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Car a and car B start from two places 120 km apart in the same direction at the same time. Car B is in front of car a. the speed of car a and car B is 60 km / h and 40 km / h respectively. After a few hours, the speed of car a and car B is 60 km / h and 40 km / h respectively A can catch up with B Solving linear equation with one variable
Let a catch up with B after X hours. That is to say, if a catches up with B, then their distances are equal
40X+120=60X
The solution is x = 6
AB starts from a and B in the same direction with a distance of 40km, (a is behind b) a starts from B 1H later. As a result, it takes 3H and B to reach the destination at the same time. It is known that the speed of a is 3 / 2 of that of B, so the speed of two vehicles can be calculated
If the speed of B is x km / h, then the speed of a is 3 / 2 x km / h
4X + 40 = 3 * (3 / 2) x ("*" means "multiply")
The speed of B is x = 80 km / h
The speed of a is (3 / 2) x = 120 km / h
The distance between a and B is 180 km. The speed of a car is 60 km / h, and that of a truck is 40 km / h
I took the entrance examination immediately. I used to do this kind of questions, but I forgot later. I didn't get any points. Sorry
If the car starts from place a and the truck starts from place B at the same time“
First find the time of meeting
60t+40t=180 => t=1.8
Then calculate the driving distance s of the car when it meets
s=60*1.8=108
(you can also calculate the distance of a large truck)
Then the meeting place is 108 kilometers away from Jiadi
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