Elementary school itinerary question f, look forward to detailed explanation, primary school itinera

Analyse. Passenger car A c b Van.

Passenger car a d b van.

Suppose d is the point of encounter.

At 8 a.m., the truck arrived at Station C --- the truck traveled for 2 hours.

The bus arrives at station C at 3 p.m. - the bus travels for 9 hours.

If the speed of the passenger car is 4x, the speed of the truck is 3x

Distance 4x 9+3x 2 42x

So the time it took to meet was 42x (4x+3x) 6 hours.

So the time of the encounter is (6 + 6) = 12 points.

Let the speed of the passenger car be s

The distance is 2*3 4s+9*s=

The time required for the encounter is hours).

So at 12 noon (6+6) the two cars meet.

Let the speed of the passenger car be x, then the speed of the truck is, and the passenger car and the truck meet in t hours.

t = 6 The passenger car and the truck meet after 6 hours after departure, that is, they meet at 6 + 6 = 12.

Solution: Speed = distance time.

Truck speed vh=cb (8-6)=cb 2;

passenger car speed vk=ac 9;

Truck speed vh = (3 4) bus speed vk, i.e. CB 2 = = (3 4) (AC 9), that is, AC = 6BC;

If the encounter is set at d, AB=7BC=AD+DB=VK*T+VH*T, that is, 7BC=(6BC 9)*T+(BC 2)*T, and the solution t=6 hours, so DB==(BC 2)*6=3BC is the meeting at a place 3 times the distance between the two stations of BC from station B; The passenger car and the truck meet after 6 hours after departure, that is, they meet at 6 + 6 = 12 hours.

In fact, it is very simple, assuming that the speed of the passenger car is x, then the truck is 3 4x, and the total distance length is (x+3 4x) * 2 + (15-8) x = 21 2x, so the total need for the encounter is (21 2) x divided by (x + 3 4) x is equal to 6 (hours), so the time of the encounter is (6 + 6) = 12 points.

Set C to walk x km.

Total distance c7x=

x = km.

Is this a primary school topic? Formula Three!

Too lazy to count! Don't do it!

Since the distance is constant, only the average speed of the march needs to be calculated, assuming that the total length of the road is l (km h.

B's velocity is (km/h).

So B is faster on average, so Class B wins.

Since the distance remains the same, the total length of the road is L

The speed of A is l (kilometers per hour.

B's velocity is (km/h).

So B is faster on average, and Class B wins.

Let the distance be s

So shift A takes (s 2) hours.

Shift B time s (hours.

Since 20s 99 s 5

Therefore, Class B takes less time, and Class B wins.

Assuming that the sum of the speeds of the two people is x kilometers per hour, the length of the journey is kilometers x = 30

So the length of the journey is 30*km.

Hey, hey, I'll give it a try.

Hours: Fewer lines per hour.

A decrease of 3km per hour is 3*

Walked an extra hour yes.

It took me five hours, I guess.

Suppose the two places are separated by x kilometers so that the previous velocity and x and the later velocity and x 5 have x

x 135 km.

Let the original velocity sum be x, then:

x=3030*

Let the original velocity be v and the original time be t, then:

vt (v=2 km/h(km.)

Let the original velocity be v, the original time is t, and the distance length is l, and there is: vt=l, (1).

v+1/2)*t*4/5=l (2)(v-1/2)* 3)

From (2): v= (4).

From (3): v=l 4+ (5).

Substitute (5) for (1), (l 4+ (6).

4)-(5), there is (7).

l = 6 km.

Let the original velocity be v

Then there is a distance of 1 2 kilometers per hour, and this section of the road only takes 4 5 times the original time.

So 4 (5V) 1 (V+1 2).

4v+2＝5v

v 2 so the distance is km.

The original time taken to check was hours.

It takes 1 2 kilometers per hour to travel faster hours.

In line with the topic.

That is to say, when A is in the middle of B and C, it fits the topic.

Let's say it takes x minutes.

If A has caught up with and overtaken B by this time, then there is.

120-100）x-400=800-(120-90)x20x-400=800-30x

50x=1200

x = 24 minutes.

At this time, A walked a total of 2880 meters, B walked 2400 meters, C walked 2160 meters, and A was 2880-400-2400=80 meters before B.

A is still 800 + 2160-2880 = 80 meters away from C, so the first solution is 24 minutes.

The second case is that B catches up with C, the distance between A and these two is naturally equal, this is very easy, B catches up with C It takes 400 (100-90) = 40 minutes and B catches up with C, dinoflagellates have already surpassed B and C, so there will be no more C behind, A in the middle, and B in front, so to sum up, there are two solutions in total, which are 8:24 and 8:40 respectively.

When B C meets, A and B C are at an equal distance.

400 (100-90) = 40 minutes, 8 o'clock plus 40 minutes, that is, the distance between A and B C is equal at 8:40.

When A is between B and C and does not exceed C, the distance between A and B and C is equal in x minutes. Then:

120x-(400+100x)=90x+800-120xx=24

So at 8:24 a.m., the distance between A and B C is equal.

Let B exceed C, and when A is in the middle of B and C, A and both are 400 meters apart when B and C meet, and the distance between A and B and C is equal in x minutes. Then:

400+100x-120x=120x-400-90xx=16

So 40 + 16 = 56 minutes.

That is, at 8:56, A and B C are at an equal distance.

There are three times in total, namely: 8:40, 8:24, and 8:56.

In the first case, A is between B and C.

Let x minutes be equal for the distance between A and B C.

120x-(400+100x) (400+400+90x)-120xx 24 minutes.

In the second case, when B and C meet, A is in front of B and C.

Time taken: 400 (100-90) 40 minutes, because after B exceeds C, A is in front of B and C, and it is impossible to be in the middle of B and C, so there is no third situation: The distance between A and B and C is equal.

Therefore, the distance between A and B and C is equal at 8:24 and 8:40 respectively.

Solution: Make the distance between A and B and C equal, there are two and only two cases.

At this time, the distance between A and B is 400 meters, and the distance between A and C is 800 meters.

1. When A is located between B and C. This happens after X minutes. Then there are 120x (400 100x) 90x 800 120x solution to this equation.

x 24 is available at 8:24 a.m.

2. When B and C meet. This happens after y minutes. Then there are 100y 90y 400

Solve this equation.

y 40 is available at 8:40 a.m.

At this point, A has surpassed B and C. Due to the difference in speed of the three people, B and C will not meet again, and A will not return to between B and C, provided that they go straight from west to east instead of going in circles, so there are only two cases.

Answer: At 8:24 and 8:40, the distance between A, B and C is equal.

I'm not a teacher!

The itinerary question is actually very simple! The key is to understand two points:

1.s=vt, i.e., distance = speed * time, and its deformation is speed = distance time.

Time = distance speed.

2.It is to grasp the equivalence relation, list the equations, when time is equal! When the velocities are equal.

It is recommended to draw a sketch often when solving such a problem, so that it is clear at a glance!

So you can understand the topic above!

Set the speed of the river x km h, the distance between the port and y km, get:

1: y x+y (x-20)=8 (this equation is equal in time and 2): y+(4-y x)*(x-20)=(x-20)*4+60 (this equation is equal in distance).

Calculated: x=50

y=150 port distance of 150 km.

What grade is elementary school, this should be at the level of junior high school.

Set the speed of the river x km h, the distance between the port and y km, get:

1：y/x+y/(x-20)=8

2: y+(4-y x)*(x-20)-(x-20)*4=60 is calculated: x=50

y=150 port distance of 150 km.

Just draw a picture and you'll find out. The first 4 hours are 60 kilometers longer than the next 4 hours, which is twice the distance traveled between the port and the next 4 hours. Then it can be calculated that the distance of driving against the water in the first 4 hours is 30, and the rest of the time is in the water.

The current is 10 x boat speed. +=4

Can elementary school students solve such problems??

The first case: car A arrives at station B to wait for the train.

Yes: Car A takes 95 to 50 hours.

Car B takes (110+15) 60 25 12 hours, so the waiting time is 25 hours, minutes, and 11 minutes.

The second case: car B arrives at station C to wait for the train.

Car A takes (95+15) 50 hours.

Car B takes 110 60 11 6 hours.

So the waiting time is hours and minutes and 22 minutes.

Therefore, choose to wait at point B, the waiting time is the shortest, and the time for A to wait for B is 11 minutes.

Cars A and B depart from A and B respectively, and continue to travel back and forth between A and B. When it is known that the speed of car A is 15 kilometers and the speed of car B is 25 kilometers, the difference between the third meeting place and the fourth meeting place of car A and B is 100 kilometers. Find the distance between A and B.

Solution: When A and B met for the third time, they had gone through 5 whole journeys, and when they met for the fourth time, they had already gone through 7 full journeys. From the speed ratio of A and B, it can be seen that B walked 25 (15+25) 5 8 full courses for the first time, then walked 25 8 3 1 8 full courses when they met for the third time, and walked 35 8 4 3 8 full courses for the fourth time.

Then the distance between the two places is 1 2 whole journeys, and it can be seen that A and B are 200 kilometers apart.

According to the speed, car B ran a round trip, and car A couldn't finish a one-way trip (speed B was greater than 2 times the speed of A), that is, when they met for the third time, the two cars ran three full journeys. , the two cars ran five full courses when they met each other four times. According to the speed ratio 3:

7 can be known, car B when returning to B, car A is running in the whole process of 6 7, the two cars will meet for the third time in this 1 7 distance, the meeting point is 1 7 * 7 10 = 1 10 in the whole journey from B place, in order to facilitate your analysis of the problem, here we divide the whole journey into 10 parts, four encounters from three encounters, two cars ran two whole journeys, both 20 equal parts, car A 20 * 3 10 = 6 parts, car B 20 * 7 10 = 14 parts, because the three meeting points are 1 10 places away from point B, Therefore, when the four encounters are 14 10-9 10=1 2 or 6 10-1 10=1 2, and the corresponding components are known, the problem is understood: 100 (1 2-1 10)=250

To solve this kind of problem, it is recommended that you draw a diagram for analysis, and the diagram is the kind of simple line segment diagram.

The upstairs idea is correct, but the value is wrong, it should be.

Solution: When A and B met for the third time, they walked a total of 5 whole courses, and when they met for the fourth time, they walked a total of 7 whole courses. From the speed ratio of A and B, it can be seen that B walked 35 (15+35) 7 10 full courses for the first time, then walked 35 10 full courses when they met for the third time, and walked 49 10 full courses for the fourth time.

Then the distance between the two places is a whole journey, and it can be seen that A and B are 250 kilometers apart.