Application Question: A and B go from A to B

In the first third of the journey, A and B both travel at 5 kilometers per hour, which takes the same time; In the second third of the journey, A and B both travel at 4 km/h in the same time. In the middle third of the way, A travels kilometers per hour. B travels at an average speed of 40 9 km/h in the first half of the journey and 4 km/h in the second half of the third of the journey.

The speed ratio of A and B is: 40 9 = 81:80, and the time ratio is 80:

81, the difference is 1 part, exactly 30 seconds, that is, the time taken by this section of the journey (one-third of the whole journey) is: 30 * 80 = 2400 seconds = 2 3 hours, the distance is a kilometer, and the whole journey is 9 kilometers.

I've already done it for you!

Total distance x x (3*5)+x (

x = 9 km.

Let the walking speed be V, the motorcycle speed is 6V, the distance from A to B is S, and the walking time of C is T (that is, the time from the departure of 3 people to A and return to meet C), and in the change time, the distance traveled by C is Vt, and the distance traveled by A is 6vt

The sum of the distances of the two is exactly 2 times (2s) of the distance from A to B, and there is vt + 6 vt = 2s from which vt 2s = 2 7

Tu Zai, give points

The velocity of the first half of the time is a, and the velocity of the second half of the time is b; The speed of the first half of the journey is A, and the speed of the second half of the journey is B, and the two of them go side by side and arrive at the same time.

The velocity of the first half of the time is a, and the velocity of the second half of the time is b;

Suppose the total time is 2t total distance = at+bt

The speed of the first half of the journey is a, the speed of the second half of the distance is b (a is not equal to b), and the total time = ( at+bt ) a + at+bt ) b = (abt+b t + a t+abt) ab

2abt+b²t + a²t ) /ab=t( 2ab+b²+ a² )/ab=t( a+b ) /ab

Because ( A+B ) 2ab so A comes first.

This problem should be a junior high school question, and if the distance between A and C is x, then there is 160 (x-10)-180 x=10 solution: x=18

That is, AC = 18 km, BC = 8 km, the closest distance of AB is 18-8 = 10 km, and the farthest is 18 + 8 = 26 km.

then 10 d 26

The distance between A and C is S1

B to C S2

A speed v1, B speed v2

1/v1-1/v2=-10

s1=v1*3*60

s1-s2=10

s2=v2*(2*60+40)

v1=17/18,s2=160

Finally, s1=170

AB is the smallest on the C side at 10

The largest on both sides is 330

i.e. 10=

If the ac distance is x km, then the bc distance is (x-10) km.

Let the speed of A be y km/min, then B's velocity is 1 (1 y +10) km/min.

Columnable Equations:

x=x-10=（1/（1/y +10））.160: x 2-8x-180 = 0

x1 = 18, x2 = -10 (rounded) i.e., the distance between a and c is 18 km.

2) BC is (X-10) = 8 km, AB is the closest distance of 18-8 = 10 km (A and B are driving in the same direction), and the farthest is 18 + 8 = 26 km (A and B are driving in the opposite direction), that is: 10

If the distance between the two places is x, A has gone all, it is 1, and it will take four hours for B to complete the journey, so B only walked three-quarters of the whole journey, that is, the two of them walked a total of times the whole journey.

then x=20, so the distance between the two places is 40 km.

The ratio of A and B speed is 4:3

A walked 4*5 20 kilometers to talk about the mountain, that is, the two places of the road group to serve the year.

If you set the distance between ab and b x then in the first case, A takes x 80 and B takes x 60, so A arrives (x 60 - x 80) minutes earlier than B.

In the second case, A takes (x 80 + 7) and B takes x 60, so B arrives (x 80 + 7 - x 60) minutes earlier than A;

If the two cases are said to arrive at the same time earlier, then there is the equation: x 60 - x 80 = x 80 + 7 - x 60, and the solution is x = 840 (meters).

The answer is: A and B are 10 kilometers apart.

This question does not seem to tell you the distance, speed, and time, only 6 kilometers and 8 kilometers, but in fact, you can first determine that the speed of A and B is different, and at the same time, you can also know that the time used by A and B when they meet twice is the same, and the question can be calculated according to this point alone, and the specific method is as follows:

First of all, the town is set up to mark the distance between the two places x kilometers, the speed of A is Sakura V A, and the speed of B is V B.

1. According to the fact that A and B are 6 kilometers away from place B when they meet for the first time, it is obtained: (x-6) v A = 6 v B, and (x-6) 6 = v A v B;

2. According to the second encounter between A and B 8 kilometers away from place B, it is obtained: (2x-8) v A = (x+8) v B, and (2x-8) (x+8) = v A v B;

3. Because v A v B = v A v B, then the two formulas derived from the above can be deduced, (x-6) 6=(2x-8) (x+8), and this formula is calculated, and the final royal sum is obtained x=10 kilometers.

Just to throw bricks and stones, maybe there will be a better way.

This question should be a junior high school question, let the distance between A and Royal Type C be changed to X, then there is 160 (X-10)-180 X=10 solution: X=18

That is, AC = 18 km, Zhen Anchao BC = 8 km, the closest distance of AB is 18-8 = 10 km, and the farthest is 18 + 8 = 26 km.

then 10 d 26

It takes T minutes to complete the journey, and AB is 80t meters apart, when the two meet.

B leaves A and walks T+30+15 minutes.

A left B and walked for 15 minutes.

The combined distance is the whole journey.

60 * (t + 30 + 15) + 80 * 15 = 80tt = 195 minutes 80t = 15600 meters.

Half an hour = 30 minutes.

When A arrives at place B, B is still 60 * (30 + 15) + 80 * 15 = 3900 meters away from place B.

The time taken by A to B is 3900 (80-60) = 195 minutes.

Therefore, AB is 80*195=15,600 meters apart.

Let the distance of a b be a and the distance of b c be b

The time of A walking = a 15 + b 4, the time of B walking = a let dig 5 + b 15, the two times are equal, so.

The whole process is a+8a 11=19 code family 11a

Divide the distance by time to get the average velocity, and substitut the time of A, and get the average velocity slip disadvantage of A=(19 11a) (a 15+b 4).

Solve v=kilometers of hours.

Distance difference = 8 km.

Speed difference = 10-6 = 4 km/h.

Travel time = 8 4 = 2 hours.

A walked for 2 hours.

The distance between the two Ming widths is 10x2 = 20 thousand pounds of bright rice.