Math problems our final paper for the first year of junior high school .

If car B is regarded as stationary, then when car A chases car B, the speed of car A relative to car B is v A - v B. From the front of car A to the tail of car B to the tail of car A leaving the head of car B, the distance between the head of car A and the head of car B (regarded as stationary) changes by 180 + 120 = 300 meters (the drawing is very clear).

Let v A be x, v B be y, 300 (x-y) = 3*60 in the same way, when two cars are moving in the opposite direction, car A relative to car B speed v A + v B.

From the front of the car to the rear of the car, the distance between the front of car A and the head of car B changes by 180 + 120 = 300 meters.

300/(x+y)=20

Solving the binary system of linear equations yields x=25 3, y=20 3

Analysis, first of all, when A and B go in the same direction, within three minutes from the front of car A to the tail of car B to the tail of car A leaving the front of car B, the condition contained is the difference between the speed of car A and B, assuming that B does not move, then the speed of A leaving B is the difference between the speed of A and B, and the length of car A is more than the length of car B during this period is the length of car A and the length of car B, that is, 300 meters, and the speed difference between car A and B can be found to be 100 meters per minute within 3 minutes. Then the speed of car A passing car B is the sum of the speed of car A and car B, and the length of car A through car B during this time is also the length of car A plus the length of car B, and the sum of the speed of car A and car B is 900 meters per minute, then the speed of A is 500 meters per minute, and the speed of B is 400 meters per minute.

Solve the problem, and use the equation to explain it without such trouble.

Set the speed of the armor to x m min and the velocity to y m min.

According to the title, got.

3x-3y=180+120

1/3）x+(1/3)y=180=120

The value of the solution is the same.

I don't know if you have any binary equations.

In fact, this is a way to use mathematics after learning physics in junior high school.

Relative motion, which is analyzed by setting a stationary reference, cancels out the same part of motion.

The embodiment of this problem is speed cancellation.

I don't know if I understand it or not, maybe there's a better way.

Good luck with your studies.

Set the speed of A x and the speed of B = 180 seconds.

Car A chases car B, from the front of car A to the tail of car B to the tail of car A and leaves the front of car B. It can be known that car A travels 300 meters more than car B in 3 minutes. 180x-180y=300

Car A chases car B, from the front of car A to the tail of car B to the tail of car A and leaves the front of car B. It can be seen that the two cars travel together for 300 meters. 20x+20y=300

The final score is the score, and I don't know if it's correct.

Let's assume that the speed of A is x and the speed of B is y, when car A chases car B, the distance traveled by car A minus the distance traveled by car B is 180 + 120, which takes a total of 3 minutes = 180 seconds, and the equation is 180x-180y = 180 + 120

When A and B are traveling in the opposite direction, the distance traveled by car A plus the distance traveled by car B is 180 + 120 for 20 seconds, and the equation is 180 + 120 = 20 (x + y).

Solution: x=40 3

y=35/3

v A - v B) * 180 seconds = 300 meters (total length of two vehicles); (v A + v B) * 20 seconds = 180 meters; Simultaneous solving.

vA: 32 6 m sec=; v B: 22 6 m sec=

Let the speeds of the two cars be x and y respectively, and the equation is trained.

x-y)*3*60=180+120

x+y)*20=180+120

Let the velocity of A be x and the velocity of B y

x-y)*180=300

x+y)*20=300

The solution is x=25 3 m s.

y=20 3 m sec.

The speed of A is 25 3, and the speed of B is 20 3

Provide specific final examination papers for the first volume of mathematics in the first year of junior high school.

I don't know if the children's shoes are learning the Sujiao version or something else, say I'll pass it on to you.

It's best to ask for a teacher, that's the most representative.

I seem to have it, but SpongeBob won't tell anyone.