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Both trains run from a to B. It is known that the speed of car a is x km / h and that of car B is y km / h. After 3 hours, car B is 5 km away from B At this time, the distance between car a and ground B () km? A 3(-x+y)-5 B 3(x+y)-5 C3(-x+y)+5 D 3(x+y)+5 I want to ask why I chose C
In another way, we can see from B that 3Y + 5 should be the distance between a and B. subtracting 3x from this distance is the distance between a and B, that is, 3Y + 5-3x. Rearrangement is 3 (- x + y) - 5, which is just to confuse you with a deformation formula
The speed of truck is 80 km / h, and that of train B is 60 km / h. how many times is the speed of truck a higher than that of train B?
80 divided by 60
=Four out of three
At night, train a was moving at a constant speed of 4 m / s. at that time, train B mistakenly entered the same track and chased car a at a speed of 20 m / s. when the driver of car B found car a, the distance between the two cars was only 125 m, and car B immediately braked. It was known that the train moving at this speed needed to pass 200 m before it could stop. (1) ask if there was a collision; (2) to avoid a collision between the two cars, increase the brake of car B The speed is at least what it should be
(1) According to the meaning of the title, the acceleration of car B is: a = - v22s = - 4002 × 200 = - 1m / S2. When the speed of two cars is equal, the time experienced is: T = V1 − VA = 4 − 20 − 1 = 16S. At this time, the displacement of car a is: x a = v1t = 64M. The displacement of car B is: x B = V12 − v22a = 16 − 400 − 2 = 192m
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