A COLUMN IS 120M LONG, AND IT IS MOVING IN A UNIFORM LINEAR MOTION AT A CERTAIN SPEED BECAUSE OF THE

What is the only and the distance?

I only asked how many meters the correspondent had covered on this round trip.

Solution: Suppose the speed of the communicator is v1 and the speed of the team is v2. Then, it is a matter of catching up with the communicator from the back of the line to the front of the line, and returning from the head of the line to the back of the line is a matter of encounter.

Then we have this equation: 120 (v1-v2) +120 (v1+v2) = 288 v2; (equal time) to get v1 v2=3 2 or v1 v2= -3 2 (the ratio of velocities cannot be negative, rounded). So v1 v2=3 2 , in the same time, the ratio of the distance is equal to the ratio of the speed, and the communicator in this round trip is 288*3 2 = 432 meters.

Set up the troop speed v, the communicator speed v', the correspondent takes t from the end of the team to the head of the team, and the time from the end of the team to the head of the team is t, then there is.

t₁=120/(v'-v) (1) [Correspondent end-to-end time]t = 120 (v.)'+v) (2) [Correspondent from beginning to end time] v(t +t) = 288 (3) [distance advanced by the troops during this time] And the result we asked was.

v'(t₁+t₂)

So just ask for v'/v

Bring (1), (2) into (3).

Yes. v[120/(v'-v) +120/(v'+v)] =288 simplified. vv'/(v'²-v²)=

Divide the left fraction by v at the same time, and set k = v'v [Set this k for the convenience of writing, in fact, it is not the same].

Get. k/(k²-1)=

Solve k=v'v=3 2 (the other negative value is rounded off) so v'(t₁+t₂)=v(t₁+t₂)*v'v=288*3 2=432 (m) is the distance traveled by the correspondent.

The only one is 288 meters, and the distance is 288+120*2 meters!

Set: the speed of the communicator is v1, the speed of the team is v2, the time of the communicator from the end of the team to the head of the team is t1, and the time from the head of the team to the end of the team is t2. The team advanced 160m with a time t.

Using the team as a reference, the speed of the communicator from the end of the team to the head of the team is (v1-v2), and the speed from the head of the team to the end of the team is (v1+v2).

The time from the end of the team to the head of the team is t1=120 (v1-v2), and the time of the communication from the head to the end of the team is t2=120 (v1+v2).

The time it takes for the team to advance 160m t=160 v2t=t1+t2

160 v2=120 (v1-v2)+120 (v1+v2)The solution is: v1 v2=2

The distance traveled by the correspondent in time t.

s=v1t=v1(160/v2)

160(v1/v2)

320m

Solution: Let the speed of the team be x and the speed of the communicator be yThe first period is t, and the second time is t to chase the team, which can be regarded as a pursuit problem

y-x)t=100 can be seen as an encounter problem when returning to the team: (y+x)t=100 total time t+t=100 (y-x)+100 (y+x) Because the team advances 100 meters during this time, t+t=100 x is 100 (y-x)+100 (y+x)=100 x solves y=(2+1)x, and the ratio of the speed is 2+1, so the ratio of the distance is also 2+1, that is, the meter (" represents the root number).

Set the speed of the correspondent = x

Team Speed = y

The first period is t, and the second period is t

y*(t+t)=100

x-y)*t=100

x+y)*t=100

Three equations, four unknowns. This one.

The displacement is 200 meters and the distance is 280 meters.

The answer is as follows: This question should be solved by drawing a picture.

Tail A – Head.

Tail – Head c

1…40m…1

1…Mengpei ......200m……1 tail b - head.

The distance between the correspondent and tail A and tail B is only 200 m, and the distance from tail A to head C back to tail B is 280 m.

The distance is from the end of the team to the leader, t1=l (v2-v1)from the head to the end, t2=l (v2+v1).

Total time t=t1+t2=l (v2-v1)+l (v2+v1)=l*2v2 (v2+v1)(v2-v1).

Distance s=v2*t=2l*v2 2 (v2+v1)(v2-v1) displacement = displacement of the team = v1t = 2lv1v2 (v2+v1)(v2-v1).

Assuming that the team is stationary, the speed of the communication pawn is v2 (v2-v1), then t = 2l v2-v1 2l + (2l times v1 v2-v1).

Summary. Pro, now only see a queue lined up 120 meters long.

Dear, Happy New Year, I'm glad to help you with the lack of the first Kai debate study needs you to provide the topic, this Sun Zao can calculate for you, because a lot of conditions are on your **.

Pro, now only see a queue lined up 120 meters long.

Dear, okay, thank you for your cooperation, please be patient**The content is very rich and needs some time to see.

The end result is like this, it can't be calculated.

Pro, the problem of missing the return involves the problem of chasing the distance according to the relevant distance, and the calculation formula of the speed can be combined and calculated, and it is also a system of equations to be formed.

You can set the speed of the communicator to v1, the speed of the team's hidden guess v2, the time of the communicator from the end of the team to the head of the team is t1, and the time from the head of the team to the end of the team is t2, then the time taken by the team to hold the mold and the team to advance is t

Therefore, there is a time relationship related to Enlightenment, and Huaihui t1+t2 can be converted into 120 v2 120 (v1-v2)+120 (v1+v2) according to the distance formula

Solution v1 (1 + 2) v2

In this way, the distance traveled by the correspondent is calculated s1 v1t (1 + 2) v2t (1 + 2) s2 120 (1 + 2).

In general, distance is equal to speed multiplied by time.

Teacher, how did you calculate this answer? Can you tell me?

Pro, that is, according to the corresponding distance relationship, because the team is 120 meters away, the team is also 120

After (45) seconds to reach the end of the row; After the order was transmitted, the communications dispatcher was informed that it would take (45) seconds to immediately return to the head of the platoon from the tail of the platoon at the original speed; The correspondent ran a total of 585 meters.

Set: The speed of the communicator is V1, the speed of the team is V2, the time of the communication code Xiang Slag from the end of the team banquet to the head of the team is t1, and the time from the head of the team to the end of the team is T2. The team advanced 160m with a time t.

Using the team as a reference, the speed of the communicator from the end of the team to the head of the team is (v1-v2), and the speed from the head of the team to the end of the team is (v1+v2).

The lag time from the end of the team to the head of the team is t1=120 (v1-v2), and the time of the communication from the head to the end of the team is t2=120 (v1+v2).

The time it takes for the team to advance 160m is t=160 v2

t=t1+t2

160 v2=120 (v1-v2)+120 (v1+v2)The solution is: v1 v2=2

The distance traveled by the correspondent in time t.

s=v1t=v1(160/v2)

160(v1/v2)

320m

One column is 120 meters long. As the line marched, the communicator rushed from the end of the line to the front of the line. Then he immediately returned to the team.

If the team has advanced 288 meters during this time, and the speed of the team and the communicator remains the same, then what is the distance traveled by the communicator during this time?

Answer: It is best to rely on the relativity of motion and rest. During this time, the communicator advanced 288 meters with the team, and at the same time the communicator walked two more lengths than the team.

So during this time, the distance traveled by the correspondent was.

s=288m+120m*2=528m。