During the military training of freshmen at a certain school, a column with a length of 120 meters w

400 meters. It's actually quite easy to understand.

Lee was originally at the opposite end, and at the end he was still at the opposite end, which meant that he was going to run 160m before the team advanced.

But in the process of marching, he ran from the end of the team to the head of the team, and from the head of the team to the back of the team.

In this way, he has to run two more teams of lengths.

Then the formula is.

120*2+160=400m

Set: the speed of the communicator is v1, the speed of the team is v2, the time of the communicator from the end of the team to the head of the team is t1, and the time from the head of the team to the end of the team is t2. The team advanced 160m with a time t.

Using the team as a reference, the speed of the communicator from the end of the team to the head of the team is (v1-v2), and the speed from the head of the team to the end of the team is (v1+v2).

The time from the end of the team to the head of the team is t1=120 (v1-v2), and the time of the communication from the head to the end of the team is t2=120 (v1+v2).

The time it takes for the team to advance 160m t=160 v2

t=t1+t2

160/v2=120/(v1-v2)+120/(v1+v2)120/(y-x)+120/(y+x)=160/x

120(y+x)+120(y-x)]/(y²-x²)=160/x

240y/(y²-x²)=160/x

240xy=160(y²-x²)

3xy=2(y²-x²)

2y²-2x²-3xy=0

2y+x)(y-2x)=0

It can be concluded that y=2x.

Solution: v1 v2=2

The distance traveled by the correspondent in time t.

s=v1t=v1(160/v2)

160(v1/v2)

320m

Let the speed of the correspondent be V1, the speed of the team is V2, the time of the correspondent from the end of the line to the head of the line is t1, and the time of the correspondent from the head of the line to the end of the line is t2, which is derived from the title, t1=120m (v1-v2) 1.

t2=120m (v1+v2) 2.

t1+t2=160m v2 3 formula.

From the formula 1,2,3 v1=2v2,v1=-v2 2 (discarded if it does not fit the topic), let the correspondent walk a total distance of s in time t1+t2, then s v1=160m v2 then the correspondent ran a total of 320m in the process.

Package is correct!! Beg!!

The length of the train is x meters, and the stove is good.

Students travel at a constant speed along the road parallel to the straight Iron Sock Route, traveling 4,500 meters per hour, as a train approaches them at a speed of 120 kilometers per hour.

How many meters can a train run in a minute relative to a student:

2075 meters, a minute train can run 2075

Rice. And the fire hides the old lead head.

It takes 60s to meet the head of the team and leave the tail of the train and the tail of the team, that is, one minute, 500 + x = solution x = 1575, the length of the train should be.

2075m-500m=1575m, so choose b

1575 meters.

First of all, 60 seconds equals 1 minute.

Then change the speed of the two to minutes.

Students walk 75 meters per minute The train travels 2,000 meters per minute, and it takes 1 minute to meet at the front of the train and then meet at the end.

then both traveled for 1 minute.

The length of the student queue plus the length of the train is equal to 1 (75+2000) equals 2075, and the length of the student queue is known to be 500 meters.

Then the length of the train 2075-500 is equal to 1575 meters.

Assuming that the train is x meters long, then when the locomotive and the leader of the team meet, the tail of the train and the students at the end of the team meet (5000 + x) meters, and because the two are traveling in opposite directions, the total speed is 124500 meters per hour, that is, meters and seconds, so (5000 + x), x = meters.

First of all, calculate the speed of the train and the team per second, the team travels 4500 meters per hour, and the speed per second is: 4500 (60 60) = meters.

The train travels 120 kilometers per hour at a speed of 120 1000 (60 60) meters per second.

After 60 seconds of meeting the students at the front of the team and the students at the end of the team, the distance they have covered is: (meters.

The distance they have traveled minus the length of the line is the length of the train: meters.

Well, this is a very typical pursuit problem, and there is a skill in doing pursuit problems, that is, you must draw a picture first, which will be very helpful for analyzing similar problems in the future. This is what my dad taught me when I was studying (my dad is very good at math, and he really thinks it's useful after trying it, so you might as well try it).

First of all, we set the time it takes to catch up with the team in x minutes. There is a common force between the team and the communicator, that is, the route taken is the same, that is, the distance is the same; Then the correspondent has to walk 14*x kilometers to catch up with the team, and the team should be 5*(18+x), well, the equation should be listed as 14x=5*(18+x), and the equation will be solved in x=10 minutes.

The unit is not the same, an hour and a minute, it should be.

Solution: It takes x hours for a correspondent to catch up with the student team.

14x＝5（18/60 x）

x＝1/6

To convert, it is eighteen minutes, which should be three-tenths of an hour, and solve x = one-sixth. You're all mistaken.

Let the planned velocity v, and the equation is: 60 v 60 (1 1 5)v 1; 10／v＝1 ；

v 10 (kmh).

Solution: 60 seconds = 1 minute = 1/60 of an hour.

120 km = 120,000 m.

120000+4500) 1/60-500=124500 1-500/60

1575 meters.

A: The train is 1575 meters long.

Set the time to x hours. Because the distance traveled is equal.

5*（x+18/60）=14x

x=1/6

One of the students went to the school for military camp training, and they did it at a speed of 5 kilometers per hour. After walking for 18 minutes, the school has to send an emergency notice to the team leader, and the correspondent sets off from the school and rides a bicycle at a speed of 14 kilometers an hour to catch up with the original road, how many hours does it take for the correspondent to catch up with the student team?

Solution: It takes x hours for a correspondent to catch up with the student team, according to the question, 5 (x+ ) 14x

5x+ =14x

9x= x=

A: It takes hours for the correspondent to catch up with the student ranks.

Students: If you have any questions about the above answers, please ask them in the discussion, and I wish you progress in your studies!

You can just swap the number and you're good to go.

I sent you a ** to see.

10 minutes The correspondent student walked in total The student walked first So before the correspondent left, so the student walked first 10 minutes.

Assuming the team is not moving, the Communicator's repentance speed relative to the team is:

From the end of the line to the head of the line, 11-7 = 4 kilometers per hour.

From the beginning of the platoon to the end of the platoon, 11 + 7 = 18 kilometers per hour.

From the end of the line to the head of the line, the time taken is 18 (4+18) small traces.

Team length: 4,000 Bizhou closed meters = 720 meters.