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The length of the train is l, and the length of the railway bridge is also L. the train accelerates evenly along the straight track to cross the bridge. The speed of the front passing the bridge head is V1, and the speed of the front passing the bridge tail is V2, so the speed of the rear passing the bridge tail is () A. v2 B. 2v2-v1 C. v12+v22 two D. 2v22−v12
For the process from the front axle to the front axle to the rear axle, according to the speed displacement formula, V22 − V12 = 2Al,
For the process from the front axle to the rear axle, according to the speed displacement formula, V32 − V12 = 2A • 2L,
The solution of simultaneous two equations is V3 =
2v22 − V12. Therefore, D is correct and a, B and C are wrong
Therefore: D
A train departing from the station moves in a straight line at a constant speed. It is known that the length of the train is L. when the locomotive passes a road sign, the speed is V1, and when the tail passes the road sign, the speed is v2. Calculate the size of the train acceleration a, the speed v when the train passes the road sign at the midpoint, and the time t for the whole train to pass the road sign,
1)v ²－ v0 ²= 2ax v2 ²－ v1 ²= 2al a=v2 ²－ v1 ²／ 2l
2) V X / 2 = V0 under the root sign ²＋ v ²／ 2 = V1 under the root sign ²＋ v2 ²／ two
3)v=v0+at t=v-v0/a=2l/v1+v2
The train leaves the station for uniform acceleration. If the resistance is proportional to the speed, then () A. The power of the train engine must be greater and greater, and the traction force must be greater and greater B. The power of the train engine is constant and the traction is getting smaller and smaller C. When a train accelerates to a certain speed, the power of the engine is greater than that when it keeps moving at a constant speed D. When the train reaches a certain speed, if it wants to keep this speed moving at a uniform speed, the power of the engine must be proportional to the square of the speed at this time
A. The train starts from the station and accelerates uniformly. The resistance is directly proportional to the speed, f = kV. With the increase of speed, the resistance increases and the acceleration remains unchanged. According to a = f − F
M shows that the traction force increases. According to P = FV, the power is also increasing, so a is correct and B is wrong;
C. When the train accelerates to a certain speed, the engine power P = FV. At this time, f > F. when the train keeps moving at a constant speed, the engine power p '= f ′ v. at this time, f' = f, so p > p ', so C is correct;
D. If the train moves at a constant speed, f = F. according to P = FV and F = kV, P = Kv2 is obtained, that is, the power of the engine is directly proportional to the square of the speed at this time, so D is correct
Therefore, ACD is selected
A train starts from a stationary station and accelerates evenly A train starts from a stationary station in a uniformly accelerated straight line. An observer stands in front of the first carriage of the train 0 | solution time: 22:52, January 22, 2007 | questioner: pyc777 A train starts from a stationary station and moves in a straight line with uniform acceleration. An observer stands at the front of the first carriage of the train. After 2S, the first carriage passes through the observer's position; All the carriages passed by him for 6 s. if the length of each carriage is equal and the distance between carriages is not counted, the train has_____ Two carriages; The cars passing by him in the last 2S are_____ Section; What is the time required for the last carriage to pass through the observer_____ s.
9.5.6-4√2
Fill in the math blank: the length of the two trains a and B is 160 meters and 200 meters respectively. Car a travels 15 meters more per second than car B The two trains run in opposite directions, and it takes 8 seconds from meeting to staggering, then the speed of car a is ___, The speed of car B is ___
The two trains run in opposite directions, from meeting to staggering,
It can be seen that the two vehicles have driven 160 + 200m in total
Vehicle B speed x, then x * 8 + (x + 15) * 8 = 360
X = 15m / S
Car a: 15 + 15 = 30m / S
The two trains, car a and car B, are 200 meters long and 280 meters long, running opposite each other on parallel tracks. It is known that it takes 18 minutes from the front to the rear, and the speed ratio of car a and car B is 5:3, (1) Find the speed of two vehicles (2) How many minutes does the fast train take to overtake the slow train if the two trains run in opposite directions If two trains go in the same direction, how many minutes does it take for the express train to overtake the slow train Use equation
（1）
The sum of two vehicle speeds is:
(200 + 280) ÷ 1000 ÷ 18 / 60 = 1.6km/h
The speed ratio of vehicle a and vehicle B is 5:3,
Then the speed of a is:
1.6÷（5+3） × 5 = 1km / h
The speed of Party B is:
1.6-1 = 0.6 km / h
（2）
The two trains go in the same direction, and the fast train overtakes the slow train,
Then the fast train is 200 + 280 = 480m = 0.48km more than the slow train
The time taken is:
0.48÷（1-0.6）
=0.48÷0.4
=1.2 hours
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