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The normal running speed of the train from station a to station B is 60km / h. There is a train leaving from station A. due to the delay of 10s, the driver raised the speed The normal running speed of the train from station a to station B is 60km / h. There is a train leaving from station A. due to the delay of 10s, the driver increased the speed to 72km / h and just arrived at station B on time. Calculate the distance between station a and station B and the normal running time
Normal driving time x
72（x-10/3600)=60x
x=1/60
sixty × 1/60=1
The distance between a and B is 1km and the normal driving time is 1 / 60H
The normal running speed of the train from station a to station B is 60km / h. once the train started from station a 3 min late, and the driver increased the speed to 75KM / h, Finally arrive at station B on time. Ask for the distance between stations a and B. what is the normal speed?
It takes X minutes from station a to station B
60X=75(X-3）
X=15
Then the distance between Party A and Party B is 60 * (15 / 60) = 15
The distance between the two stations is 15km
The normal running speed of the train from station a to station B is 60km / h. once it was delayed for 20 minutes for some reason. In order to arrive at station B on time, the train driver needs to increase the speed To 72km / h (1) : distance between stations a and B (2) : the normal travel time of the train from station a to station B. for physical problems, don't use X and y, thank you
Hello: time ratio 72:60 = 6:520min = 1 / 3 hour, normal driving time 1 / 3 ÷ (6-5) X6 = 2 hours, distance between two stations 60x2 = 120 km. If you don't understand this question, you can ask. If you are satisfied, please click "adopt as satisfactory answer" in the lower right corner. If you have other questions, please adopt this question and send it separately and click me
The normal running speed from station a to station B is 60km / h. When a train leaves from station a, it leaves 5min late for some reason The normal speed of the train from station a to station B is 60km / h. When a train starts from station a, it leaves 5min late for some reason. In order not to be late, the driver increases the speed of the train to 72km / h just to arrive. Find the distance between Party A and Party B
If the first time is a minute, the second time (a-5) is used
60a=(a-5)72
A = 30, i.e. 1 / 2 hour
Distance = speed x time
＝60x1/2
＝30（㎞）
Give me some points. I gave it three times
The speed of the train from a to B is 60km / h. The train left 5 minutes late. The driver increased the speed to 72km / h. as a result, 1 arrived at B on time. What is the distance between a and B fast [write the calculation process]
5/12km.
Suppose both trains a and B go from a to B
The speed of a is 60km / h and that of B is 72km / h
If a drives five minutes earlier, it is 60 ahead of B × (5÷60)=5km
B drives 72-60 more per hour than a = 12km
If they are to arrive at the same time, the distance is 5 / 12km
The train is 60 percent faster than the car, and the car is several percent slower than the train
If the steam speed is V, the train speed is V (1 + 60%)
Therefore, the car speed is slower than the train = V (1 + 60%) - V / V (1 + 60%) = 6 / 16 = 37.5%
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