Nanjing Yangtze River Bridge is 1600m long, and a 250m long train passes through the main bridge at a constant speed for 3min5s. Calculate the time when the train runs on the main bridge The format is 1. Known 2. Ask 3.4. Answer

Nanjing Yangtze River Bridge is 1600m long, and a 250m long train passes through the main bridge at a constant speed for 3min5s. Calculate the time when the train runs on the main bridge The format is 1. Known 2. Ask 3.4. Answer

First, let's ask the train speed:
The actual length of the train from entering the main bridge to leaving the main bridge is:
1600 + 250 = 1850m, the time is 185s, so the train speed is (1600 + 250) / 185 = 10 m / s
Let's find the travel time of the train on the main bridge:
The train runs on the main bridge, and its actual running length is 1600-250 = 1350m
Using length / speed = time, you can get
The time when the train runs on the main bridge is:
t=1350/10=135s

The total length of the railway line from Zhengzhou to Shanghai is about 1080km. A 200m long train from Zhengzhou to Shanghai passes through Nanjing Yangtze River Bridge. The total length of the lower railway is 6772m, of which the river main bridge is about 1600m long. It takes 2min for the train to pass through the river main bridge. How many hours does it take from Zhengzhou to Shanghai?

Speed (1600 + 20) ÷ (2x60) = 1800 ÷ 120 = 15m / S = 48km / h
It takes 1080 ÷ 48 = 22.5 hours to get to Shanghai
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8. Someone estimated the acceleration of the train with a watch. He observed it for 3 minutes and found that the train was moving 540 meters. After 3 minutes, he observed it for 1 minute and found that the train was moving 3 meters

For problems, we should be good at finding out the starting point, that is, the conditions
Method 1
Direct utilization condition
Since time and displacement are given for uniformly accelerating motion, the time displacement formula is considered
Let the observed initial velocity V0 and acceleration be a, then
Start T1 = 3min = 180s, yes
S1 = v0t1 + at1m2 / 2,
After the interval T2 = 3min 180s, observe T3 = 1min = 60s, and then
V=V0＋at2,
S3 = VT3 + at3 square / 2,
It can be solved by simultaneous
Method 2
Using the inference of uniform variable speed motion, the velocity at the midpoint of time is equal to the average velocity
Within T1 = 3min = 180s,
v1=s1/t1,
V1 is the speed at 1.5min;
Within T3 = 1min = 60s,
v3=s3/t3,
V3 is the speed at the midpoint of time within 1min (i.e. the speed at 6.5min)
From V1 to V3, the intermediate time is t = 5min = 300s
And V3 = V1 + at,
It can be solved by substituting the data

A train starts and leaves the station at a constant acceleration from a standstill. It is assumed that the length of each carriage is the same. Regardless of the gap distance between carriages, an observer stands in the front of the first carriage. Through time measurement, he estimates that the speed when the tail of the first carriage passes him is V0, and the speed when the tail of the nth carriage passes him is () A. nv0 B. n2v0 C. nv0 D. 2nv0

Let the length of each train be l, which is obtained by V2 = 2aX,
v02＝2al，
v2=2a•nl
The solution of simultaneous two equations is v=
Nv0. Therefore, C is correct and a, B and D are wrong
Therefore: C

Someone estimated the acceleration of the train with a watch. He observed it for 3 minutes and found that the train moved 540 meters. After 3 minutes, he observed it for 1 minute and found that the train moved 360 meters. If the train moves in a straight line with uniform acceleration within 7 minutes, the acceleration of the train is () A. 0.03m/s2 B. 0.01m/s2 C. 0.5m/s2 D. 0.6m/s2

Instantaneous speed V1 = 540 at the middle time in the first 3 minutes
180m / S = 3m / s, instantaneous speed V2 = 360 at the middle time in 1 minute
60m / S = 6m / s, and the time interval between two times is 300s, so a = V2 − v1
t=6−3
300 = 0.01m/s2. Therefore, B is correct and a, C and D are wrong
Therefore: B

Someone estimated the acceleration of the train with a watch. He observed it for 3 minutes and found that the train moved 540 meters. After 3 minutes, he observed it for 1 minute and found that the train moved 360 meters. If the train moves in a straight line with uniform acceleration within 7 minutes, the acceleration of the train is () A. 0.03m/s2 B. 0.01m/s2 C. 0.5m/s2 D. 0.6m/s2

Instantaneous speed V1 = 540 at the middle time in the first 3 minutes
180m / S = 3m / s, instantaneous speed V2 = 360 at the middle time in 1 minute
60m / S = 6m / s, and the time interval between two times is 300s, so a = V2 − v1
t=6−3
300 = 0.01m/s2. Therefore, B is correct and a, C and D are wrong
Therefore: B