How many seconds does it take for two trains to meet from the front of the train and the rear to be

Solution: The problem of trains crossing bridges.

Formula: (car length + bridge length) train speed = train crossing the bridge time speed is a train traveling kilometers per hour, the speed per second is 18 meters per second, a train through a 250-meter-long tunnel takes 25 seconds, through a 210-meter iron bridge takes 23 seconds, then the train speed is: (250-210) (25-23) = 20 meters per second distance difference divided by the time difference is equal to the train speed.

The length of the train is: 20 * 25-250 = 250 (m) or 20 * 23-210 = 250 (m).

So the train is on the wrong side of another 320-meter-long train with a speed of kilometers per hour.

320+250) (18+20)=15 (seconds) Wrong train is the process when the front of two trains meet and the tails of two trains are separated.

"It took 25 seconds for train A to pass through a 250-meter-long tunnel, and 23 seconds to pass through a 210-meter-long railway bridge" assumes that car A is x meters long, and the equation is.

250+x) 25=(210+x) 23, the solution is x=250, so that the speed of car A is 20 meters per second. The speed of vehicle B is kilometers per hour = 18 meters per second.

In this way, it takes (250+320) (20+18)=15 (seconds) for two trains to meet from the front of the train and leave the rear of the train

According to the relative motion, it can be considered that B is not moving, then the velocity of A is adding, let the length x velocity of A be v m s has (250+x) v=25 (210+x) v=23

x = 250 m v = 20 m s = 72 km h and the speed is.

The total length is 320+250=570m, t=570, 38=15s, that is, it takes 15 seconds.

Set a train length x meters and a speed of y has:

x+250）/y=25

x+210）/y=23

The solution is x=250m

y=20m/s

t=（320+250）/（18+20）=15s

Summary. A train passes through a kilometer-long bridge and leaves the bridge in 66 seconds from the front to the rear. Trains travel kilometers per second. How many meters is the length of the train? Rice.

A train passes through a kilometer-long bridge and leaves the bridge in 66 seconds from the front to the rear. Trains travel kilometers per second. How many meters is the length of the train?

A train passes through a section of the bridge in Changgong Fuqili, and leaves the bridge in 66 seconds from the front to the tail. Trains travel kilometers per second. How many meters is the length of the train for the lack of debate? Rice.

The total distance from the front of the train to the rear of the train from the bridge includes the length of the train and the bridge, so the answer can be solved.

From the question, we know: He Fan 1, the driving distance between the two cars is L when they meet for the first time, and the driving distance is 2L when they meet for the second time

Therefore, if the opening time of the first encounter is t, the operating time of the second encounter is 2t, and the total operating time is 3t

The travel time from train A is simple

The distance of train A: the first time we met the mountain, it was 75, and the second time we met, it was l+55 from the meaning of the question to know that l+55=3*75, so l=225-55=170 testifies the evidence:

Calculation of the travel time from train B:

Train B travel distance: L-75 at the time of the first encounter, 2L-55 at the time of the second encounter, so 2L-55 = 3L-225, so L=170

There are two trains, one 280 meters long and traveling 18 meters per second and the other 320 meters long and traveling 22 meters per second. Now two trains are moving in opposite directions, how many seconds does it take to meet each other?

15 (seconds).

It depends on the speed of the train, but also the length of the train. If the train speed is 200 and the train length is 500, then it takes 1000 divided by 400 to meet and leave from the phase, and this is without acceleration. If there is acceleration, consider acceleration as well.

Algorithm omitted...

The distance between A and B is 360 km, and an express train departs from Station B and travels 72 km per hour; A slow train departs from Station A and travels 48 kilometers per hour

1) If two trains leave at the same time and travel in opposite directions, how many hours does it take for the two trains to meet?

2) If the express train runs for 25 minutes and the two cars are moving in the opposite direction, how long does it take for the slow train to meet each other?

1) Set two cars to drive in the opposite direction at the same time, after x hours the two cars meet, that is, 72x+48x=360, the solution: x=3, answer: after 3 hours the two cars meet

2) Set the slow train to travel for y hours, and the two cars meet;

According to the question: 48y+72(y+25 60)=360, and the solution: y=11 4

A: The slow train traveled for 11 4 hours and the two cars met

Because the total distance traveled by the train from the front bridge to the parking space macro and departure, is the length of the bridge plus the length of the train. It has been sold for 1 minute, the speed is 156 meters per second, so the total distance is 60 156 = 9360 meters, and the bridge is 1,000 meters long, so the train is 9360-6900 = 2460 meters.

v=156m/s

t=1min=60s

S = S car refers to the shed Qichang + bridge only as long as the posture = S car length + 6900ms = VTS car length + 6900 = 156x60

Vehicle length = 2460m

The first car went 230 meters and the second 150 meters.

Since the two cars are 380 meters long and travel in opposite directions at a speed of 38 meters per second, each car travels for 10 seconds when they are apart, and then multiplied by the speed

102+83)forward touch (20+17),185 37,5 (s);

Answer: It takes a total of 5 seconds for the two cars to leave from the old man to the rear of the car

Let the speed of vehicle A be 5x msec and car B be 3xmsec.

12 (5x 3x) = 200 cavity 280

x = 5 car A: 5x 25 m sec.

Car B: 3x Wu Xian 15 meters second to block.