A math problem for a junior high school itinerary problem

The first method.

Let the starting velocity of A and B be: v A v B The speed after acceleration is v1 and the circumference is ss=(v1-v A) (22-18)=4(v1-v A), and the distance between B and A is s1 when it is 15 minutes

s1 = 15 (v B - v A) = (18-15) (v1 - v B), launch v1 - v B = 5 (v B - v A).

Set 15 clocks, B is driving S2 first over A.

s2 = s-s1 = 4 (v1 - v B) - 3 (v1 - v B) = v1 - v B.

Set 15 minutes at the elapsed time t B first super A.

t=s2 (vB-vA)=(v1-vB) (vB-vA) and v1-vB=5(vB-vA).

Find t=5, so B overtakes A at 15+5=20 minutes.

Method 2 (simpler).

At 15 minutes, B leads A: s1 = (18-15) (v1-v A) = 3 (v1-v A).

Circumference s=(22-18)(v1-vA)=4(v1-vA) binding s1=3 4 s

B took 15 minutes to lead A by 3 4 s

Now it is still necessary to drive 1 4 s to catch up with A, and the same relative speed can be seen: it still takes 15 1 3 = 5 minutes to catch up with A.

In addition to the first 15 minutes, it catches up with A at the 15th + 5 = 20 minutes.

After A accelerates the speed, it takes 4 minutes to surpass B, 18-15 = 3 minutes, which means that A has been 3 4 laps behind B when speeding up, and B has taken 15 minutes to lead A 3 4 laps, so if A does not speed up, B will use 15 (3 4) = 20 minutes to surpass A by one lap.

Let the unknowns be solved by equations.

1 A and B are 360 kilometers apart, car A starts from A to the ground, driving 72km per hour, car A departs 25 minutes later, car B departs from B to A, driving 48km per hour, after the two cars meet, they continue to drive at the original speed, so when the two cars are 120km apart after meeting, how long does car A share from the start?

2 A and B stations are 240 kilometers apart, a bus leaves station A and travels 48 kilometers per hour, and a car leaves station B and travels 72 kilometers per hour. 1 hour after the car leaves station B, the bus leaves station A, the two cars go in the opposite direction, how many hours later the two cars meet?

3 A tractor is about to pull goods, 30 kilometers per hour, 30 minutes after departure, the family has something to send a car 50 kilometers an hour speed to chase the tractor, ask how much time the car can catch up with the tractor?

4 A and B run on a 10km ring road, A runs 230m per minute, and B runs 170m per minute

1) If two people set off in the same direction at the same time, how long does it take for them to meet for the first time?

2) If A runs for 10 minutes and B starts in the same place and direction, how long will it take for the two to meet for the first time?

3) How long does it take for people to meet for the second time when they set off in the same place and in the same direction?

5 If two aircraft fly between two cities, it takes 4 hours to return downwind and 5 hours to return against wind, and the speed of the aircraft in a calm wind is 360 km h to find the wind speed and the distance between the two cities.

6 The steamer goes down the river from place A for 8 hours to reach place B, and it takes 12 hours to return to place A The speed of the water flow is 3km per hour, and the distance between A and B is found

If the speed of the motorcycle is V, then the speed of the car is 30 V

The solution is v = 40 km

The speed of the car is kilometers per hour.

1.Let the student's speed be x and the teacher be (x+10).

45x=(x+10)15 x 5, i.e. the teacher's speed is 5+10 152(1) Set x seconds after meeting.

That is, 48x+72x=360 solution x 3

2) Set up a slow train to meet after driving for x hours.

72 (5 12 + x) + 48x 360 x hours (3) Set the slow train to run again after starting x hours and meet the fast train 48x+72x=180 x, so the slow train travels another 180+( x 48)=252km

4) The second distance of 180 means that the distance between the two cars after the encounter is 180, so the total distance of the two cars is s=360+180=540km;

t=540/（48+72）=

Question 1: If the speed of the teacher's car is x, then the speed of the student is x-10; 15 min = , so, solution is: x = 15km h

Question 2: (1) (48+72)*t=360 solution: t=3h

2) 25 minutes = 5 12 h, 48 + 72) * t + 5 12 * 72 = 360

Solution: t=11 4 h

3) The time it takes for the two cars to meet on half of the journey is t=180 (48+72)=;

The distance traveled by slow train is s=180+

4) The second distance of 180 means that the distance between the two cars after the encounter is 180, so the total distance of the two cars is s=360+180=540km;

t=540/（48+72）=

1. If the speed of the teacher's bicycle is x km h, then the speed of the student is (x-10) km h, and the distance taken by the teacher when catching up is s, then:

15 minutes = solve an equation and you get it.

2. (1) Set t hours to meet:

48t+72t=360.

2) Set the slow train to meet in t hours

72*（25/60+t）+48t=360.

3) Let the slow train meet the fast train after the departure of the fast train, s-360 2) 48=(360-s) 72 (using the equation of equal time).

4) 180km apart again, that is, 180km more walked in the original journey, set the second distance between the two cars 180km after the departure of t hours, 48t + 72t = 360 + 180

Solution: Let the speed of A be 2x km/h, and the speed of B be 3xkm/h3 2 3x-6=7 4 2x

9/2x-7/2x=6

x = 6, then A's speed is 12 km/h, B's speed is 18 km/h, A's path becomes 12 7 4=21 km, B's distance is 18 3 2=27 km, and the distance between A and B is 21+27=48km.

Analysis: A starts 15 minutes earlier than B, and meets B after 1 hour and 45 minutes, so A walks for a total of 1 hour and 45 minutes, that is, 7 4 hours, and B walks for a total of 1 hour and 30 minutes, that is, 3 2 hours to solve the equation.

The speed of car A and car B departs from the same station, the speed of car A is 165 km h, and the speed of car B is 185 km trillion h. If car A starts 2 hours earlier than car B, how long does it take for a car to catch up with car A? Solve with equations.

Setup time x hours.

The key to the pursuit of the problem is.

The distance traveled by the two vehicles A and B is the same (the same starting point).

Rule. The speed of car A Car A time = the speed of car B The time of one car.

So 165(x+2)=

185x is 20x = 330

x=So car B catches up with A.

Travel problem: The motion of an object, in the simplest case, is a uniform linear motion, let the speed of the object move in a uniform linear motion is. v at the time.

The distance of motion within t is. S has.

s=vt, which is the mathematical model of the object moving in a uniform linear line

Because the actual object motion may not be uniform, for example, in the journey of the car, the speed changes from small to large when driving, and the speed changes from large to small when stopping, we ignore these and regard the speed of the car as the "average speed", so that the situation that the actual family is not moving at a uniform speed is simplified, and it is approximately regarded as a uniform motion In the primary and junior high school stages, we study the travel problem, and we are all carried out in the uniform motion model

In the problem of itinerary, catching up and encountering are the two most basic models

Catch up with the model. A and B are separated by distance. s.

ab The two places are in the same direction at the same time. Composed. a

Arrive. b direction. Walking at a speed. V A.

Greater than B speed. v

Second. Set up.

After t time, A can catch up with B. c Yes.

s=(vA. v B.

T encounter model.

A and B are separated by distance. s.

ab two places walk in opposite directions at the same time, and the speed of the first is . V A.

B speed is. v B. Set up.

After t time, the two met. c Yes.

s=(vA. v B.

The travel problems solved by using the unary linear equation and the binary linear equation system can be generally incorporated into the two models: chase or encounter

Let the plane fly as far east as possible after take-off x kilometers, and in the process of returning, the aircraft carrier will continue to sail eastward, so the distance at the time of the return of the attack plus the distance of the aircraft carrier is equal to x kilometers, so the total distance of the aircraft flight plus the distance traveled by the aircraft carrier is equal to 2x, so the equation can be listed according to the meaning of the question: 2x=1200*3+40*3, and the solution is x=1860, A: The farthest eastward distance of the aircraft after takeoff is 1860 kilometers.

36/3=12

The water flow velocity is 2km h

The hydrostatic velocity is 12-2=10(km h).

The second question, lz, did you learn the system of equations?

Four minutes = 120 seconds.

Let A be faster than B, A speed is x, and B is y

30x+30y=200

240x-240y=200

The solution yields x=35 12 y=5 4

The speeds are 35 12 m s 15 4 m s and hopefully that helps.

then the velocity of the water is 2 km/h (36-24) 3 2

The speed of the boat in still water 10 km h 36 3-2

225 m min and 175 m min. 1 encounter in 30 seconds, indicating that the speed sum is 200 meters min. 1 encounter every 4 minutes, indicating that the speed difference is 200 4 = 50 m min, so it is 225 and 175

If a boat sails 36 kilometers downstream in 3 hours and 24 kilometers against the current in 3 hours, the current speed is 2 kilometers and the speed of the boat in still water is 10 kilometers Write the process and answer).

Solution: When the speed of the ship in still water is x kilometers, and the speed of the current is y kilometers, then: {x+y) 3=36

x-y）×3=24

Solution: {x=10.)

y=2A...

A and B practice race walking on a 200-meter circular track, when they both start from a certain place at the same time and walk in the opposite direction, they meet every 30 seconds, and when they walk in the same direction, they meet every 4 minutes, then the speed of the two people is 15 4m s and B 35 12m s respectively Write the process and answer).

First of all, there is a problem with the topic. , I didn't say who is faster than whom, then I will say that A is faster than B, pay attention to catch up, A walks one more circle than B, that is, 200 meters.

Let the velocity of A be xm s and the velocity of B be ym s

30(x+y)=200

4×60（x-y）=200

The solution yields {x=15 4

y=35/12

Answer... I don't understand. Please ask. Happy o(o

Water speed + boat speed = 36 12

Boat speed - water speed = 24 3

Boat speed = 10 kmh, water speed = 2kmh.

Reverse 30 (speed A + speed B) = 200

4 x 60 in the same direction (speed A - speed B) = 200

B speed = 7 9 A speed.

A velocity = 15 4 m s.

B speed = 35 12 m sec.

You start by setting two unknowns x,y

3(x+y)=36;

3（x-y)=24

Xiao Ming was walking along the road, and a year bus came on the opposite side, and he asked the driver, "Is there a bicycle in the back?" ”

Driver him: "I overtook a bike 10 minutes ago. Xiao Ming asked again: "What is your speed?" ”

Driver: "60 km/h", Xiao Ming continued to walk for 20 minutes before he came across the bicycle, Xiao Ming estimated that his walking speed was 3 km/h.

Q: Can you calculate the speed of this bike?