determine the distance that the car traveled in one hour. And this is the speed of movement. To determine the speed, you need to divide the distance traveled by the time of movement:
180: 3 = 60 km/h
Answer: car speed is 60 km/h
Task 3. In 2 hours the car traveled 96 km, and in 6 hours the cyclist traveled 72 km. How many times faster was the car moving than the cyclist?
Solution
Let's determine the speed of the car. To do this, divide the distance he traveled (96 km) by the time he moved (2 hours)
96: 2 = 48 km/h
Let's determine the speed of the cyclist. To do this, divide the distance he traveled (72 km) by the time he moved (6 hours)
72: 6 = 12 km/h
Find out how many times the car moved faster than the cyclist. To do this, find the ratio 48 to 12
Answer: The car moved 4 times faster than the cyclist.
Problem 4. The helicopter covered a distance of 600 km at a speed of 120 km/h. How long was he in flight?
Solution
If a helicopter covered 120 kilometers in 1 hour, then finding out how many of these 120 kilometers are in 600 kilometers, we will determine how long it was in flight. To find the time, you need to divide the distance traveled by the speed of movement.
600: 120 = 5 hours
Answer: The helicopter was on the way for 5 hours.
Problem 5. The helicopter flew for 6 hours at a speed of 160 km/h. How much distance did he cover during this time?
Solution
If in 1 hour the helicopter covered 160 km, then in 6 hours it covered six times more. To determine the distance, you need to multiply the speed of movement by the time
160 × 6 = 960 km
Answer: in 6 hours the helicopter covered 960 km.
Problem 6. The car covered the distance from Perm to Kazan, 723 km, in 13 hours. For the first 9 hours he drove at a speed of 55 km/h. Determine the speed of the car in the remaining time.
Solution
Let's determine how many kilometers the car traveled in the first 9 hours. To do this, multiply the speed at which he drove for the first nine hours (55 km/h) by 9
55 × 9 = 495 km
Let's determine how much time is left to travel. To do this, subtract from the total distance (723 km) the distance traveled during the first 9 hours of movement
723 − 495 = 228 km
The car covered these 228 kilometers in the remaining 4 hours. To determine the speed of the car in the remaining time, you need to divide 228 kilometers by 4 hours:
228: 4 = 57 km/h
Answer: the vehicle speed during the remaining time was 57 km/h
Closing speed
The speed of approach is the distance traveled by two objects towards each other per unit of time.
For example, if two pedestrians start from two points towards each other, and the speed of the first is 100 m/m, and the second is 105 m/m, then the speed of approach will be 100+105, that is, 205 m/m. This means that every minute the distance between pedestrians will decrease by 205 meters.
To find the closing speed, you need to add up the speeds of the objects.
Let's assume that the pedestrians meet three minutes after the start of the movement. Knowing that they met three minutes later, we can find out the distance between the two points.
Every minute, pedestrians covered a distance of two hundred and five meters. After 3 minutes they met. This means that by multiplying the speed of approach by the time of movement, we can determine the distance between two points:
205 × 3 = 615 meters
There is another way to determine the distance between points. To do this, find the distance that each pedestrian walked before the meeting.
So, the first pedestrian walked at a speed of 100 meters per minute. The meeting took place in three minutes, which means in 3 minutes he walked 100x3 meters
100 × 3 = 300 meters
And the second pedestrian walked at a speed of 105 meters per minute. In three minutes he walked 105x3 meters
105 × 3 = 315 meters
Now you can add up the results and thus determine the distance between two points:
300 m + 315 m = 615 m
Task 1. Two cyclists rode out from two settlements towards each other at the same time. The speed of the first cyclist is 10 km/h, and the speed of the second is 12 km/h. After 2 hours they met. Determine the distance between settlements
Solution
Let's find the speed of approach of cyclists
10 km/h + 12 km/h = 22 km/h
Let's determine the distance between settlements. To do this, multiply the speed of approach by the time of movement
22 × 2 = 44 km
Let's solve this problem in the second way. To do this, we will find the distances covered by the cyclists and add up the results.
Let's find the distance covered by the first cyclist:
10 × 2 = 20 km
Let's find the distance covered by the second cyclist:
12 × 2 = 24 km
Let's add up the resulting distances:
20 km + 24 km = 44 km
Answer: the distance between settlements is 44 km.
Problem 2. From two settlements, the distance between which is 60 km, two cyclists rode towards each other at the same time. The speed of the first cyclist is 14 km/h, and the speed of the second is 16 km/h. How many hours later did they meet?
Solution
Let's find the speed of approach of cyclists:
14 km/h + 16 km/h = 30 km/h
In one hour, the distance between cyclists decreases by 30 kilometers. To determine how many hours later they will meet, you need to divide the distance between populated areas by the speed of approach:
60:30 = 2 hours
So the cyclists met in two hours
Answer: The cyclists met after 2 hours.
Problem 3. From two settlements, the distance between which is 56 km, two cyclists rode towards each other at the same time. Two hours later they met. The first cyclist was traveling at a speed of 12 km/h. Determine the speed of the second cyclist.
Solution
Let's determine the distance covered by the first cyclist. Like the second cyclist, he spent 2 hours on the road. By multiplying the speed of the first cyclist by 2 hours, we can find out how many kilometers he walked before the meeting
12 × 2 = 24 km
In two hours, the first cyclist covered 24 km. In one hour he walked 24:2, that is, 12 km. Let's depict this graphically
Subtract from the total distance (56 km) the distance covered by the first cyclist (24 km). This is how we determine how many kilometers the second cyclist has traveled:
56 km − 24 km = 32 km
The second cyclist, like the first, spent 2 hours on the road. If we divide the distance he traveled by 2 hours, we will find out at what speed he moved:
32: 2 = 16 km/h
This means the speed of the second cyclist is 16 km/h.
Answer: The speed of the second cyclist is 16 km/h.
Removal speed
Recession velocity is the distance that increases per unit time between two objects moving in opposite directions.
For example, if two pedestrians start from the same point in opposite directions, and the speed of the first is 4 km/h, and the speed of the second is 6 km/h, then the speed of removal will be 4+6, that is, 10 km/h. Every hour the distance between two pedestrians will increase by 10 kilometers.
To find the speed of removal, you need to add up the speeds of the objects.
So, in the first hour the distance between pedestrians will be 10 kilometers. In the following picture you can see how this happens
It can be seen that the first pedestrian walked his 4 kilometers in the first hour. The second pedestrian also completed his 6 kilometers in the first hour. In total, in the first hour the distance between them became 4+6, that is, 10 kilometers.
After two hours, the distance between pedestrians will be 10x2, that is, 20 kilometers. In the following figure you can see how this happens:
Task 1. A freight train and an express passenger train departed from the same station simultaneously in opposite directions. The speed of a freight train was 40 km/h, the speed of an express train was 180 km/h. What will be the distance between these trains after 2 hours?
Solution
Let us determine the speed at which trains move away. To do this, let's add up their speeds:
40 + 180 = 220 km/h
We obtained a train removal speed of 220 km/h. This speed shows that in an hour the distance between trains will increase by 220 kilometers. To find out what the distance between trains will be in two hours, you need to multiply 220 by 2
220 × 2 = 440 km
Answer: in 2 hours the distance between trains will be 440 kilometers.
Task 2. A cyclist and a motorcyclist left the point at the same time in opposite directions. The speed of the cyclist is 16 km/h, and the speed of the motorcyclist is 40 km/h. What is the distance between the cyclist and the motorcyclist after 2 hours?
Solution
16 km/h + 40 km/h = 56 km/h
Let's determine the distance that will be between the cyclist and the motorcyclist after 2 hours. To do this, multiply the removal speed (56 km/h) by 2 hours
56 × 2 = 112 km
Answer: After 2 hours the distance between the cyclist and the motorcyclist will be 112 km.
Problem 3. A cyclist and a motorcyclist left the point at the same time in opposite directions. The speed of the cyclist is 10 km/h, and the speed of the motorcyclist is 30 km/h. After how many hours will the distance between them be 80 km?
Solution
Let us determine the speed of removal of the cyclist and motorcyclist. To do this, let's add up their speeds:
10 km/h + 30 km/h = 40 km/h
In one hour, the distance between a cyclist and a motorcyclist increases by 40 kilometers. To find out after how many hours the distance between them will be 80 km, you need to determine how many times 80 km contains 40 km
80: 40 = 2
Answer: 2 hours after the start of the movement, there will be 80 kilometers between the cyclist and the motorcyclist.
Problem 4. A cyclist and a motorcyclist left the point at the same time in opposite directions. After 2 hours, the distance between them was 90 km. The cyclist's speed was 15 km/h. Determine the speed of the motorcyclist
Solution
Let us determine the distance covered by the cyclist in 2 hours. To do this, multiply its speed (15 km/h) by 2 hours
15 × 2 = 30 km
The figure shows that the cyclist walked 15 kilometers in every hour. In total, in two hours he walked 30 kilometers.
Subtract the distance traveled by the cyclist (30 km) from the total distance (90 km). This is how we determine how many kilometers the motorcyclist has traveled:
90 km − 30 km = 60 km
A motorcyclist covered 60 kilometers in two hours. If we divide the distance he traveled by 2 hours, we will find out at what speed he moved:
60: 2 = 30 km/h
This means the speed of the motorcyclist was 30 km/h.
Answer: The motorcyclist's speed was 30 km/h.
The task of moving objects in one direction
In the previous topic, we looked at problems in which objects (people, cars, boats) moved either towards each other or in opposite directions. At the same time, we found various distances that changed between objects over a certain time. These distances were either closing speeds or removal rates.
In the first case we found closing speed- in a situation where two objects were moving towards each other. Per unit time, the distance between objects decreased by a certain distance
In the second case, we found the speed of removal - in a situation where two objects were moving in opposite directions. For a unit of time, the distance between objects increased by a certain distance
But objects can also move in the same direction, and at different speeds. For example, a cyclist and a motorcyclist can leave the same point at the same time, and the speed of the cyclist can be 20 kilometers per hour, and the speed of the motorcyclist can be 40 kilometers per hour
The figure shows that the motorcyclist is twenty kilometers ahead of the cyclist. This is due to the fact that he covers 20 kilometers more per hour than a cyclist. Therefore, every hour the distance between a cyclist and a motorcyclist will increase by twenty kilometers.
In this case, 20 km/h is the speed at which the motorcyclist moves away from the cyclist.
After two hours, the distance covered by the cyclist will be 40 km. The motorcyclist will travel 80 km, moving away from the cyclist by another twenty kilometers - in total, the distance between them will be 40 kilometers
To find the speed of removal when moving in one direction, you need to subtract the lower speed from the higher speed.
In the example above, the removal speed is 20 km/h. It can be found by subtracting the speed of the cyclist from the speed of the motorcyclist. The cyclist's speed was 20 km/h, and the motorcyclist's speed was 40 km/h. The motorcyclist's speed is greater, so subtract 20 from 40
40 km/h − 20 km/h = 20 km/h
Problem 1. A car and a bus left the city in the same direction. The speed of the car is 120 km/h, and the speed of the bus is 80 km/h. What will be the distance between them after 1 hour? 2 hours?
Solution
Let's find the removal rate. To do this, subtract the lower speed from the higher speed
120 km/h − 80 km/h = 40 km/h
Every hour a passenger car moves 40 kilometers away from the bus. In one hour, the distance between the car and the bus will be 40 km. In 2 hours twice as much:
40 × 2 = 80 km
Answer: in one hour the distance between the car and the bus will be 40 km, in two hours - 80 km.
Let's consider a situation in which objects began their movement from different points, but in the same direction.
Let there be a house, a school and an attraction. From home to school 700 meters
Two pedestrians went on an attraction at the same time. Moreover, the first pedestrian went on an attraction from the house at a speed of 100 meters per minute, and the second pedestrian went on the ride from school at a speed of 80 meters per minute. What will be the distance between pedestrians after 2 minutes? How many minutes after the start of the movement will the first pedestrian catch up with the second?
Let's answer the first question of the problem - what will be the distance between pedestrians after 2 minutes?
Let us determine the distance covered by the first pedestrian in 2 minutes. He moved at a speed of 100 meters per minute. In two minutes he will travel twice as far, that is, 200 meters.
100 × 2 = 200 meters
Let us determine the distance covered by the second pedestrian in 2 minutes. He moved at a speed of 80 meters per minute. In two minutes he will travel twice as far, that is, 160 meters
80 × 2 = 160 meters
Now we need to find the distance between pedestrians
To find the distance between pedestrians, you can add the distance covered by the second pedestrian (160m) to the distance from home to school (700m) and subtract the distance covered by the first pedestrian (200m) from the result obtained.
700 m + 160 m = 860 m
860 m − 200 m = 660 m
Or, from the distance from home to school (700m), subtract the distance covered by the first pedestrian (200m), and add the distance covered by the second pedestrian (160m) to the result obtained.
700 m − 200 m = 500 m
500 m + 160 m = 660 m
Thus, after two minutes the distance between pedestrians will be 660 meters
Let's try to answer next question task: how many minutes after the start of the movement will the first pedestrian catch up with the second?
Let's see what the situation was like at the very beginning of the journey - when pedestrians had not yet begun to move
As can be seen in the figure, the distance between pedestrians at the beginning of the path was 700 meters. But within a minute after the start of movement, the distance between them will be 680 meters, since the first pedestrian moves 20 meters faster than the second:
100 m × 1 = 100 m
80 m × 1 = 80 m
700 m + 80 m − 100 m = 780 m − 100 m = 680 m
Two minutes after the start of movement, the distance will decrease by another 20 meters and will be 660 meters. This was our answer to the first question of the problem:
100 m × 2 = 200 m
80 m × 2 = 160 m
700 m + 160 m − 200 m = 860 m − 200 m = 660 m
After three minutes, the distance will decrease by another 20 meters and will already be 640 meters:
100 m × 3 = 300 m
80 m × 3 = 240 m
700 m + 240 m − 300 m = 940 m − 300 m = 640 m
We see that every minute the first pedestrian will approach the second one by 20 meters, and will eventually catch up with him. We can say that a speed of twenty meters per minute is the speed of approaching pedestrians. The rules for finding the speed of approach and distance when moving in one direction are identical.
To find the closing speed when moving in one direction, you need to subtract the smaller from the higher speed.
And since the original 700 meters decrease by the same 20 meters every minute, we can find out how many times 700 meters contain 20 meters, thereby determining how many minutes later the first pedestrian will catch up with the second
700: 20 = 35
This means that 35 minutes after the start of movement, the first pedestrian will catch up with the second. Just for fun, let’s find out how many meters each pedestrian has walked by this time. The first one moved at a speed of 100 meters per minute. In 35 minutes he covered 35 times more
100 × 35 = 3500 m
The second walked at a speed of 80 meters per minute. In 35 minutes he covered 35 times more
80 × 35 = 2800 m
The first walked 3500 meters, and the second 2800 meters. The first one walked 700 meters more because it was coming from the house. If we subtract these 700 meters from 3500, we get 2800 m
Consider a situation in which objects are moving in one direction, but one of the objects began its movement before the other.
Let there be a house and a school. The first pedestrian went to school at a speed of 80 meters per minute. Five minutes later, a second pedestrian followed him to school at a speed of 100 meters per minute. How many minutes will it take for the second pedestrian to catch up with the first?
The second pedestrian began his movement 5 minutes later. By this time, the first pedestrian had already moved some distance away from him. Let's find this distance. To do this, multiply its speed (80 m/m) by 5 minutes
80 × 5 = 400 meters
The first pedestrian moved 400 meters away from the second. Therefore, at the moment when the second pedestrian begins to move, there will be these same 400 meters between them.
But the second pedestrian moves at a speed of 100 meters per minute. That is, he moves 20 meters faster than the first pedestrian, which means that with every minute the distance between them will decrease by 20 meters. Our task is to find out in how many minutes this will happen.
For example, in just a minute the distance between pedestrians will be 380 meters. The first pedestrian will walk another 80 meters to his 400 meters, and the second will walk 100 meters
The principle here is the same as in the previous problem. The distance between pedestrians at the moment of movement of the second pedestrian must be divided by the speed of approach of pedestrians. The speed of approach in this case is twenty meters. Therefore, to determine how many minutes later the second pedestrian will catch up with the first, you need to divide 400 meters by 20
400: 20 = 20
This means that in 20 minutes the second pedestrian will catch up with the first.
Problem 2. From two villages, the distance between which is 40 km, a bus and a cyclist left at the same time in the same direction. The speed of a cyclist is 15 km/h, and the speed of a bus is 35 km/h. How many hours will it take for the bus to catch up with the cyclist?
Solution
Let's find the speed of approach
35 km/h − 15 km/h = 20 km/h
Let's determine that in a few hours the bus will catch up with the cyclist.
40: 20 = 2
Answer: the bus will catch up with the cyclist in 2 hours.
River movement problem
Ships move along the river at different speeds. At the same time, they can move both along the river and against the current. Depending on how they move (with or against the current), the speed will change.
Let's assume that the speed of the river is 3 km/h. If you lower a boat onto a river, the river will carry the boat away at a speed of 3 km/h.
If you lower a boat into still water, in which there is no current, then the boat will stand still. The speed of the boat in this case will be zero.
If a boat floats on still water in which there is no current, then the boat is said to float with own speed.
For example, if a motorboat is sailing through still water at a speed of 40 km/h, then we say that motor boat's own speed is 40 km/h.
How to determine the speed of a ship?
If a ship floats with the flow of a river, then the speed of the river current must be added to the ship’s own speed.
with the flow rivers, and the river flow speed is 2 km/h, then the river flow speed (2 km/h) must be added to the motor boat’s own speed (30 km/h)
30 km/h + 2 km/h = 32 km/h
The current of the river can be said to help the motor boat with an additional speed of two kilometers per hour.
If a ship is sailing against the flow of a river, then the speed of the river current must be subtracted from the ship's own speed.
For example, if a motorboat is sailing at a speed of 30 km/h against the stream rivers, and the river flow speed is 2 km/h, then from the motor boat’s own speed (30 km/h) it is necessary to subtract the river flow speed (2 km/h)
30 km/h − 2 km/h = 28 km/h
In this case, the river current prevents the motor boat from moving forward freely, reducing its speed by two kilometers per hour.
Problem 1. The speed of the boat is 40 km/h, and the speed of the river is 3 km/h. At what speed will the boat move down the river? Against the flow of the river?
Answer:
If the boat moves along the river flow, its speed will be 40 + 3, that is, 43 km/h.
If the boat moves against the flow of the river, its speed will be 40 − 3, that is, 37 km/h.
Problem 2. The speed of the ship in still water is 23 km/h. The river flow speed is 3 km/h. Which the path will pass ship in 3 hours along the river? Against the stream?
Solution
The ship's own speed is 23 km/h. If the ship moves along the river, its speed will be 23 + 3, that is, 26 km/h. In three hours it will travel three times as far
26 × 3 = 78 km
If the ship moves against the flow of the river, its speed will be 23 − 3, that is, 20 km/h. In three hours it will travel three times as far
20 × 3 = 60 km
Problem 3. The boat covered the distance from point A to point B in 3 hours 20 minutes, and the distance from point B to A in 2 hours 50 minutes. In which direction does the river flow: from A to B or from B to A, if it is known that the speed of the yacht has not changed?
Solution
The speed of the yacht did not change. Let's find out which path she spent more time on: the path from A to B or the path from B to A. The path that spent more time will be the path whose river flow went against the yacht
3 hours 20 minutes is more than 2 hours 50 minutes. This means that the river current reduced the speed of the yacht and this was reflected in the travel time. 3 hours 20 minutes is the time spent traveling from A to B. This means the river flows from point B to point A
Problem 4. How long does it take to move against the flow of a river?
The ship will travel 204 km if its own speed
15 km/h, and the current speed is 5 times less than its own
speed of the ship?
Solution
You need to find the time it takes for the ship to travel 204 kilometers upstream of the river. The ship's own speed is 15 km/h. It moves against the flow of the river, so you need to determine its speed during this movement.
To determine the speed against the flow of the river, you need to subtract the speed of the river from the ship’s own speed (15 km/h). The condition says that the river flow speed is 5 times less than the ship’s own speed, so first we determine the river flow speed. To do this, let’s reduce 15 km/h by five times
15:5 = 3 km/h
The river flow speed is 3 km/h. Subtract this speed from the speed of the ship
15 km/h − 3 km/h = 12 km/h
Now let’s determine the time it takes the ship to travel 204 km at a speed of 12 km/h. The ship travels 12 kilometers per hour. To find out how many hours it will take him to travel 204 kilometers, you need to determine how many times 204 kilometers contains 12 kilometers
204: 12 = 17 hours
Answer: the ship will travel 204 kilometers in 17 hours
Problem 5. Moving along the river, in 6 hours the boat
walked 102 km. Determine the boat's own speed
Solution
Let's find out how fast the boat was moving along the river. To do this, divide the distance traveled (102 km) by the driving time (6 hours)
102: 6 = 17 km/h
Let's determine the boat's own speed. To do this, from the speed at which it moved along the river (17 km/h), we subtract the speed of the river flow (4 km/h)
17 − 4 = 13 km/h
Problem 6. Moving against the flow of the river, in 5 hours the boat
walked 110 km. Determine the boat's own speed
if the current speed is 4 km/h.
Solution
Let's find out how fast the boat was moving along the river. To do this, divide the distance traveled (110 km) by the driving time (5 hours)
110: 5 = 22 km/h
Let's determine the boat's own speed. The condition says that she was moving against the flow of the river. The river flow speed was 4 km/h. This means that the boat’s own speed has been reduced by 4. Our task is to add these 4 km/h and find out the boat’s own speed
22 + 4 = 26 km/h
Answer: The boat's own speed is 26 km/h
Problem 7. How long does it take for a boat to move against the flow of a river?
will travel 56 km if the current speed is 2 km/h, and its
is its own speed 8 km/h greater than the speed of the current?
Solution
Let's find the boat's own speed. The condition says that it is 8 km/h more than the current speed. Therefore, to determine the boat’s own speed, we add another 8 km/h to the current speed (2 km/h)
2 km/h + 8 km/h = 10 km/h
The boat is moving against the flow of the river, so from the boat’s own speed (10 km/h) we subtract the speed of the river (2 km/h)
10 km/h − 2 km/h = 8 km/h
Let's find out how long it takes the boat to travel 56 km. To do this, divide the distance (56 km) by the speed of the boat:
56:8 = 7 hours
Answer: when moving against the flow of the river, the boat will travel 56 km in 7 hours
Problems to solve independently
Problem 1. How long will it take a pedestrian to walk 20 km if his speed is 5 km/h?
Solution
In one hour, a pedestrian walks 5 kilometers. To determine how long it will take him to travel 20 km, you need to find out how many times 20 kilometers contain 5 km. Or use the rule for finding time: divide the distance traveled by the speed of movement
20:5 = 4 hours
Task 2. From point A to point IN a cyclist rode for 5 hours at a speed of 16 km/h, and back he rode along the same path at a speed of 10 km/h. How long did the cyclist take on the return journey?
Solution
Let's determine the distance from the point A to point IN. To do this, multiply the speed at which the cyclist was traveling from the point A to point IN(16km/h) for driving time (5h)
16 × 5 = 80 km
Let's determine how much time the cyclist spent on the way back. To do this, divide the distance (80 km) by the speed (10 km/h)
Problem 3. A cyclist rode for 6 hours at a certain speed. After he drove another 11 km at the same speed, his distance became 83 km. How fast was the cyclist traveling?
Solution
Let us determine the distance covered by the cyclist in 6 hours. To do this, from 83 km we subtract the distance he traveled after six hours of movement (11 km)
83 − 11 = 72 km
Let's determine at what speed the cyclist rode for the first 6 hours. To do this, divide 72 km into 6 hours
72: 6 = 12 km/h
Since the condition of the problem says that the cyclist drove the remaining 11 km at the same speed as in the first 6 hours of driving, then a speed of 12 km/h is the answer to the problem.
Answer: The cyclist was traveling at a speed of 12 km/h.
Problem 4. Moving against the flow of a river, a motor ship travels a distance of 72 km in 4 hours, and a raft travels the same distance in 36 hours. In how many hours will the motor ship cover a distance of 110 km if it floats with the flow of the river?
Solution
Let's find the speed of the river flow. The condition states that the raft can travel 72 kilometers in 36 hours. The raft cannot move against the flow of the river. This means that the speed of the raft with which it covers these 72 kilometers is the speed of the river flow. To find this speed, you need to divide 72 kilometers by 36 hours
72: 36 = 2 km/h
Let's find the ship's own speed. First, let's find the speed of its movement against the flow of the river. To do this, divide 72 kilometers by 4 hours
72: 4 = 18 km/h
If the speed of the ship against the river flow is 18 km/h, then its own speed is 18+2, that is, 20 km/h. And along the river its speed will be 20+2, that is, 22 km/h
By dividing 110 kilometers by the speed of the ship along the river (22 km/h), you can find out how many hours it will take the ship to travel these 110 kilometers
Answer: The ship will travel 110 kilometers along the river for 5 hours.
Problem 5. Two cyclists left the same point at the same time in opposite directions. One of them was driving at a speed of 11 km/h, and the second at a speed of 13 km/h. What will be the distance between them after 4 hours?
21 × 6 = 126 km
Let us determine the distance traveled by the second ship. To do this, multiply its speed (24 km/h) by the time it takes to reach the meeting (6 hours)
24 × 6 = 144 km
Let's determine the distance between the piers. To do this, add up the distances traveled by the first and second ships
126 km + 144 km = 270 km
Answer: The first ship traveled 126 km, the second - 144 km. The distance between the piers is 270 km.
Problem 7. Two trains left Moscow and Ufa at the same time. 16 hours later they met. The Moscow train was traveling at a speed of 51 km/h. At what speed did the train leave Ufa travel if the distance between Moscow and Ufa is 1520 km? What was the distance between the trains 5 hours after they met?
Solution
Let's determine how many kilometers the train leaving Moscow traveled before the meeting. To do this, multiply its speed (51 km/h) by 16 hours
51 × 16 = 816 km
Let's find out how many kilometers the train left Ufa traveled before the meeting. To do this, from the distance between Moscow and Ufa (1520 km) we subtract the distance traveled by the train leaving Moscow
1520 − 816 = 704 km
Let's determine the speed at which the train left Ufa was traveling. To do this, the distance he traveled before the meeting must be divided by 16 hours
704: 16 = 44 km/h
Let's determine the distance that will be between the trains 5 hours after they meet. To do this, find the speed of the trains moving away and multiply this speed by 5
51 km/h + 44 km/h = 95 km/h
95 × 5 = 475 km.
Answer: The train leaving Ufa was traveling at a speed of 44 km/h. 5 hours after the trains meet, the distance between them will be 475 km.
Problem 8. Two buses departed from one point simultaneously in opposite directions. The speed of one bus is 48 km/h, the other is 6 km/h more. After how many hours will the distance between the buses be 510 km?
Solution
Let's find the speed of the second bus. It is 6 km/h more than the speed of the first bus
48 km/h + 6 km/h = 54 km/h
Let's find the speed at which the buses move away. To do this, let's add up their speeds:
48 km/h + 54 km/h = 102 km/h
In an hour, the distance between buses increases by 102 kilometers. To find out after how many hours the distance between them will be 510 km, you need to find out how many times 510 km contains 102 km/h
Answer: 510 km between buses will be in 5 hours.
Problem 9. The distance from Rostov-on-Don to Moscow is 1230 km. Two trains left Moscow and Rostov towards each other. The train from Moscow travels at a speed of 63 km/h, and the speed of the Rostov train is the same as the speed of the Moscow train. At what distance from Rostov will the trains meet?
Solution
Let's find the speed of the Rostov train. It is the speed of a Moscow train. Therefore, to determine the speed of the Rostov train, you need to find from 63 km
63: 21 × 20 = 3 × 20 = 60 km/h
Let's find the speed of approach of trains
63 km/h + 60 km/h = 123 km/h
Let's determine how many hours later the trains will meet
1230: 123 = 10 hours
Let's find out at what distance from Rostov the trains will meet. To do this, it is enough to find the distance traveled by the Rostov train before the meeting
60 × 10 = 600 km.
Answer: The trains will meet at a distance of 600 km from Rostov.
Problem 10. From two piers, the distance between which is 75 km, two motor boats simultaneously departed towards each other. One was moving at a speed of 16 km/h, and the speed of the other was 75% of the speed of the first boat. What will be the distance between the boats after 2 hours?
Solution
Let's find the speed of the second boat. It is 75% of the speed of the first boat. Therefore, to find the speed of the second boat, you need 75% of 16 km
16 × 0.75 = 12 km/h
Let's find the closing speed of the boats
16 km/h + 12 km/h = 28 km/h
Every hour the distance between the boats will decrease by 28 km. After 2 hours it will decrease by 28×2, that is, by 56 km. To find out what the distance between the boats will be at this moment, you need to subtract 56 km from 75 km
75 km − 56 km = 19 km
Answer: in 2 hours there will be 19 km between the boats.
Problem 11. A car with a speed of 62 km/h is catching up with a truck with a speed of 47 km/h. After how much time and at what distance from the start of movement will a passenger car catch up with a truck, if the initial distance between them was 60 km?
Solution
Let's find the speed of approach
62 km/h − 47 km/h = 15 km/h
If initially the distance between the cars was 60 kilometers, then every hour this distance will decrease by 15 km, and eventually the passenger car will catch up with the truck. To find out how many hours later this will happen, you need to determine how many times 60 km contains 15 km
Let's find out at what distance from the start of movement the passenger car caught up with the truck. To do this, multiply the speed of the car (62 km/h) by the time it takes to reach the meeting (4 hours)
62 × 4 = 248 km
Answer: A passenger car will catch up with a truck in 4 hours. At the moment of the meeting, the passenger car will be at a distance of 248 km from the start of movement.
Problem 12. Two motorcyclists left the same point in the same direction at the same time. The speed of one was 35 km/h, and the speed of the other was 80% of the speed of the first motorcyclist. What will be the distance between them after 5 hours?
Solution
Let's find the speed of the second motorcyclist. It is 80% of the speed of the first motorcyclist. Therefore, to find the speed of the second motorcyclist, you need to find 80% of 35 km/h
35 × 0.80 = 28 km/h
The first motorcyclist moves 35-28 km/h faster
35 km/h − 28 km/h = 7 km/h
In one hour, the first motorcyclist covers 7 kilometers more. Every hour she will get closer to the second motorcyclist by these 7 kilometers.
After 5 hours, the first motorcyclist will cover 35×5, that is, 175 km, and the second motorcyclist will travel 28×5, that is, 140 km. Let's determine the distance between them. To do this, subtract 140 km from 175 km
175 − 140 = 35 km
Answer: after 5 hours the distance between motorcyclists will be 35 km.
Problem 13. A motorcyclist whose speed is 43 km/h catches up with a cyclist whose speed is 13 km/h. In how many hours will the motorcyclist catch up with the cyclist if the initial distance between them was 120 km?
Solution
Let's find the speed of approach:
43 km/h − 13 km/h = 30 km/h
If initially the distance between the motorcyclist and the cyclist was 120 kilometers, then every hour this distance will decrease by 30 km, and eventually the motorcyclist will catch up with the cyclist. To find out how many hours later this will happen, you need to determine how many times 120 km contains 30 km
So in 4 hours the motorcyclist will catch up with the cyclist
The figure shows the movement of a motorcyclist and a cyclist. It can be seen that 4 hours after the start of movement they leveled off.
Answer: The motorcyclist will catch up with the cyclist in 4 hours.
Problem 14. A cyclist whose speed is 12 km/h catches up with a cyclist whose speed is 75% of his speed. After 6 hours, the second cyclist caught up with the first cyclist. What was the original distance between the cyclists?
Solution
Let's determine the speed of the cyclist riding ahead. To do this, let’s find 75% of the speed of the cyclist riding behind:
12 × 0.75 = 9 km/h - speed of the person driving ahead
Let's find out how many kilometers each cyclist traveled before the second one caught up with the first one:
12 × 6 = 72 km - passed by the one driving behind
9 × 6 = 54 km - the person in front passed
Let's find out what the distance was between the cyclists initially. To do this, from the distance covered by the second cyclist (who was catching up), we subtract the distance covered by the first cyclist (who was caught up)
It can be seen that the car is 12 km ahead of the bus.
To find out how many hours later the car will be 48 kilometers ahead of the bus, you need to determine how many times 48 km contains 12 km
Answer: 4 hours after leaving, the car will be 48 kilometers ahead of the bus.
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First, let's remember the formulas that are used to solve such problems: S = υ·t, υ = S:t, t = S: υ
where S is the distance, υ is the speed of movement, t is the time of movement.
When two objects move uniformly at different speeds, the distance between them for each unit of time either increases or decreases.
Closing speed– this is the distance by which objects approach each other per unit of time.
Removal speed is the distance that objects move away per unit time.
Moving closer oncoming traffic And chasing after. Motion to remove can be divided into two types: movement in opposite directions And lagging movement.
The difficulty for some students is to correctly place “+” or “–” between speeds when finding the speed of approaching objects or the speed of moving away.
Let's look at the table.
It shows that when objects move in opposite directions their speeds add up. When moving in one direction, they are deducted.
Examples of problem solving.
Task No. 1. Two cars are moving towards each other at speeds of 60 km/h and 80 km/h. Determine the speed of approach of the cars.
υ 1 = 60 km/h
υ 2 = 80 km/h
Find υ sat
Solution.
υ sb = υ 1 + υ 2– approach speed in different directions)
υ sat = 60 + 80 = 140 (km/h)
Answer: closing speed 140 km/h.
Task No. 2. Two cars left the same point in opposite directions at speeds of 60 km/h and 80 km/h. Determine the speed at which the machines are removed.
υ 1 = 60 km/h
υ 2 = 80 km/h
Find υ beat
Solution.
υ beat = υ 1 + υ 2– removal rate (the “+” sign since it is clear from the condition that the cars are moving in different directions)
υ beat = 80 + 60 = 140 (km/h)
Answer: the removal speed is 140 km/h.
Task No. 3. First a car leaves one point in one direction at a speed of 60 km/h, and then a motorcycle leaves at a speed of 80 km/h. Determine the speed of approach of the cars.
(We see that here is a case of chasing movement, so we find the speed of approach)
υ av = 60 km/h
υ motor = 80 km/h
Find υ sat
Solution.
υ sb = υ 1 – υ 2– approach speed (the “–” sign since it is clear from the condition that the cars are moving in one direction)
υ sat = 80 – 60 = 20 (km/h)
Answer: approach speed 20 km/h.
That is, the name of the speed - approaching or moving away - does not affect the sign between the speeds. Only the direction of movement matters.
Let's consider other tasks.
Task No. 4. Two pedestrians left the same point in opposite directions. The speed of one of them is 5 km/h, the other is 4 km/h. What will be the distance between them after 3 hours?
υ 1 = 5 km/h
υ 2 = 4 km/h
t = 3 h
Find S
Solution.
in different directions)
υ beat = 5 + 4 = 9 (km/h)
S = υ beat ·t
S = 9 3 = 27 (km)
Answer: after 3 hours the distance will be 27 km.
Task No. 5. Two cyclists simultaneously rode towards each other from two points, the distance between which is 36 km. The speed of the first is 10 km/h, the second is 8 km/h. In how many hours will they meet?
S = 36 km
υ 1 = 10 km/h
υ 2 = 8 km/h
Find t
Solution.
υ сб = υ 1 + υ 2 – approach speed (the “+” sign since it is clear from the condition that the cars are moving in different directions)
υ sat = 10 + 8 = 18 (km/h)
(meeting time can be calculated using the formula)
t = S: υ Sat
t = 36: 18 = 2 (h)
Answer: we will meet in 2 hours.
Task No. 6. Two trains departed from the same station in opposite directions. Their speeds are 60 km/h and 70 km/h. After how many hours will the distance between them be 260 km?
υ 1 = 60 km/h
υ 2 = 70 km/h
S = 260 km
Find t
Solution .
1 way
υ beat = υ 1 + υ 2 – removal rate (the “+” sign since it is clear from the condition that pedestrians are moving in different directions)
υ beat = 60 + 70 = 130 (km/h)
(We find the distance traveled using the formula)
S = υ beat ·t ⇒ t= S: υ beat
t = 260: 130 = 2 (h)
Answer: after 2 hours the distance between them will be 260 km.
Method 2
Let's make an explanatory drawing:
From the figure it is clear that
1) after a given time, the distance between the trains will be equal to the sum of the distances traveled by each of the trains:
S = S 1 + S 2;
2) each of the trains traveled the same time (from the problem conditions), which means
S 1 =υ 1 · t— the distance traveled by 1 train
S 2 =υ 2 t— the distance traveled by the 2nd train
Then,
S= S 1 + S 2= υ 1 · t + υ 2 · t = t (υ 1 + υ 2)= t · υ beat
t = S: (υ 1 + υ 2)— time during which both trains travel 260 km
t = 260: (70 + 60) = 2 (h)
Answer: the distance between trains will be 260 km in 2 hours.
1. Two pedestrians simultaneously set out towards each other from two points, the distance between which is 18 km. The speed of one of them is 5 km/h, the other is 4 km/h. In how many hours will they meet? (2 hours)
2. Two trains left the same station in opposite directions. Their speeds are 10 km/h and 20 km/h. After how many hours will the distance between them be 60 km? (2 hours)
3. From two villages, the distance between which is 28 km, two pedestrians simultaneously walked towards each other. The speed of the first is 4 km/h, the speed of the second is 5 km/h. How many kilometers per hour do pedestrians approach each other? What will be the distance between them after 3 hours? (9 km, 27 km)
4. The distance between the two cities is 900 km. Two trains left these cities towards each other at speeds of 60 km/h and 80 km/h. How far apart were the trains 1 hour before the meeting? Is there an extra condition in the problem? (140 km, yes)
5. A cyclist and a motorcyclist left at the same time from one point in the same direction. The speed of a motorcyclist is 40 km/h, and that of a cyclist is 12 km/h. What is the speed at which they move away from each other? After how many hours will the distance between them be 56 km? (28 km/h, 2 h)
6. Two motorcyclists left at the same time from two points 30 km apart from each other in the same direction. The speed of the first is 40 km/h, the second is 50 km/h. In how many hours will the second one catch up with the first one?
7. The distance between cities A and B is 720 km. A fast train left A for B at a speed of 80 km/h. After 2 hours, a passenger train left B to A to meet him at a speed of 60 km/h. In how many hours will they meet?
8. A pedestrian left the village at a speed of 4 km/h. After 3 hours, a cyclist followed him at a speed of 10 km/h. How many hours will it take for a cyclist to catch up with a pedestrian?
9. The distance from the city to the village is 45 km. A pedestrian left the village for the city at a speed of 5 km/h. An hour later, a cyclist rode towards him from the city to the village at a speed of 15 km/h. Which of them will be closer to the village at the time of the meeting?
10. An ancient task. A certain young man went from Moscow to Vologda. He walked 40 miles a day. A day later, another young man was sent after him, walking 45 miles a day. How many days will it take for the second one to catch up with the first one?
11. An ancient problem. The dog saw a hare in 150 fathoms, which ran 500 fathoms in 2 minutes, and the dog ran 1300 fathoms in 5 minutes. The question is, at what time will the dog catch up with the hare?
12. An ancient problem. 2 trains left Moscow for Tver at the same time. The first passed at the hour 39 versts and arrived in Tver two hours earlier than the second, which passed at the hour 26 versts. How many miles from Moscow to Tver?
IN movement tasks formulas expressing the law of uniform motion are usually used, i.e.
s = v t.
When composing equations in such problems, it is convenient to use a geometric illustration of the motion process.
When moving in a circle, it is convenient to use the concept of angular velocity, i.e. the angle through which a moving object rotates around its center per unit time. It happens that to complicate a problem, its condition is formulated in different units of measurement. In such cases, to compose equations, it is necessary to express all given values in terms of the same unit of measurement.
The source of compiling equations in motion problems is the following considerations:
1) Objects that start moving towards each other at the same time move for the same amount of time until they meet. The time after which they will meet is found using the formula
t = s/(v 1 + v 2) (*).
2) If one body catches up with another, then the time after which the first will catch up with the second is calculated by the formula
t = s/(v 1 – v 2) (**).
3) If the objects have traveled the same distance, then it is convenient to take the value of this distance as the general unknown of the problem.
4) If, when two objects move simultaneously in a circle from one point, one of them catches up with the other for the first time, then the difference in the distances they have traveled up to this moment is equal to the length of the circle
5) For time new meeting when moving in opposite directions, we get formula (*), if in one direction, then formula (**).
6) When moving along a river, the speed of an object is equal to the sum of the speeds in still water and the speed of the current. When moving against the current, the speed of movement is the difference between these speeds.
Analytical solution of motion problems
Task 1.
Two pedestrians walked towards each other at the same time and met each other 3 hours and 20 minutes later. How long did it take each pedestrian to cover the entire distance, if it is known that the first arrived at the point from which the second left, 5 hours later than the second arrived at the point from which the first left?
Solution.
There is no distance traveled data in this problem. This is its main feature. In such cases it will be convenient to take the entire distance as a unit, then the speed of the first pedestrian will be equal to
v 1 = 1/x, and the second – v 2 = 1/y, where x hours is the travel time of the first, and y is the travel time of the second pedestrian.
The conditions of the problem allow us to create a system of equations:
(3⅓ 1/x + 3⅓ 1/y = 1,
(x – y = 5.
Solving this system, we find that y = 5, x = 10.
Answer: 10 o'clock and 5 o'clock.
Task 2.
A cyclist rode from point A to point B. After 3 hours, a motorcyclist drove towards him from point B, at a speed 3 times greater than the speed of the cyclist. The meeting of the cyclist and the motorcyclist takes place in the middle, between points A and B. If the motorcyclist left 2 hours later than the cyclist, their meeting would take place 15 kilometers closer to point A. Find the distance AB.
Solution.
Let's make an illustration of the problem (Fig. 1).
Let AB = s km, v km/h is the speed of the cyclist, 3v km/h is the speed of the motorcyclist.
t 1 = 0.5 s/v hours – time until the cyclist meets,
t 2 = 0.5 s/3v hours – time until the motorcyclist meets.
By condition, t 1 – t 2 = 3, which means 0.5 s/v – 0.5s / 3v = 3, whence s = 9v.
If the motorcyclist left 2 hours later than the cyclist, they would meet at point F.
AF = 0.5s – 15, BF = 0.5s + 15.
Let's create an equation: (0.5s – 15)/v – (0.5s + 15)/3v = 2, from which s – 60 = 6v.
We get a system of equations:
(s = 9v,
(s = 60 + 6v.
(v = 20,
(s = 180.
Answer: v = 20 km/h, s = 180 km.
Graphic method for solving motion problems
There is also a graphical method for solving problems. Let's consider the application of this method to solve motion problems. Graphic representation of functions that describe the condition of a problem is often a very convenient technique that allows you to visualize the situation of the problem. It also allows you to create new equations or replace the algebraic solution of a problem with a purely geometric one.
Task 3.
The pedestrian left point A to point B. Following him, a cyclist left point A, but with a delay of 2 hours. After another 30 minutes, a motorcyclist left in the direction of point B. The pedestrian, cyclist and motorcyclist moved to point B without stopping and uniformly. Some time after the motorcyclist left, it turned out that by that moment all three had covered the same part of the path from A to B. How many minutes before the pedestrian did the cyclist arrive at point B if the motorcyclist arrived at point B 1 hour before the pedestrian?
Solution.
For algebraic solution it requires the introduction of many variables and the creation of a cumbersome system. Graphically, the situation described in the problem is presented in Figure 2. 
Using the similarity of triangles AOL and KOM, as well as triangles AOP and KON, you can create a proportion:
x = 4/5 hours = 48 minutes.
Answer: 48 minutes.
Task 4.
Two messengers left the two cities to meet each other at the same time. After the meeting, one of them was on the road for another 16 hours, and the second - 9 hours. Determine how long each messenger was on the road.
Solution.
Let the travel time before meeting each messenger be t. Based on the conditions of the problem, we build a graph (Fig. 3). 
Similar to problem 3, it is necessary to use the similarity of triangles.
This means that 12 + 16 = 28 (hours) – the first one was on the way, 12 + 9 = 21 (hours) – the second one was on the way.
Answer: 21 hours and 28 hours.
So we have looked at the main methods for solving motion problems. They appear very often in the Unified State Exam, so be sure to practice solving these problems.
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It is certainly not easy to determine what place the poetry of Vladimir Gommerstadt occupies in great Russian literature; however, this may not need to be done at all, because it is clear in advance that what is quite modest - nothing can be done about it, that’s the kind of literature we have. It is much more interesting to see and understand what place it occupies in great Russian art.
The author's primary area of interest is fine arts, mainly graphics. It is impossible to hide it - and why? - because the poems of V. Hommerstadt are, so to speak, of a pronounced pictorial nature: before you understand what they are about, you need to see the object image described in them - namely, an image, and not a real landscape or interior.
The author seems to be well aware that in his graphics he is close to the artistic principles of Japan and China. There may be great artistic meaning hidden in this, but for us here it is important that a Japanese or Chinese drawing without a predominantly poetic inscription is considered not only unfinished, but simply non-existent. That is why we can say that the poems of V. Gommerstadt represent a counter movement from word to image and are frankly incomplete without the visual images behind them. One can also say that most of them simply tend to be considered as a classic poetic inscription - but as an undrawn picture.
Closeness to art Far East manifests itself in a purely poetic aspect. Here, for example, is a poem that (with one caveat) could appear on a Japanese engraving:
God, thank you, who gave me autumn:
light ripples on the water, a handful of unharvested ears of corn,
the first early graying, frost on a withered leaf...
The promised disclaimer concerns the fact that gods can hardly be mentioned in a Japanese inscription, and not because there is a prohibition, but for purely aesthetic reasons. But temples can be mentioned quite freely, but not as shrines or even as cultural objects, but primarily as details of the landscape, which also have certain semantic nuances.
As for the poem itself, it very clearly illustrates the artistic principle in question here. The reader is entrusted with a certain aesthetic work: to see the ripples in the water, the ears of corn in the field (maybe, since we are mostly not Japanese after all, to remember “ Uncompressed strip” Nekrasov), try, identifying yourself with the lyrical hero, to feel and experience his early gray hair, to see that not only the leaves have already withered, but also the air moisture freezes and settles on them with frost - such a late autumn - and only after all this immerse yourself in the atmosphere of autumn... and Thanksgiving.
And all this can be done with a thoughtful reader by three modest lines.
Here lies precisely what gives us the right to assert that the poetry of V. Hommerstadt is original and not imitative, because in the great world art it is almost impossible to do without echoes and even artistic borrowings, and they should be distinguished from imitations and be able to see, where is the actual creative principle, and where is the combination of borrowed techniques.
For Japanese art, autumn is a theme of sublime sadness when contemplating falling asleep nature, an occasion for reflection on the transience of all living things. And in the poetry of V. Hommerstadt, the expressive details of autumn lead the thought to gratitude to the Almighty. The personal aspect of poetry, which makes it lyrical, consists precisely in this, and in this connection the following seems essential: when do we have the right to talk about Christian art? You can understand in advance that we won’t answer this question here, but you can think about it...
I am afraid that according to the generally accepted opinion, in Christian art the objects are only objects and events of religious significance, understood rather strictly and narrowly. This is partly true, for example, when it comes to the construction of churches, the artistic details of their interiors, icon painting and church music. But don’t a person’s thoughts and feelings, his attitude and actions have such significance?
This topic is too vast to attempt here; it can only be designated as follows: Christian art occurs when the artist looks at the world through the eyes of a Christian. Otherwise it is possible procession describe it as a parade of athletes, paying primary attention to the number of participants and the clarity of the organization.
I think that we can call the poetry of V. Gommerstadt Christian not only (and not so much) because scenes of monastic life (“Midnight Office”) appear in it or are mentioned church holidays (“Palm Sunday"), but because the author looks at the whole world through the eyes of a Christian, that is, as at the world of God. And if we talk about the scenes in the monastery, then there is nothing formally monastic in the short poem about the cowgirl Martha, but the spiritual mood of this Martha is, as it were, intended to justify her “Marfa” efforts, to show that Maria, which is hidden in her simple life.
But here is a poem that is called “Engraving”. And again we are faced with going beyond purely aesthetic problems into problems of a different dimension:
Some area. Russia? Japan?
A lover of engravings knows what to prefer,
But this hardly counts as righteousness.
Because righteousness is not just higher than aesthetics, but, without in any way canceling it, refers to another world.
Finally, another poem seems to demonstrate the asserted pattern in full:
In the noise of autumn and night, I will go out into the garden.
Night autumn garden- what an exquisite subject for a Chinese design! And so, such a drawing (maybe in the form of a poem - for the author this is not so important) is considered completed:
I'll write it down. And I will sign: Hommerstadt.
And this is where the closure of more than two layers occurs: not only is a poem written about the night garden, which is intended to be depicted in the drawing, but then, when the drawing (verse?) has already been signed, another reality enters:
Wang Wei shook his yellow finger at me.
Wang Wei is a great Chinese draftsman. Where did his threat come from or in best case scenario reproachful gesture? Obviously, the point is not only that “Wang Wei”, as the personification of Far Eastern graphics, calls for greater purity of the genre, but also that he believes that individual authorship here could be challenged.
...But there is, there is an area of art that is very close to us, in which authorship is completely naturally can be determined with some degree of convention - iconography. Yes, every outstanding icon painter has his own “handwriting”, which cannot be confused with anyone else, and Reverend Andrew he depicted the Holy Trinity like no one before or after him, but he didn’t “invent” this “picturesque subject”! IN in a high sense the authorship of the Old Testament Trinity belongs to the Church. More precisely, the Church created an icon-painting canon, which any icon painter follows, as long as he recognizes himself as such. So we can say with a clear conscience that in art the phenomenon of “incomplete authorship” is known and is far from new.
But the combination of image and text in highest degree inherent in the icon, so that an icon without an inscription cannot be considered completed to such an extent decisive degree that she, strictly speaking, is not yet an icon.
And how can we not mention the painting and graphics of Elena Cherkasova, who includes inscriptions in the artistic space of her works (while constantly emphasizing that they are not of an iconographic nature)! These inscriptions can be the names of the characters depicted, plot explanations, or even extensive passages of Scripture. The latter case turned out to be so fascinating for the artist that, so to speak, it grew into a different genre, and Elena created manuscripts of two biblical books - Ruth and Tobit - in which the text and illustrations form an indissoluble whole.
Finally, having considered all these things from this angle, we see that the counter-movement from word to image and from image to word is much more widespread in art than one might think. This is a very old memory.
Once upon a time, when I was a young employee of the Academy of Sciences, I had to take a group of foreign scientists and guests to the Tretyakov Gallery scientific conference. Reaction to Russian painting of the 19th century. was rather lethargic, although within the limits of politeness. However, Bryullov’s “Horsewoman” caused great excitement: “Real English painting!” And then something clicked in my head, and I realized that here, so to speak, the plot (or rather, the absence of a literary plot) was obvious, but in other cases... and I began not only to translate the names of the paintings, but tell. Great interest immediately arose, and we walked through the halls again: from “Alyonushka” to “ Unequal marriage” and “We didn’t expect it” - everything became interesting, everything was examined and praised.
Later, I discussed this case with very qualified art historians in Moscow and Prague, and they confirmed what I came across in such a rather random way; the literary, narrative character of Russian painting was no secret to them.
And so it turns out that the border between types of art... is not that insignificant, but they themselves strive to overcome it and come out to meet each other.
...This is how far you can go in your thoughts, starting from a small selection of poems by a modest poet.