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A passenger and freight car and a freight train run on two parallel tracks respectively, with a passenger car of 150m and a train of 250m Weak 2 trains run towards each other and leave the supply and demand for 10s from the corner of the front to the rear. If the passenger and freight vehicles catch up with the freight train from the rear, it takes 100s from the front to the rear of the freight train to completely exceed the freight train, try to find the speed of 2 trains
Set the speed of passenger car x m / s and the speed of freight car Y M / s
（X+Y） × 10=150+250
（X-Y） × 100=150+250
X+Y=40
X-Y=4
The solution is x = 22, y = 18
A man walked along the railway at the speed of 60 meters per minute. A 144 meter long bus came from behind him and passed by him. It took 8 seconds to get the train Equations cannot be used
（144＋8*60／60）＝17m／s
A man walked along the railway at the speed of 60 meters per minute. A 144 meter long bus came opposite him and passed him. It took 8 seconds to find the speed of the train
(1) Solution 1:
Distance traveled by car in 8 seconds = body length - distance traveled by person in 8 seconds
（144-60÷60 × 8) ÷ 8 = 17 (M / s);
(2) Solution 2:
60 m / min = 1 m / s,
144 ÷ 8 = 18 (M / s),
18-1 = 17 (M / s);
A: the train speed is 17 meters per second
A man walked along the railway at the speed of 60 meters per minute. A 144 meter long bus came opposite him and passed him. It took 8 seconds to find the speed of the train
(1) Solution 1:
Distance traveled by car in 8 seconds = body length - distance traveled by person in 8 seconds
（144-60÷60 × 8) ÷ 8 = 17 (M / s);
(2) Solution 2:
60 m / min = 1 m / s,
144 ÷ 8 = 18 (M / s),
18-1 = 17 (M / s);
A: the train speed is 17 meters per second
One walked along the railway at the speed of 60m per minute, opposite a 144m long bus, passing by him in 8 seconds, What is the speed of the bus per minute?
144 * 60 / 8-60 = 1020m / min
A freight train was traveling on a flat and straight road at the speed of V1 = 28.8km/h. Due to dispatching error, a passenger train behind it came on the same road at the speed of V2 = 72km / h. at a distance of S0 = 600m, the driver of the passenger train immediately braked after discovering the freight train. What is the braking acceleration of the passenger train at least in order to prevent the two vehicles from colliding?
Take the running direction as the positive direction, and set the acceleration of the bus after braking as A2. From the above non collision conditions, v2t − 12a2t2 ≤ v1t + S0 ①； V2-a2t ≤ V1 ② when the braking acceleration takes the minimum value, the two inequalities can be changed into an equation. From equation ②, it is obtained that the speed of the passenger car is reduced to the time equal to the speed of the freight car
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