Trip Chase Problem 5, Trip Chase Problem

Catch-up: Speed difference Catch-up time = catch-up distance Catch-up distance Speed difference = catch-up time (in the same direction) Encounter: Encounter distance Speed sum = Encounter time Speed sum Encounter time = Encounter distance Example A and B start at the same time and run around a 300-meter loop, A is 6 meters per second, B is 4 meters per second, When catching up with B for the second time, how many laps did A run?

Basic equiquantitative relationship: chasing time Speed difference = chasing distance The speed difference in this question is: 6-4=2 After A catches up with B for the first time, the catching distance is 300 meters around the circumference of the circular runway After catching up for the first time, the two can be regarded as starting at the same time and in the same place, so the problem of catching up for the second time is transformed into solving the problem of chasing for the first time.

The first time A catches up with B is: 300 2=150 seconds A catches up with B for the first time and runs: 6*150=900 meters At this time, B runs:

4*150=600 meters This shows that A catches up with B at the starting point, therefore, the second catch-up problem can be simplified to multiply the distance run when catching up the first time by two, so that A catches up with B for the second time and runs a total of 900 + 900 = 1800 B runs a total of 600 + 600 = 1200 then A runs 1800 300 = 6 laps B runs 1200 300 = 4 laps.

Pursue the solution to the problem.

The conventional method of solving the problem is to list the equations according to the equal displacement, and the displacement formula of linear motion with uniform velocity is a one-dimensional quadratic equation, so the properties and discriminants of the quadratic trinomial equation (y=ax2+bx+c) are often used in solving the linear motion problem. In addition, when there are two (or several) objects in motion, one of them is often used as a reference, that is, if it becomes "stationary", only the other (or others) object is moving. In this way, the research process is simplified, so that the problem is often solved by changing the method of the reference.

At this time, it is necessary to determine the initial velocity of other objects relative to the reference object and the acceleration relative to it, so as to determine the motion of other objects

Steamer A travels 16 kilometers per hour, Steamer B travels 14 kilometers per hour, it is known that Steamer B is 10 kilometers in front of Steamer A at the beginning, how long will it take for Steamer A to catch up with Steamer B.

t=s/(v1-v2)

The catch-up time is equal to the distance divided by the speed difference between the two.

Travel issues. In some of the questions, the question stem is often set to be someone walking around the park or on the track and field, and the route of the movement can form a closed loop, which we call the circular encounter and chase problem.

Let's take a look at the problem of circular encounters and find ways to solve it.

1. Circular encounters.

A circular encounter is when two people walk in opposite directions on a circular runway, and one person goes clockwise.

Movement, the other person moves counterclockwise, and after a period of time, the two meet at a certain point on the runway. If two people start at the same time and in the same place, the sum of the distances traveled by the two people is equal to the circumference of the runway when they meet for the nth time, and the sum of the distances traveled by the two people is equal to the circumference of the runway n times when they meet for the nth time. Written as:

Example: A and B are walking by a circular pond with a circumference of 400 meters. A walks 9 meters per minute, and B walks 16 meters per minute. Now two people are walking in opposite directions from the same point, so how many minutes after departure do they meet for the second time?

Analysis: From the meaning of the title, it can be seen that A and B walk in the opposite direction from the same point at the same time, and when they meet for the second time, the total distance they have traveled is twice the circumference of the circular pond, that is, 400 2 = 800 meters, and the time spent is t = 800 (9 + 16) = 32 minutes, so option B is chosen.

Second, the leakage of suspicion is chased in a ring.

Circular chase refers to two people walking in the same direction on a circular runway, both moving clockwise or counterclockwise, and after a period of time, the faster person catches up with the slower person. If two people start at the same time and in the same place, the difference between the distance between the two people is equal to the circumference of the runway when they catch up for the first time, and the difference between the distance between the two is equal to the circumference of the runway n times when they catch up for the nth time. Written as:

Example 2: A circular runway is 400m long, Xiao Zhang and Xiao Wang start from the same point at the same time, go in the same direction, Xiao Zhang's speed is 6 meters per second, Xiao Wang's speed is 4 meters per second, when Xiao Zhang catches up with Xiao Wang for the fourth time, Xiao Zhang runs a few laps?

Analysis: From the meaning of the title can be known late, A and B at the same time from the same point and the same direction, when Xiao Zhang caught up with Xiao Wang for the fourth time, Xiao Zhang and Xiao Wang walked the distance difference should be 4 times the runway circumference, that is, 4 400 = 1600 meters, according to the travel formula, the equation can be listed 6t-4t = 1600, and the solution is t = 800 seconds. At this time, the road traveled by Xiao Zhang became 6 800 = 4800 meters, 4800 400 = 12 laps, so choose item D.

(1) The issue of encounters.

Two moving objects moving in opposite directions or in opposite motion on a circular runway, with the development of time, will inevitably meet face to face, this kind of problem is called the encounter problem. It is characterized by two moving objects traveling together for the entire distance.

The itinerary problem in primary school mathematics textbooks generally refers to the encounter problem.

Encounter problems can be divided into three types according to the quantitative relationship: finding the distance, finding the meeting time, and finding the speed.

Their basic relationship is as follows:

Total distance = (speed A + speed B) Time of encounter.

Encounter time = total distance (speed A + speed B).

Another velocity = A and B velocity and - one known velocity.

2) Catch up on the problem.

The location of the chase problem can be the same (e.g. the catch up problem on a circular runway) or it can be different, but the direction is generally the same. Due to the difference in speed, the problem of fast catching up and slow happens.

According to the relationship between the velocity difference, the distance difference and the catch-up time, the following formula is commonly used:

Distance difference = speed difference catch-up time.

Catch-up time = distance difference speed difference.

Speed difference = distance difference catch-up time.

Speed difference = fast - slow.

The key to solving the problem is to find out the two of the three that are related and correspond to each other, such as distance difference, speed difference, and chasing time, and then use the formula to find the third party to achieve the solution.

3) 2. The problem of separation.

When two moving objects move apart due to their opposite motion, they are separated from each other. The key to solving the distance problem is to find the distance (velocity sum) of the common tendency of two moving objects.

The basic formulas are:

Distance between two places = speed and time apart.

Separation time = distance between two places, speed sum.

Speed sum = distance between two places and time apart.

Running water problems. The problem of going down the river and going up the river is often called the flow problem, which is a travel problem, and it is still solved by using the relationship between speed, time, and distance. When answering, pay attention to the meaning of each speed and the relationship between them.

When a boat travels in still water, the distance traveled per unit of time is called rowing speed or stroked force; The speed of the boat traveling along the river is called the speed of the river; The speed of the boat against the current is called the countercurrent velocity; The boat releases the middle stream, and does not rely on power to travel along the water, and the distance traveled per unit time is called the current velocity. The relationship between the various speeds is as follows:

1) Paddle speed + water velocity = downstream speed.

2) Paddle speed - water velocity = countercurrent speed.

3) (Downstream velocity + Countercurrent speed) 2 = Paddle speed.

4) (Downstream velocity - Countercurrent velocity) 2 = Water velocity.

The quantitative relationship of the flow problem is still the relationship between speed, time and distance. i.e.: velocity time = distance; Distance: Speed = Time; Distance Time = Velocity.

However, the river water flows, so there is a difference between going with and against the current. When calculating, it is necessary to clarify the relationship between the various velocities.

Whatever the travel problem, always keep in mind that speed x time = distance.

Do everything you can to find out the three elements you need, namely time, distance, speed, distance before departure, speed difference, speed and, and the multiple relationship between the distance traveled in the same time and the speed.

Distance, speed, time; Distance Time = Speed; Distance Speed = Time [this paragraph] Key Question.

Determine the location during the journey Distance Encounter Distance Speed Sum = Encounter Time Encounter Distance Encounter Time = Speed Sum.

Encounter Problems (Straight Lines).

A's distance + B's distance = total distance.

Encounter Problems (Rings).

A's distance + B's distance = circumference of the ring.

This paragraph] to follow up on the issue.

Catch-up time Distance difference Speed difference Speed difference Distance difference Catch-up time Catch-up time Speed difference Distance difference.

Catch up on the problem (straight line).

Distance difference = chaser's distance - chaser's distance = speed difference x chasing time and chasing problem (ring).

Fast distance - slow distance = circumference of the curve.

This paragraph]

Downwater stroke (boat speed water speed) Downstream time Reverse water stroke (boat speed Water speed) Reverse water time Downstream speed = boat speed Water speed Reverse water speed Boat speed Water speed Hydrostatic speed = (Downstream speed Reverse water speed) 2 Water speed: (Downstream speed Reverse water speed) 2

Encounter Problems (Straight Lines).

A's distance + B's distance = total distance.

Encounter Problems (Rings).

A's distance + B's distance = circumference of the ring.

Basic formula. Distance, speed, time; Distance Time = Speed; Distance Speed = Time to catch up with the problem.

Catch-up time Distance difference Speed difference Speed difference Distance difference Catch-up time Catch-up time Speed difference Distance difference.

Catch up on the problem (straight line).

Distance difference = chaser's distance - chaser's distance = speed difference x chasing time and chasing problem (ring).

Fast distance - slow distance = circumference of the curve.

Running water problems. Downwater stroke (boat speed water speed) Downwater time Reverse water stroke (boat speed Water speed) Reverse water time Downstream speed = boat speed Water speed Reverse water speed Boat speed Water speed Hydrostatic speed = (Downstream speed Reverse water speed) 2 Water speed: (Downstream speed Reverse water speed) 2 Boat speed: (Downstream speed + Reverse water speed).

360 2=180 speed and.

360 10 36 Speed difference.

180+36) 2 108 A speed.

180-36) 2 72 B speed.

The sum of the speed is 360 2 = 180 meters, the speed difference is 360 10 = 36 meters, the first speed (180 + 36) 2 = 108 meters, and the second speed (180-36) 2 = 72 meters.

Go hand in hand, and the pursuit distance is the length of the express train.

Express train length: (22-16) 30 = 180 meters.

Tail in unison, the pursuit distance is the length of the slow train.

Slow train length: (22-16) 26=156 meters.

Catch-up: Speed difference Catch-up time = catch-up distance.

Catch up distance Speed difference = catch up time (in the same direction).

Speed difference = chase distance catch up time.

The distance traveled by A - the distance traveled by B = the distance between the time of pursuit.

Encounter: Encounter distance Speed and = Encounter time.

Speed and time of encounter = distance of encounter.

Encounter distance Encounter time = speed and.

The distance traveled by A + the distance traveled by B = the total distance.

Opposite time distance and speed and.

Catch up with the time, the distance difference, the speed difference.

Landlord, don't you see that the person above is fooling you? A sucker! You've got a problematic question.

The time it takes for the train to arrive at the terminal is what you want! At this point, the express train has stopped at the terminal.

The question of pursuit is to figure out whether it is in the same direction or in the opposite direction, and who is chasing whom. Then grasp the distance traveled by the previous person plus the original distance equal to the distance traveled by the pursuer, list the equation, and solve it.

For example, the first question. Let B be y kilometers per hour. That's 8 x 2+10=y x 2 , y=13.

Question 2. Set to y minutes. That's 610 x y+1500=660 x y. y=30。

Try to answer the third question on your own.

1.Let B be y kilometers per hour.

8 x 2+10=2y

y=132.Set to y minutes.

610y+1500=660y

y=303.Let us take x hours, and we get the equation 21x=15(2+x).

Solution x=5