Tom was floating down the river on a raft when 1km lower down, michael took to the water in a rowing boat.? - michael row esupplystore
Michael rowed on the river at its fastest pace. Then he turned and went back to its starting point as a tramp available from Tom. If the speed of Michael paddle in calm water ten times the speed of the flow in the river, was to what extent, is covered before Michael realized his boat around?
Thursday, February 11, 2010
Michael Row Esupplystore Tom Was Floating Down The River On A Raft When 1km Lower Down, Michael Took To The Water In A Rowing Boat.?
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2 comments:
Let x = distance traveled before Michael had their boat
around
m = Michael rowing speed in calm water
r = velocity of the river
Depending on the problem:
m = 10r
1 km = distance traveled by Tom meet Michael '
Starting Point
Therefore,
Michael Speed Down = 10R + r = 11r
Michael upload speed = 10r - 4 = 9r
Tom downstrea speed = r
Time for Tom to reach the starting point of Michael Michael = time to return to your starting point
Tt = + TMT TMU, where n = time Tom down
Michael TMD = time downstream
Michael TMU = time
But time = distance / speed, so
1 / r = x/11r + x/9r
x/11r + x/9r = 1 / r
x (1/11r + 1/9r) = 1 / r
x = (1 / r) / [1 / (11r) + 1 / (9R)]
x = (1 / r) / (1 / r) [1 / 11 + 1 / 9]
x = 1 / [1 / 11 + 1 / 9]
x = 4.95 kilometers RESPONSE
Hope this helps.
Teddy Boy
Under the assumption that the river is as a constant, we see the opposite can be treated. Treat the problem as a whole and in relation to the movement of water in the river as a country.
This simplifies things a little Michael covers a certain distance, then on the return journey the route plus1km (compared to drift to the flow, followed by Tom). Michael Speed is 10 times the speed of the drift Tom. Distance = rate times time. You can use the time constant of 1 unit (though long trips taken).
So, if c is the speed at which the rate of 10c, and Michael is a distance 2d + 1, the speed of Tom C. and covered the clearance 1
2D + 1 km = 10c * Travel 1 hours
1km = c * 1 Time:
Please note that the device requires 1 kmthat the answer lies at km.
2D = 9C
d = 4.5 kilometers
Time: If Michael lines 4.5 kilometers downstream (relative to today) that the 5.5 kilometers upstream line again (compared to the current) because the electricity to be transferred quickly enough water below 1 km at the time. So, the 10 km in connection with the river stretches when you move the current 1 km, is the rate shown by his oars.
It is of course greatly simplified the whole problem. View the source for a discussion.
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