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Abhisar
 one year ago
A man wishes to swim across a river 0.5 m wide. If he can swim at the rate of 2 Km/ h in still water and the river flows at the rate of 1 km/hr. The angle (wrt flow of water) along which he should swim so as to reach a point exactly opposite his starting point, should be?
Abhisar
 one year ago
A man wishes to swim across a river 0.5 m wide. If he can swim at the rate of 2 Km/ h in still water and the river flows at the rate of 1 km/hr. The angle (wrt flow of water) along which he should swim so as to reach a point exactly opposite his starting point, should be?

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VincentLyon.Fr
 one year ago
Best ResponseYou've already chosen the best response.0Hint: Draw a picture in order to add up the velocities.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0the velocity of the swimmer with respect to the water is 2 Km/hour, whereas the velocity of the current is 1 Km/hour, so the situation of your problem is like in my drawing: dw:1437510950238:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0Now, I stay on the bank of the river and I use my stopwatch to measure the time needed to the swimmer to cross the river. within that time the swimmer travels the distance AB: dw:1437511307871:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0and the water has traveled the distance CB, so we can write: \[\begin{gathered} A{C^2} + C{B^2} = A{B^2} \hfill \\ \hfill \\ {d^2} + {u^2}{t^2} = {v^2}{t^2} \hfill \\ \end{gathered} \] where t is the time measured by my stopwatch, u is the speed of the water, and v is the speed of the swimmer with respect to the water

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0so we have: \[t = \frac{d}{{\sqrt {{v^2}  {u^2}} }} = \frac{{0.5}}{{\sqrt {5  1} }} = 0.25hours\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0the space traveled by the water is therefore: \[s = ut = 1 \times 0.25 = 0.25Km\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0oops.. we have: \[\begin{gathered} t = \frac{d}{{\sqrt {{v^2}  {u^2}} }} = \frac{{0.5 \times {{10}^{  3}}}}{{\sqrt {5  1} }} = 2.5 \times {10^{  4}}hours \hfill \\ \hfill \\ s = ut = 1 \times 2.5 \times {10^{  4}} = 2.5 \times {10^{  4}}Km = 0.25meters \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437511991612:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0so we have: \[0.25 = 0.5\tan \theta \] then: \[\tan \theta = \frac{1}{2}\]

radar
 one year ago
Best ResponseYou've already chosen the best response.1At only .5 meter wide, why not just jump across it. lol

radar
 one year ago
Best ResponseYou've already chosen the best response.1@Abhisar Is it copied correctly or should the width be .5 km ?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0@radar , lol sorry it's 0.5 Km

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0@michele_laino , I tried it the same way but the answer given is 120°

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437545407226:dw

radar
 one year ago
Best ResponseYou've already chosen the best response.1I was on a different method, determining time, then distance and still got the same answer (120 degrees) same as IrishBoy123.dw:1437569372484:dw So in effect the width of the river was irrelevant. Or was it?

radar
 one year ago
Best ResponseYou've already chosen the best response.1@Abhisar Good luck with your studies.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1yes @Radar, width irrelevant. [unless it is 0.5m wide as per OP's question; because, as you say, he should just jump :)) ]

VincentLyon.Fr
 one year ago
Best ResponseYou've already chosen the best response.0Radar's sketch is ok, but can be simplified if you use velocities instead of displacements. Since \(\sin \theta =\dfrac{v_{river}}{v_{swimmer}}=\dfrac 1 2\), then \(\theta=30°\) Breadth of river is only needed if you have to work out the time it mill take to cross. I think the river is 0.5 km wide, (not 0.5 m; a mere jump would do the trick) The velocity wrt ground is \(2\cos 30°=2\times 0.87=1.74 \) km/h So it will take him \(\dfrac {0.5}{1.74}=0.29\) h = 17,5 min to get across.

radar
 one year ago
Best ResponseYou've already chosen the best response.1Yes, I've reached that conclusion now, but at first I thought the time it took to cross would have to be determined, but now I know that the velocities were the determining criteria, Thanks to all who worked this.

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0WoW ! Thank you for all the work, all these discussions gave me lots of insights. thank you all :)
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