solving equations by algebratic method請列式

distance / time = speed
--> time = distance / speed

1km/h = 1000 / 60 (m/min.) = 50 / 3 (m/min.)

Let v (m/min.) be the rowing speed in still water

[400 / (v – 50/3)] – [400 / (v + 50/3)] = 6

(Note: 400 / (v – 50/3) = upstream time,
400 / (v + 50/3) = downstream time)

((120000) / ((3v - 50)(3v + 50))) = 6
9v^2 – 22500 = 0
v^2 = 2500
v = 50 or v = -50(rejected)

Therefore his rowing speed in still water is 50m/min. or 3km/h.

let t hr be the time taken for the downstream journey
so upstream journey takes (t + 0.1) hr

let v km/hr be the relative speed of the rowing speed at still water
so the actual speed of upsteam journey is (v - 1) km/hr
and that of downstream journey is (v + 1) km/hr

now, both journeys are 400m long.

(v - 1) x (t + 0.1) = 400m = (v + 1) x t

iff

t = 400 / (v + 1)

and

v = sqrt (8001) km/hr

2007-05-19 11:43:07 補充：
sorry,it should be (v - 1) x (t + 0.1) = 0.4m = (v + 1) x tifft = 2 / [5 x (v + 1)]and (v - 3) x (v + 3) = 0however, v > 0As a result, v = 3 km/hr