River is running at 2 kmph. if lakes a man twice as long to row up as to row down the river. the rate of the man in still water is:
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River is running at 2 km/h.
It look a man twice as long to row up as to row down the river.
the rate(in km/h)of the man in still water is?
:
let s = the rate in still water
then
(s+2) = effective speed down stream
and
(s-2) = effective speed up stream
:
let d = the one way distance
:
Write a time equation; time = dist/speed
time up = twice time down
![\frac{d}{(s - 2)} = 2( \frac{d}{s + 2}) \\ \frac{d }{(s - 2)} = \frac{2d}{(s + 2)} \frac{d}{(s - 2)} = 2( \frac{d}{s + 2}) \\ \frac{d }{(s - 2)} = \frac{2d}{(s + 2)}](https://tex.z-dn.net/?f=+%5Cfrac%7Bd%7D%7B%28s+-+2%29%7D++%3D+2%28+%5Cfrac%7Bd%7D%7Bs+%2B+2%7D%29++%5C%5C++%5Cfrac%7Bd+%7D%7B%28s+-+2%29%7D++%3D++%5Cfrac%7B2d%7D%7B%28s+%2B+2%29%7D+)
cross multiply
2d(s-2) = d(s+2)
Divide both sides by d
2(s-2) = s + 2
2s - 4 = s + 2
2s - s = 2 + 4
s = 6 km/hr his rate in still water
It look a man twice as long to row up as to row down the river.
the rate(in km/h)of the man in still water is?
:
let s = the rate in still water
then
(s+2) = effective speed down stream
and
(s-2) = effective speed up stream
:
let d = the one way distance
:
Write a time equation; time = dist/speed
time up = twice time down
cross multiply
2d(s-2) = d(s+2)
Divide both sides by d
2(s-2) = s + 2
2s - 4 = s + 2
2s - s = 2 + 4
s = 6 km/hr his rate in still water
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