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The two trains run at a uniform speed in the same direction. When they start, the slow train is 60 kilometers in front of the fast train. The speed of the slow train is 2 / 5 of that of the fast train. After driving for half an hour, the fast train catches up with the slow train. Q: what are the speeds of the two trains?
Set the fast train speed x, then the slow train 2 / 5x = 0.4x
1/2X=1/2*0.4X+60
X=200
0.4X=80
Is train 1282 an express
It means the general speed. It's a little faster than the slowest. Trains with K and t prefix are express trains
A freight train was running on the railway at the speed of 28.8km/h. Due to the dispatching accident, an express train behind it was coming in the same direction on the same track at the speed of 72km / h. When the express driver found that the freight train was 600m away, he immediately closed the brake, but the express train had to slide 2km to stop. Please judge whether the two cars would collide and explain the reasons In the answer, the acceleration is -0.1m/s2, that is, the final speed is zero, but can the express train not collide as long as it is at the same speed as the truck?
If the braking distance is 2000m and the acceleration is a, then 2As = 0-20 ^ 2, a = -0.1m/s ^ 2
The acceleration is calculated from here. The meeting problem is the following idea:
They move in the same direction. As long as the speed is the same, they can not hit. At this time, the displacement difference of their respective movements is 600m
Then, the displacement is calculated by using their respective conditions to judge whether there is a collision
A freight train is running on the railway at the speed of 28.8km/h. Due to the dispatching accident, an express train is running at the speed of 72km / h at 700m behind the freight train. When the express driver finds out, he immediately closes the brake, but the express train has to slide 2000m to stop, then: (1) Try to judge whether the two cars will collide and write the analysis process (2) If there is no collision, find the distance between the two vehicles at the nearest time. If there is a collision, find how long it takes for the express to collide with the truck after braking?
(1) After braking, the fast train moves in a straight line with uniform deceleration. The initial speed is V0 = 72km / h = 20m / s, the final speed is v = 0, and the displacement is x = 2000m
Then from v2-v02 = 2aX:
a=v2−v
two
0
2x=0−202
two × 2000m / S2 = -0.1m/s2, so the acceleration of fast train braking is 0.1m/s2
The speed of freight car v = 28.8km/h = 8m / s
When the speed of the express train is reduced to be equal to the speed of the freight train, set the time for movement as t,
Then t = V cargo − v0
a=8−20
−0.1s=120s
During this period, the truck displacement x cargo = V cargo t = 8 × 120m = 960m, displacement of express x fast = V0 + V fast
2t=1680m
Because x fast > x goods + 700,
So the two cars will collide
(2) If the fast train collides with the truck after braking for time t ', there are:
X fast t '+ 1
2at ′ 2 = 700 + X cargo t ′
The substitution solution shows that t ′ 1 = 100s, t ′ 2 = 140s
It can be seen from the above question that t ′ 2 is unreasonably omitted, so the express collided with the truck after braking for 100s
A: (1) the two cars will collide. (2) the express car will collide with the truck after braking for 100s
A freight train runs on a straight railway at the speed of 28.8km/h. Due to a dispatching accident, an express train runs in the same direction at the speed of 72km / h at 700m behind. When the express driver finds out, he immediately closes the brake, but the express train has to slide 2000m to stop. Try to judge whether the two vehicles will collide through calculation
The speed of the freight train V1 = 28.8km/h = 8m / s, and the original speed of the express train V2 = 72km / S = 20m / s. according to the question, the express train has to slide for 2000m to stop. According to the law of uniform speed change movement: 0-v22 = 2As: the acceleration of the express is a = − v222s = - 2022 × 2000m/s2=-0.1m/s2； Get: a = -0.1m/s2 set express
Five trains stop on five parallel tracks at a station. If express train a cannot stop on track 3 and freight train B cannot stop on track 1, the five trains will stop Methods A,78 B,72 C,120 D,96
78 species
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