A travel question, a travel question, to answer the process

A arrives first. A runs half the time and walks half the time, so A runs half the time more than half the time.

Because two people run the same distance, more than half of A's distance is spent in the run, and because the two people run at the same speed, A takes less time, so A arrives first.

Running is supposed to save time, but A spends half the time running, which must be short. B, on the other hand, spends most of his time walking. So A comes first.

The solution process is as follows:

Since there is only one answer, let's assume that:

Assuming that the distance is 2km, the speed of walking is 1km, and the distance of A is a, it is certain that a>1.

Then: the time for B to complete the journey = 1 1 + 1 hours.

Time for A to complete the course = a 1 + (2-a).

A is the distance that A runs, so it must be greater than 1, so 4-A<3 so A arrives first.

According to the question Serie A travel time is a definite value, set to t

Suppose the speed of running is p, the speed of walking is z, and the total distance s=p*t 2+z*t 2 assumes that B's travel time is x

then x=(s 2) p + (s 2) z

t=2s/(p+z)

x=s/(2p)+s/(2z)

To determine whether t-x is greater than or less than or equal to 0, t-x=2s-(pz+zs) 2p-(ps+zs) 2z--this is the p+z>0 obtained after (t-x)*(p+z).

Multiply the above equation by 2pz(p>0, z>0, equal to 0, you don't need to count, ->2pz>0)>2pzs-z 2s-p 2s

s(2pz-z^2-p^2)

s((p-z)^2)

When s > 0, t-x<=0

So A is fast, and p-z=0 is just as fast.

A arrives first. A must run for half an hour than walk for half an hour, so compared with A and B, A can run a longer distance, and of course it will save time.

Let the running speed be x and y distance 2

So B comes first.

Actually, the title is:

Catch up on the problem. Therefore, the formula used is: chasing time = chasing distance

Difference in speed.

Your equation is suitable for encounters, that is, they move in opposite directions from different places.

But this topic is a question of pursuit.

Subtraction is to find out how many meters Zhang walked more than Zhao per minute.

Their encounters are no longer what you think they are.

Instead, Zhang led Zhao to see Zhao in a circle.

Hope you understand.

It's a bit hard to say. You can just run around and see for yourself.

I won't write about the analysis process of this question, because you are an expert in Olympiad mathematics:

Quite simply, it is to use the basic formula: distance = speed The distance of time is certain, and the speed is inversely proportional to time.

Planned speed = 60 1 = 60 km h.

Scheduled time = 2 (5-3) 3+1 = 4 hours.

So the whole journey is 60 4 = 240 kilometers.

Have fun.

Assuming that the whole journey is x kilometers and the planned speed is v kilometers per hour, then 1 + x - 1 v) (3v 5) = x v + 2x = 4v

x + 60)÷v = x/v + 1

v = 60

x = 240

A: The planned speed is 60 km/h, and the whole journey is 240 km/h.

If the distance traveled at the planned speed is increased by 60 km, the time to reach the destination is only 1 hour later than the planned time, and it can be known that 60 meters more, and if you travel for 1 hour, you can get the speed v=60 1=60kmh, and the distance can be set by the previous conditions s s 60=1+(s-60) 36--2 s=240

Let me be fair, don't lie anymore, you are an expert in primary school Olympiad, and you have Olympiad questions every day, what do you mean? Stand up and explain.

The distance traveled by train in 1 minute is 1

Each tram meets an oncoming tram every 6 minutes, indicating that the distance between the two trains is 12.

Zheng Zheng encounters an oncoming tram every 5 minutes, indicating that the distance of 5 minutes of cycling = tram travel (12-5) = 7

In the same way, Hao Hao's distance of riding for 6 minutes = (12-6) = 6, then Zheng Zheng's speed is 7 5, Xiao Wang is 6 6 = 1, and the total distance is 56

So when Zheng Zheng and Hao Hao met on the way, they had walked for 56 (7 5+1) = 70 3 minutes.

Have fun. If you don't understand, please ask!

From the inscription, it can be seen that Wu Wu's speed is the same as that of the tram. It only requires Zheng Zheng and the tram to travel from the two places at the same time, and it takes a long time to meet.

Assuming that Zheng Zheng and the tram start at the same time, the distance from the tram on the opposite side is 6*2 56=3 14

The time it takes for Zheng Zheng and the tram to meet each other from A and B at the same time is 5 (3 14) = 70 3 (minutes).

The arithmetic synthesis is: 5 (6*2 56)=70 3 (minutes) The problem can be solved in three steps.

The first sentence of the question should say that two people are ...... at the same time

The time for the dog to run is equal to the time it takes for Honghong and Lanlan to meet

36 (4+5) = 4 hours.

The dog ran: 8 4 = 32 km.

Blue butterflies fly lightly, hello:

Analysis: The time when the two meet is the time of the dog's walk.

The time of the encounter is:

36 (4+5) = 4 hours.

The hounds walked together:

4 8 32 (km).

The time when the dog walks is the time when Honghong and Lan Lan meet.

Encounter time 36 9=4

The dog walked 32 kilometers.

The shared time from the turtle to the finish line is 1500 25 = 60 minutes, and the distance of the rabbit run is 1500-200 = 1300 meters.

The time for the rabbit to run is 1300 325 = 4 minutes.

The rabbit sleeps for 60-4 = 56 minutes.

The speed at which the car returns is 1+ times the speed of the large truck.

The section of the road where the large truck travels in 2 hours, and the car returns in 2 2=1 hours.

So it takes 2+1=3 hours to make a round trip by car.

Method 1, the car returns to the 1 2 distance to the midpoint of A and B, if it is still driving at the original speed, it can only go 1 2 (1 + 50) = 1 3, so when it goes, the speed ratio of the car and the large truck (1 + 1 3): 1 = 4: 3

When the car returns and the speed ratio of the large truck is 4 (1 + 50): 3 = 2:1 method two, if the car also speeds up when it goes, then it will go to the whole journey of 1 (1 + 50 ) = 3 2, so the speed ratio of the car when returning to the large truck is (3 2 + 1 2): 1 = 2: 1 so the section of the road where the big truck goes for 2 hours, the car returns 2 2 = 1 hours, so the car needs 2 + 1 = 3 hours to go back and forth.