The two piers are 150 km apart

<> explanation: (1) According to the image, the distance of ship A sailing down the river for 6 hours is 150 kilometers, so the speed of sailing along the river is 1506=25 (kilometers per hour).

Boat B travels 150 kilometers in 10 hours, so the speed of sailing against the current is 15010 = 15 (kilometers hour) (2 points).

Since the velocity of the two passenger ships in the still water is the same, and the speed of the current is not the same, so when the speed of the passenger ship in the still water is v1 km and the speed of the current is v2 km, the equation system of the column gives v1+v2=25v1-v2=15

The solution is v1 = 20 v2 = 5 (4 points).

Answer: The speed of the passenger ship in still water is 20 km, and the current speed is 5 km (5 minutes).

2) According to the title, the speed of the freighter sailing downstream is 10 + 5 = 15 (kilometers per hour).

It is also known that the freighter departs two hours in advance, so the image is (0,30) and (8,150) two points, and the image is as shown in the line segment de (6 points) in the figure on the right

Let the analytic formula of DE be y=k1x+b1

So 30 = b1150 = 8k1 + b1, and the solution is k1 = 15b1 = 30

So the analytic formula for de is y=15x+30 (7 points).

Let the analytic formula of BC be y=k2x+b2

So 150 = b20 = 10k2 + b2, and the solution is k2 = -15b2 = 150

So the analytic formula for BC is y=-15x+150 (8 points).

Solving the system of equations y=15x+30y=-15x+150 gives x=4y=90 (9 points).

A: The distance between the freighter and passenger ship B at Pier A is 90 km (10 minutes).

Let the water flow velocity v:

26-v）* =（26 + v）* 3

Solution v = 2 m?

Alpha. alpha

Find the speed of boats A and B in still water.

Let the speed of A be x km-h, and the speed of B be ykm-h, and the system of equations, 1)4 (x+y)=200

2）20×（x-y）=200

Solving the equation system shows that x = 30 km h and y = 20 km h, and the speed of ships A and B in the old still water section is 30 km h and 20 km h respectively.

According to the two ships, A and B set sail from the two piers of AB at the same time and met 4 hours later. It can be seen that the speed of the two ships is as follows

200 4 50 (kmh).

Then according to the two ships of A and B at the same time, they set sail from the AB pier in the same direction, and A catches up with B after 20 hours, and the speed difference between A and B can be seen as:

200 20 10 (kmh).

B's velocity base roll is as follows:

20 (kmh).

The velocity of A is:

50-20 30 (kmh).

Let the velocity of ship A in still water be v1, the velocity of ship B in still water is v2, and the velocity of the current is v0

According to the title: 3 (v1 + v0 + v2 - v0) = 90 v1 + v2 = 30

v1-v2)×15=90

v1-v2=6

Solution: v1 = 18, v2 = 12

Answer: The speed of the two vessels in still water is 18 km/h and 12 km/h.

90 3=30 (kmh).

90 15 = 6 (kmh).

Ship A speed: (30+6) 2=18 (kmh) B's speed: (30-6) 2=12 (kmh).

Let the velocities be x, y, and the water velocity v, and let the water flow from a to bx+v+y-v]x3=90

x+v-(y+v)]x15=90

x=18, y=12

Let the time for ship A to leave port A and turn around to chase the item overboard is t1, and the time from the beginning of the pursuit of the item to the time of chasing the item is t2, and the speed of the water is v.

6V*T2=5Vt1+ Vt1 Therefore, T1=T2, and because of the jujube 5Vt1+60km=7Vt2+(T1+T2)*5V, it is 12vt=60 km and vt=5 km.

The cover is dismantled with S A = 5VT1 = 25 km.

Let the water speed be high, then the speed of the ship is 6v, and the running time is t The distance traveled by B + the distance traveled by the potato = 5vt + vt = 60 The distance of vt = 10 and B is 5vt = 50, and the distance of A and B is the same, so it is also 50

The conditions are not enough, is it a mistake?

Topic: Ships A and B depart at the same time from piers A and B, which are 150 kilometers away, and go back and forth between A and B. Upstream. The speed of the two ships in still water is 20 kilometers per hour, and the speed of the current is 5 kilometers per hour.

So how many kilometers away from A and B are the places where the two ships A and B meet for the second time?

Solution: According to the title, after departure, it takes 150+ (20+5) = 6 (hours) for ship A to arrive at pier B, and it takes 150+ (20-5) = 10 (hours) for ship B to arrive at pier A, so when ship B arrives at pier A, ship A is 150- (10-6) x (20-5) = 90 (km) from pier A, and if it takes x hours for ship B to return from pier A and meet ship A, then 25x+ 15x= 90, and the solution is x 9 4

Therefore, the place where the two ships A and B meet for the second time is from pier A.

9 4x (20+5) = (km).

It takes 150 25 6 hours to walk 150 kilometers down the river.

It takes 150 15 10 hours to walk 150 kilometers against the water.

The difference is 10 6 4 hours, and the distance traveled against the water in 4 hours is 4 15 60 km.

There are 150 60 90 km left.

At a distance of 90 km, the two ships began to move opposite each other.

Duration of the encounter is 90 (15 25) hours.

Then the hour is the driving distance along the water.

km) second approach.

Let the second encounter be at a distance of X kilometers from point A.

10＋x／25＝6＋（150－x）／15

Solution: x km).

Hello: Go time 120 30 = 4 hours.

Time to return 120 20 = 6 hours.

Average per hour 120x2 (4+6)=24 km, if you don't understand anything about this question, you can ask, if you are satisfied, remember to adopt, if there are other questions, please adopt this question, click to me for help, it is not easy to answer the question, please understand, thank you.

Good luck with your studies!

The two wharves are 120 kilometers apart, and the average distance of the ship from A to B is 30 kilometers per hour, and the average travel from B to A is 20 kilometers per hour. Find out how many kilometers per hour a boat travels on average?

Solution, get: The time when you go is: 120 30 == 4 hours.

The time of return is: 120 20 == 6 hours.

So the average round-trip speed is: 120*2 (4+6)==240 10==24 km/h.

120\30=4

6+4=10 hours.

120 10 = 12 km hours.

Because it is a round trip, it takes 12x2=24 km hours.

Xiao Xinyue, hello:

When the passenger ship traveled the whole 3 7, the freighter traveled the whole way:

The distance between the two piers of A and B:

36 3 10 120 (km).

At 1 7 o'clock in the passenger ferry's journey, the freighter traveled 36 3 = 12 km.

When the passenger ship had completed its journey, the freighter had traveled (12 7) kilometers, and another freighter had traveled 7 10 kilometers at this time

Then, the distance between the two piers is 12 7 7 10 = 120 kilometers.

Analysis: When the passenger ship had completed its journey, the freighter had traveled 36 (1 3 7) kilometers, equivalent to seven-tenths of the entire journey. Seven-tenths of a known number is 36 (1 3 7), find this number and divide it.

Solution: 36 (1 3 7) 7 10

120 (km).

A: The two piers are 120 kilometers apart.

Solution: If the speed of the passenger ship is v1, the speed of the cargo ship is (7 10) v1, and the distance between A and B is x

Then there is: (3 7) x v1 = 36 (7 10) v1 solution x = 120

A: The two piers are 120 kilometers apart.

36 divided by three-sevenths gives 84 84 divided by seven-tenths gives 120km

Set the speed of A to slip early and dress up for X, and the speed of B for Y

It can be known from the title.

3*（x+3）+3(y-3)=240

24*(x-y)=240---A chases B from A and travels 230 kilometers in 24 hours, regardless of the water speed.

The solution is x=45 y=35

240 45 = hours.

Let the ship speed be the focal slow tung mold x, and the water speed be y, and the equation is calculated according to the conditions:

x+y)*10=1000

x-y)*12=1000

Solve the equation with x=275 3, y=25 3

Untie; Let the velocity of the ship in the still water be x and the current velocity be y

10（x+y)=1000。。。Grinding Min carrying. （1）12 (x-y) =1000。。。2) Blind ambush.

Because (1) = (2) Na Chun.

10x+10y=12x-12y

x=y=

a+y).10=1000

a-y).12=1000

Transform into one and two.

a+y=100

a-y=1000 is calling 12

One type plus two traces of the reed style.

2a=2200/12

a=275 Juzhou Kai3

Substitute a into the formula.

y=25/3

a=275/3

y=25/3