The speed of the ship in still water is 24 km / h, and the current speed is 2 km / h. When the ship goes upstream from a wharf and then downstream, how far should it go to return to the wharf within 6 hours?

If the ship leaves at most x km, it should return to the dock within 6 hours. Then, there is: x24 − 2 + x24 + 2 = 6, and the solution is: x = 71.5. Therefore, the ship leaves at most 71.5 km, it should return to the dock within 6 hours

It takes 4 hours for a person to go downstream from a to B, and then upstream to C. It is known that the speed of the ship in the water is 7.5 km, and the current speed is 2.5 km. If the distance between a and C is 10 km, the distance between a and B can be calculated. Binary linear equation

Let AB distance be x km and BC distance be y km
x=y+10 …… ①
x/(7.5+2.5)+y/(7.5-2.5)=4…… ②
So, x + 2Y = 40 ③
Substituting ① into ③, y + 10 + 2Y = 40, y = 10
∴x=10+10=20,
That is, the distance AB is 20 kilometers

A. B is SKM away from the two places. Someone originally planned to start from a at the speed of vkm / h. After walking 1 / 3, he was informed to arrive in advance, so he increased the speed by 1km per hour
How early did he arrive at B?
If 3a-2b = 4, then 2 (2b-3a) square + 1
2x-y = 3, then 1-2x + y=

The original time T1 = s / V
Present time T2 = s / 3V + 2S / 3 (V + 1)
So the lead time △ t = T1-T2 = 2S / 3v-2s / 3 (V + 1) = 2S / 3 * [1 / V (V + 1)]
3a-2b = 4 then 2 (2b-3a) ^ 2 + 1 = 2 * 16 + 1 = 33
2x-y = 3, then 1-2x + y = 1 - (2x-y) = - 2

The distance between Party A and Party B is s km, and the speed of the car from Party A to Party B shall not exceed C km / h. We know the transportation cost of the car per hour (in Yuan)

The distance between a and B is s kilometers. The speed of a car from a to B should not exceed C kilometers per hour. It is known that the hourly freight cost of a car is composed of variable part and fixed part. The variable part is proportional to the square of speed V, and the proportion coefficient is B, and the fixed part is a yuan. (a ∠ BC ^ 2). In order to minimize the whole transportation cost, how fast should the car run
The freight cost per hour is a + bv & # 178; (V ≤ C);
The total transportation cost is (a + bv & # 178;) (s / V) = (as / V) + (BSV) ≥ 2S √ (AB);
When as / v = BSV, i.e. v = √ (A / b), the equal sign holds
Given a ＜ BC & # 178;, we can get: √ (A / b) ＜ C, satisfying V ≤ C
Therefore, in order to minimize the transportation cost, the car should drive at √ (A / b)

The speed of the ship in still water is 24 km / h, and the current speed is 2 km / h. When the ship goes upstream from a wharf and then downstream, how far should it go to return to the wharf within 6 hours?