Solving for Angle at Which a Girl Swims Across a River with Given Velocities

[kalupahana]

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

 

[tiny-tim]

hi kalupahana!

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

 

[kalupahana]

hi kalupahana!

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

Then it come as

1/2 = u/v
v/2u = tanα
α = tan^-1(v/2u)

is it come like this

 

[pongo38]

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.

 

[tiny-tim]

hi pongo38!

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? ​

 

[kalupahana]

hi pongo38!

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? ​

To horizontal 4km/h. V_RG is constant. Therefore The V_RG of both triangles are equal.
V_RM is 60^o inclined downwards from V_MG to opposite direction.

tan60^o = x/4
4√3 = V_RG
cos60 = 4/y
1/2 = V_RM
Now its okay?

 

[tiny-tim]

hi kalupahana!

(just got up :zzz: …)

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

you should have got an equilateral triangle … did you?

 

[kalupahana]

hi kalupahana!

(just got up :zzz: …)

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

you should have got an equilateral triangle … did you?

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