If two cars are from A and B at the same time, A and B are respectively from A and B and two cities

The speed of the passenger car is 3x, and the speed of the truck is 2x.

3*（3x-2x）＝60

3x＝60x＝20

Passenger car speed: 20*3 60 (km).

The distance between cities A and B: 60*3*2 360 (km)Answer: The distance between cities A and B is 360km

Let the distance between the two cities of A and B be x, then the distance of the two cars is x 2, and the speed of the passenger car is x 2 3 = x 6

The speed of the wagon is (x 2-60) 3

Since the speed ratio of passenger cars to trucks is 3:2, then the equation is:

x 6) ((x 2-60) 3)=3 2 solution x=360

Let the velocity of A and B be 32 respectively, then 3v=2v+60 (starting from the midpoint), v=60 and the distance s=3*60*2=360

Let the distance between cars A and B be x

Then (x 2) 3:(x 2-60) 3=3:2

Solution is available.

Passenger car speed 3x Truck 2x

3（3x-2x）＝60

x 20 bus speed 60

Let the speed of A and B be 3V and 2V respectively

Time t = 3 hours.

3v*t - 2v*t=60

vt=60s=3v*t+2v*t+60

s=360km

Solution: Let B travel kilometers per hour, then A travels +4 kilometers per hour and the solution of 2(6 -16)=(6(+4)) is obtained by the equation of the problem column =18

18×6-16）×2=184(km)

A: Cities A and B are 184 kilometers apart.

The distance between the two cities is x km, and the speed of car A is y km h

6y=6(y-4)=

Solve the equation to find x=160

So the two cities of ABL are 160 kilometers apart.

Let the distance between the two cities be x, the speed of car A is a, and the speed of car B is b. Then the column equation 6a = 3 x 4, 6b = 16 + x 2, a-b = 4

It is known that the distance between the two places is 160 kilometers.

Let B's velocity be x

kmh, then A's velocity is (x+4) km/h.

6(x+4)÷

x = 16 (kilometer hours).

x+4=20 (km-h).

The distance between A and B is 6(x+4) km).

That 24 meters is faster than the speed of Chi Xi Cavity B, so: 24 divided by 6 = 4 (this code can be known to be 4 kilometers faster than him per hour) Lu Feng.

Then directly: 56-4 = 52 kilometers (B per hour).

The first encounter was 52 kilometers away from Kojo.

In the total distance of the whole journey, the car from A travels 52 kilometers.

The second meeting point is 44 kilometers away from Kojo.

The limb excitation was three whole courses, and the one who set off from A walked 52 3 = 156 thousand lead Xun meters, and the whole journey was (156 + 44) 2 = 100 kilometers.

The third encounter, 5 full courses, the one who started from A walked 52 5 = 260 km, and the place of meeting was 260-100 2 = 60 km away from A.

The first socks meet four times, 7 full courses, A sets off 52 7 = 364 km, and the meeting place is 364-100 3 = 64 km.

First of all, you have to understand that the time required for the two encounters is the same, because they did not stop, and each time they met, the two cars traveled together equivalent to the distance between AB. Therefore, the distance traveled by car A is the same for the two encounters. The first time car A is 120 kilometers away from point B, and the second time is 100 kilometers away from point A, indicating that A is faster than B.

Assuming they are as fast, then the second meeting after meeting at a distance of 120 km from point B should be 120 km from point A. Now it is 100 kilometers, which means that car A has to share the distance of car B for every distance that is equivalent to AB, and car A has to share the distance of car B for 20 kilometers, and it can be concluded that the midpoint of AB should be 140 kilometers away from point A, so the total length is 280 kilometers.

The solution to the equation has been given upstairs, so I'll give the arithmetic solution.

120*3-100=260 km.

The two cities are located at a distance of x kilometers.

x-120：120=2x-100：x+100

x = 260 km.

Encounter time = 1 (1 4 + 1 6) = 1 5 12 = 12/5 (hour).

After reading it, I wish you progress in learning!

The velocity of A is equal to s divided by 4, and the velocity of B is equal to s divided by 6, and the sum of their velocities is 5/12s. So the time it takes for them to meet is divided by s by 5/12s, which is equal to hours.

Let the distance be s, then the velocity of A is s 4, the velocity of B is s 6, and the time is x when the two meet, then, s 4 * x + s 6 * x = s, and x= is calculated

Hello, it is known that car A travels 3 kilometers per hour more than car B.

After 4 hours of multiple trips, 3*4=12 km.

Car A traveled the whole 80

Car A exceeds 3 10 in the middle of the whole course

Car B is 13 km past the midpoint.

Car A exceeds the midpoint by 13 + 12 = 25 kilometers.

Full journey = 25 3 10 = 250 3 km.

B exceeds the midpoint by 13 kilometers, then A should exceed the midpoint by 13 + 4 * 3 = 25 meters.

25 meters is (80-50)% of the total distance

AB The distance between the two cities is 250 3 km.

Solution: Let A and B meet after driving for x hours.

60x=45x+30

If x=2 is solved, then the distance traveled by A is: 60 2=120 (km) when they meet, the distance between B and A is equal to the distance traveled by A.

B is 120 kilometres from place A.

A travels 60 kilometers per hour, and B travels 45 kilometers per hour.

Speed difference between A and B: 60-45 = 15 km/h.

When we meet, A travels 30 kilometers more than B.

Time taken at the time of encounter: 30 15 = 2 hours.

The distance between B and A at the time of the encounter is the distance traveled by A at the time of the encounter

60 2 = 120 km.

Integer: 30 (60-45) 60=120 km.

The speed of car A is 60, the speed of car B is 45, car A travels 15 kilometers more than car B, it takes two hours to drive 30 kilometers more, and the two cars travel for 2 hours to meet, and the distance from car B to A is the same as the distance from car A to A, that is, 60x2 = 120 kilometers.