Direction of swim

  • Thread starter kalupahana
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  • #1
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Homework Statement


A girl swim one bank of a river with 100m when she swims perpendicular to the water current. She reaches the other bank 50m down the stream. The angle at which she should swim ti reach a point directly opposite on the other bank of the river is?
(Velocity of girl=u, velocity of river=v)


Homework Equations





The Attempt at a Solution


I have a problem, please tell why the question give the velocities to find the angle.
It can easily taken by applying tan value to the displacements.

then it can be gain as
tanα = 50/100 = 1/2
α = tan-(1/2)

Please tell me is this right
 

Answers and Replies

  • #2
tiny-tim
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hi kalupahana! :wink:

no, you're misunderstanding the question

first she swims directly across, but the current makes her go at an angle which (as you say) is tan-1(1/2)

ok, so that means that the ratio of her speed and the speed of the current is … ?

now use that ratio to find the angle she must swim upstream to land exactly opposite :smile:
 
  • #3
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hi kalupahana! :wink:

no, you're misunderstanding the question

first she swims directly across, but the current makes her go at an angle which (as you say) is tan-1(1/2)

ok, so that means that the ratio of her speed and the speed of the current is … ?

now use that ratio to find the angle she must swim upstream to land exactly opposite :smile:
Then it come as

1/2 = u/v
v/2u = tanα
α = tan-1(v/2u)

is it come like this
 
  • #4
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It's a good question, and I think tiny tim is right and kalupahana is not (yet). I don't know how you people can solve these problems without drawing vector addition triangles.
 
  • #5
tiny-tim
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hi pongo38! :smile:

yup, kalupahana, you should have drawn a second triangle by now …

you seem to be still on the first one.

what does your second triangle look like? :smile:
 
  • #6
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hi pongo38! :smile:

yup, kalupahana, you should have drawn a second triangle by now …

you seem to be still on the first one.

what does your second triangle look like? :smile:
To horizontal 4km/h. VRG is constant. Therefore The VRG of both triangles are equal.
VRM is 60o inclined downwards from VMG to opposite direction.

tan60o = x/4
4√3 = VRG
cos60 = 4/y
1/2 = VRM
Now its okay?
 
  • #7
tiny-tim
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hi kalupahana! :smile:

(just got up :zzz: …)

i think you've got it, but i'm finding it difficult to understand what you've written :confused:

you should have got an equilateral triangle … did you?
 
  • #8
36
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hi kalupahana! :smile:

(just got up :zzz: …)

i think you've got it, but i'm finding it difficult to understand what you've written :confused:

you should have got an equilateral triangle … did you?

:eek:

oh, sorry tiny tim, i posted two posts at one time. When I check them, I reply a 2nd questions tags in here.

okk, i can understand it

thnx a lot
 

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