2題數學題~~~~

(1)
Let the speed of the ship be x km/h, and let the speed of the current be y km/h.
x+y=48/3---(1)
x-y=48/4---(2)
(1)+(2):
2x=28
x=14---(3)
Sub (3) into (1):
14+y=16
y=2
Therefore the speed of the current is 2 km/h.
(2)
Let the speed of A be x km/h, and let the speed of B be y km/h.
x=2y---(1)
x+y=30/2---(2)
Sub (1) into (2):
2y+y=15
y=5---(3)
Sub (3) into (1):
x=2(5)=10
Therefore the speed of A is 10 km/h, the speed of B is 5 km/h.
Remark:
The relative speed (without considering the theory of relativity) of two objects A and B motioning with the speed x and y is the difference of the magnitude of their vector of velocity.
If they are motioning in opposite direction, the relative speed is x+y.
If they are motioning in same direction, the relative speed is |x-y|.









圖片參考：http://xs513.xs.to/xs513/07090/fdklsaaaaaaaaaaaaa.png

Let vec. A denotes vector A.
The relative speed of vec. AB and vec. AC is
|vec. AC-vec. AB|=|vec. AC+vec. BA|=|vec. BC|=BC
We can apply cosine law:
BC²=AC²+AB²-2(AB)(AC)cos ∠BAC
BC²=x²+y²-2xy cos ∠BAC
BC=(x²+y²-2xy cos ∠BAC)sqrt.
The relative speed of x and y is BC=(x²+y²-2xy cos ∠BAC)sqrt.

1.The total distance the chip go =3+4=7 hours,
the total time used = 48*2 = 96 km
The speed of the current = 96/7 km/h

2.Let x be the speed of A and y be the speed of B
2y=x...(1)
2x-2y=30...(2)

(1)+(2)
2x=x+30
x=30

Sub. x=30 into (1)
2y=30
y=15

A's speed is 30km/h and B's speed is 15km/h