Relative Velocity of two swimmers

[JayKo]

Homework Statement

Two swimmers, Alan and Beth, start together at the same point on the
bank of a wide stream that flows with a speed v. Both move at the
same speed c (c > v), relative to the water. Alan swims downstream a
distance L and then upstream the same distance. Beth swims so that her
motion relative to the Earth is perpendicular to the banks of the stream.
She swims the distance L and then back the same distance, so that both
swimmers return to the starting point. Which swimmer returns first?
(Note: First guess the answer.)

Homework Equations

relative velocity and vector

The Attempt at a Solution

let time taken for Alan swim for a distance Ta= L / (V+C)

let time taken for Beth swim for a distance Tb= L / \sqrt{}c^{}2+v^{}2

compare this 2 equation, its obvious Alan will took the less time to finish a distance, L. but the question ask distance of 2L. so i am not sure the vector calculation for Beth to swim back to the origin point.

hope someone can help me out.this is my 1st year physics tutorial question.thanks

 

[rootX]

let time taken for Beth swim for a distance Tb= L / \sqrt{}c^{}2+v^{}2

compare this 2 equation, its obvious Alan will took the less time to finish a distance, L. but the question ask distance of 2L. so i am not sure the vector calculation for Beth to swim back to the origin point.

hope someone can help me out.this is my 1st year physics tutorial question.thanks

you forgot to take in account the return distance?
because they returned back to the same place.

it's from Irodov lol

 

[JayKo]

you forgot to take in account the return distance?
because they returned back to the same place.

it's from Irodov lol

i aware of the return distance which i mention it as 2L.
nonetheless, thank for the tip for Irodov. shall take a look at it carefully.thanks again

 

[rootX]

Ta= L / (V+C) + L / (V-C)
Tb=2*L/\sqrt{}c^{}2+v^{}2

 

[JayKo]

Ta= L / (V+C) + L / (V-C)
Tb=2*L/\sqrt{}c^{}2+v^{}2

yupe, i got it.thanks :D