How should I answer the itinerary question? What to do if you have a travel problem?

I hope my examples and analysis are helpful to you.

When driving, boating, and walking, according to the dependence between speed, time and distance, two of the quantities are known, and the third quantity is required. It's also called a travel problem.

This question is an encounter problem in the itinerary 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.

Example: A and B travel opposite each other from A and B at the same time, and they meet for the first time at a distance of 40 kilometers from place A, and they return immediately after reaching the end point, and they meet again at a distance of 20 kilometers from place B.

Let AB and the two places be separated by x kilometers.

The first encounter takes 40 km for A and X - 40 km for B, taking the same amount of time.

In the second encounter, A travels x+20 km, and B walks x+(x-20)=2x-20 km, taking the same amount of time.

i.e. (x+20) 40=(2x-20) (x-40).

x²-100x=0

Solution x=100

A: The distance between AB and the two places is 100 kilometers.

Example: The two cities are 138 kilometers apart, and two people, A and B, start from the two cities by bicycle and travel in opposite directions. A travels 13 kilometers per hour, B travels 12 kilometers per hour, B is delayed for one hour due to car repairs, and then continues to travel and meets A.

How many hours does it take from departure to meeting?

1. Arithmetic solution: B is delayed for 1 hour due to car repair and waiting in the process, which can be regarded as B retreating 12 1 = 12 kilometers to set off, then A and B set off at the same time, and the distance before departure is 138+12 = 150 kilometers, and it can be obtained: 150 (13+12) = 6 hours from departure to meeting.

2. Equation solution: Let x hours elapse from departure to encounter.

Then the process takes x hours for line A and line B for x - 1 hour, and the equation can be listed: 13x+12(x-1) = 138, and the solution is: x = 6, answer: 6 hours have passed from departure to meeting.

Can you be specific? It's too abstract.

Let us know the unknowns, and then it's the most basic distance = speed multiplied by time.

The formula for going in the opposite direction: time of encounter = distance and velocity sum (velocity of A time + velocity of B time = distance).

The formula for going against each other: Distance against each other = speed and time. (A's velocity time + B's velocity time = opposing distance).

The formula for moving in opposite directions: (the slow one is in front, the fast one is behind) Pursuit time = pursuit distance Speed difference.

If you are on a circular runway, (the fast one is in front, the slow one is behind) the pursuit distance = the time of the speed difference. Pursuit distance Time = speed difference.

The journey time is simple!

Draw graphs, column equations, and find equiquantity relationships.

According to the title.

A 40-minute journey by van is 30 kilometers.

Vans travel per hour: 30 40 60 = 45 km, passenger cars travel per hour: 45-6 = 39 km.

When the van arrived in City B, the bus was still missing:

39 40 60 = 26 km.

The whole journey time of the van trip: 26 6 = 13 3 hours.

The distance between A and B: 45 13 3 = 195 km.

Solution: It can be known from the question.

The speed of the van: 30 40

From the question, the van drove more than 30 kilometers in 40 minutes, and calculated that the van drove 45 kilometers per hour, that is, the passenger car drove 45-6 = 39 kilometers per hour, assuming that the distance between A and B is x kilometers, then use the equation: x 45 + 40 minutes 60 minutes = x 39, which means that the time of the whole journey of the van + 40 minutes early should be equal to the time of the bus to complete the journey, and calculate x, x = 5 * 39 = 195 kilometers.

If two cars travel the same amount of time from departure to meeting, then the distance traveled ratio = speed ratio = 7:8

A car traveled 7 to 15 in full length

Overall length = 9 (1 2-7 15) = 270 km.

Three unknowns and two equations can be solved.

Two methods:1At the time of the meeting, Xiaoling had 120 more lines than Xiaoping 2 = 240 meters.

Per minute, Xiaoling travels 100-80=20 meters more than Xiaoping.

Encounter time: 240 20 = 12 minutes.

Distance from school to Children's Palace: (hail 100+80) 12=2160 meters 2Let's say they're gone for x minutes.

100x-80x＝120*2

20x＝240

x 12 So the distance between the school and the Slip Shouting Children's Palace is 12 * 100 + 12 * 80 2160 meters mountain sail.

When they met and laughed, Xiaoling had 120 2 = 240 meters more than Xiaoping.

Per minute, Xiaoling touches her fingers more than Xiaoping to change the comic line by 100-80 = 20 meters.

Encounter time: 240 20 = 12 minutes.

Distance from school to Children's Palace: (100+80) 12=2160 meters.

Let's say they rent and nucleate and go for x minutes.

100x-80x＝120*2

20x＝240

x 12, so the distance between Jiansan School and the Children's Palace is 12*100+12*80 2160 meters.

Let them go for x minutes.

100x-80x=120x2

x=12100+80)x12=2160m.

Solution: If B's velocity is x, then the distance is 20x

According to the conditions in the question, it can be known.

The time it took for the two cars to meet was.

20x/(105+x)

105×20x/(105+x)=(1+1/4)x×20x/(105+x)，105=(1+1/4)x，x=84

B's speed is 84 km/h.

84 20 = 1680 km.

The distance between the two places is 1,680 km.

b[Analysis] There are many known quantities in the problem, and all the data given are proportional data, and the assignment method is used to solve the problem. If the speed of cycling is 100, the speed of walking is 25, and the speed of the bus is 200, then the speed of walking Cycling speed Bus speed = 1 4 8, so the time ratio is 8 2 1, then a walking and a bus sharing time is hours, that is, 90 minutes, then 9 parts represent 90 minutes, it can be known that 1 part is 10 minutes, then it takes 2 parts of time to ride a bicycle, then it is 20 minutes, therefore, the answer to this question is option B.

Itinerary application problemStroke problem is the problem of studying the movement of objects under certain conditions, environments and ranges, and this kind of problem mainly involves the relationship between three quantities: distance, speed and time. The more complex travel problem should also pay attention to the understanding of "speed and", "speed difference" and the departure time, departure place, direction of movement and movement result of the two cars in the trip, and other four elements, the travel problem can be divided into three categories according to the different direction of motion: First, the encounter problem Two objects meet due to the opposite motion in the opposite direction, which is the encounter problem.

The key to solving the encounter problem is to find the sum of the velocities of two moving objects, and its basic formula is: meeting time = distance between two places Speed and speed sum = distance between two places Encounter time and distance between two places = speed and meeting time Second, the separation problem Two moving objects are separated from each other due to their opposite motion, which is the separation problem. The key to solving the distance problem is to find the distance (velocity sum) of the common tendency of two moving objects

The basic formula is: distance between two places = speed and time away from each other time distance = distance between two places speed and speed sum = distance between two places distance from time 3. Catch up with the problem Two moving objects travel in the same direction, one fast and one slow, after the fast train, before the slow train, after a certain amount of time, the fast catches up with the slow is the chase and the problem. According to the different conditions given, there are two types:

1) directly to catch up with the distance (at the same time in different places); (2) Indirect pursuit distance (when the same place is different).The key to solving the chase problem is to determine or find the chase distance and speed difference, the basic formula is: chase time = chase distance speed difference chase distance = speed difference chase time speed difference speed difference = chase distance chase time.

The distance between the two places is 1680 meters.

Columnable equation: let ae be the chain x and be be y

x/60=y/80

2y type Xiaosun 60 = 2x 80 + 14

The solution is x=720m, y=960m.

Set the speed of A x the speed of B y ab distance of 10s

The speed ratio of A and B is 3:7

The first encounter A walks 10s (3 10) = 3s and a 3s distance

The second encounter A walks 30s (3 10) = 9s and is 9s away from A 9s

The third encounter A walks 50s (3 10) = 15s and is 5s away from A

The 4th encounter, A walks 70s (3 10) = 21s and is 1s away from a (the number of tens is even, and the distance from a is equal to a single digit).

The 5th encounter A walks 90s (3 10) = 27s and is 7s away from A

In the 6th encounter, A walks 110s (3 10) = 33s and is 7s away from a (the number of tens is odd, and the distance a is equal to ten minus a digit).

The 2007th encounter A walks 40130s (3 10) = 12039s and is 1s away from A

The 2008th encounter A walks 40150s (3 10) = 12045s and is 5s away from A

4s=120

10s = 300 km.

Let AB distance 10s, the distance ratio of A and B is 3:7, and the first encounter A walks 10s (3 10) = 3s and the distance A 3s

The second encounter, A walks 30s(3 10)=9s, the third encounter, A walks 50s(3 10)=15s, the distance A 5s, the 4th encounter, A walks 70s(3 10)=21s, the distance A 1s, the 5th encounter, A walks 90s(3 10)=27s, the 6th encounter A walks 110s(3 10)=33s, the distance from A 7s, the 7th encounter, A walks 130s(3 10)=39s, the distance from A 1s, the 8th encounter, A walks 150s(3 10)=45s, and the distance A 5s 9th encounter A walks 170s (3 10) = 51s from A 9s 10th encounter A walks 190s (3 10) = 57s from A 3s 11th encounter A walks 210s (3 10) = 63s from A 3s 12th encounter A walks 230s (3 10) = 69s from A 9s 13th encounter A walks 250s (3 10) = 75s from A 5s from the 1st to the 10th is a cycle, so, 2007 10 = 200....7, the distance a is 1s

2008÷10=200…8. The distance a is 5s, 4s = 12010s = 300 kilometers.

Reference ( zhaoyi 628|Level 6).